Interview Q&A

public class MyQueue {

private Stack<int> stackIn = new Stack<int>();

private Stack<int> stackOut = new Stack<int>();

// Enqueue: push into stackIn (O(1))

public void Enqueue(int x) {

stackIn.Push(x);

// Dequeue: if stackOut empty, pour all from stackIn to

stackOut, then pop (amortized O(1))

public int Dequeue() {

if (stackOut.Count == 0) {

while (stackIn.Count > 0) {

stackOut.Push(stackIn.Pop());

return stackOut.Pop();

public int Peek() {

if (stackOut.Count == 0) {

while (stackIn.Count > 0) {

stackOut.Push(stackIn.Pop());

return stackOut.Peek();

public bool IsEmpty() {

return stackIn.Count == 0 && stackOut.Count == 0;

Follow on:

Explanation:

Two stacks are used: stackIn for enqueue, stackOut for dequeue. Elements are

transferred only when needed, making both operations amortized O(1).

begins

public class ListNode {

public int val;

public ListNode next;

public ListNode(int x) { val = x; next = null; }

ListNode DetectCycle(ListNode head) {

if (head == null) return null;

ListNode slow = head, fast = head;

Follow on:

// Detect cycle using Floyd's Tortoise and Hare

while (fast != null && fast.next != null) {

slow = slow.next;

fast = fast.next.next;

if (slow == fast) { // cycle detected

ListNode ptr = head;

while (ptr != slow) {

ptr = ptr.next;

slow = slow.next;

return ptr; // start node of cycle

return null; // no cycle

Explanation:

First detect cycle meeting point with two pointers. Then find start node by moving one

pointer from head and one from meeting point until they meet.

int Fibonacci(int n)

if (n <= 1) return n;

int a = 0, b = 1;

for (int i = 2; i <= n; i++)

int temp = a + b;

a = b;

b = temp;

return b;

Explanation:

Iterative DP approach; each Fibonacci number is sum of two previous.

Follow on:

substring (needle) in a string (haystack)

int StrStr(string haystack, string needle)

if (needle == "") return 0;

int n = haystack.Length, m = needle.Length;

for (int i = 0; i <= n - m; i++)

int j;

Follow on:

for (j = 0; j < m; j++)

if (haystack[i + j] != needle[j])

break;

if (j == m) return i; // found

return -1;

Explanation:

Simple sliding window check each position; returns starting index or -1.

public class TreeNode {

public int val;

public TreeNode left, right;

public TreeNode(int x) { val = x; }

TreeNode prev = null;

TreeNode head = null;

TreeNode ConvertToDLL(TreeNode root) {

if (root == null) return null;

ConvertToDLL(root.left);

if (prev == null) {

head = root; // first node becomes head

} else {

root.left = prev;

Follow on:

prev.right = root;

prev = root;

ConvertToDLL(root.right);

return head;

Explanation:

Inorder traversal connects nodes as doubly linked list by linking current with previous node.

void BFS(Dictionary<int, List<int>> graph, int start) {

var visited = new HashSet<int>();

var queue = new Queue<int>();

queue.Enqueue(start);

visited.Add(start);

while (queue.Count > 0) {

int node = queue.Dequeue();

Console.WriteLine(node);

foreach (var neighbor in graph[node]) {

if (!visited.Contains(neighbor)) {

visited.Add(neighbor);

queue.Enqueue(neighbor);

Explanation:

Classic BFS using a queue and visited set.

int MergeSortAndCount(int[] arr, int[] temp, int left, int right)

int invCount = 0;

if (right > left)

int mid = (right + left) / 2;

invCount += MergeSortAndCount(arr, temp, left, mid);

invCount += MergeSortAndCount(arr, temp, mid + 1, right);

invCount += Merge(arr, temp, left, mid + 1, right);

return invCount;

int Merge(int[] arr, int[] temp, int left, int mid, int right)

int i = left, j = mid, k = left;

int invCount = 0;

while (i <= mid - 1 && j <= right)

if (arr[i] <= arr[j])

temp[k++] = arr[i++];

else

temp[k++] = arr[j++];

invCount += (mid - i); // Count inversions

while (i <= mid - 1)

temp[k++] = arr[i++];

Follow on:

while (j <= right)

temp[k++] = arr[j++];

for (int idx = left; idx <= right; idx++)

arr[idx] = temp[idx];

return invCount;

Explanation:

Using a modified merge sort to count pairs where arr[i] > arr[j] for i < j efficiently.

int UniquePaths(int m, int n) {

int[,] dp = new int[m, n];

for (int i = 0; i < m; i++) dp[i, 0] = 1;

for (int j = 0; j < n; j++) dp[0, j] = 1;

for (int i = 1; i < m; i++) {

for (int j = 1; j < n; j++) {

dp[i, j] = dp[i - 1, j] + dp[i, j - 1];

return dp[m - 1, n - 1];

Follow on:

Algorithm

int GCD(int a, int b)

Follow on:

while (b != 0)

int temp = b;

b = a % b;

a = temp;

return a;

Explanation:

Repeatedly replace (a, b) with (b, a mod b) until b is 0; then a is the GCD.

public List<int> PrimeFactors(int n) {

List<int> factors = new List<int>();

// Print the number of 2s that divide n

while (n % 2 == 0) {

factors.Add(2);

n /= 2;

// n must be odd at this point

for (int i = 3; i * i <= n; i += 2) {

while (n % i == 0) {

factors.Add(i);

n /= i;

Follow on:

// If n is a prime number > 2

if (n > 2) {

factors.Add(n);

return factors;

Explanation:

We repeatedly divide by 2, then check odd factors up to √n. If leftover n > 2, it's prime.

public List<int> FindAnagrams(string s, string p) {

List<int> result = new List<int>();

if (p.Length > s.Length) return result;

int[] pCount = new int[26];

int[] sCount = new int[26];

Follow on:

for (int i = 0; i < p.Length; i++) {

pCount[p[i] - 'a']++;

sCount[s[i] - 'a']++;

if (Enumerable.SequenceEqual(pCount, sCount))

result.Add(0);

for (int i = p.Length; i < s.Length; i++) {

sCount[s[i] - 'a']++;

sCount[s[i - p.Length] - 'a']--;

if (Enumerable.SequenceEqual(pCount, sCount))

result.Add(i - p.Length + 1);

return result;

Explanation:

Sliding window with frequency count arrays for the pattern and current window.

void QuickSort(int[] arr, int low, int high)

if (low < high)

Follow on:

int pi = Partition(arr, low, high);

QuickSort(arr, low, pi - 1);

QuickSort(arr, pi + 1, high);

int Partition(int[] arr, int low, int high)

int pivot = arr[high];

int i = low - 1;

for (int j = low; j < high; j++)

if (arr[j] < pivot)

i++;

(arr[i], arr[j]) = (arr[j], arr[i]);

(arr[i + 1], arr[high]) = (arr[high], arr[i + 1]);

return i + 1;

Explanation:

Choose last element as pivot, partition array so left < pivot < right, recursively sort

subarrays.

element appears twice

public int SingleNonRepeated(int[] nums) {

int result = 0;

foreach (int num in nums) {

result ^= num;

return result;

Explanation:

XOR of a number with itself is 0; XOR with 0 is the number. So duplicates cancel out,

leaving the unique number.

Follow on:

void MergeSort(int[] arr, int left, int right)

if (left < right)

int mid = (left + right) / 2;

MergeSort(arr, left, mid);

MergeSort(arr, mid + 1, right);

Follow on:

Merge(arr, left, mid, right);

void Merge(int[] arr, int left, int mid, int right)

int n1 = mid - left + 1;

int n2 = right - mid;

int[] L = new int[n1];

int[] R = new int[n2];

Array.Copy(arr, left, L, 0, n1);

Array.Copy(arr, mid + 1, R, 0, n2);

int i = 0, j = 0, k = left;

while (i < n1 && j < n2)

arr[k++] = (L[i] <= R[j]) ? L[i++] : R[j++];

while (i < n1) arr[k++] = L[i++];

while (j < n2) arr[k++] = R[j++];

Explanation:

Divide array, sort left & right halves, then merge sorted halves.

int Knapsack(int[] weights, int[] values, int W)

int n = weights.Length;

int[,] dp = new int[n + 1, W + 1];

for (int i = 1; i <= n; i++)

for (int w = 1; w <= W; w++)

if (weights[i - 1] <= w)

dp[i, w] = Math.Max(dp[i - 1, w], values[i - 1] +

dp[i - 1, w - weights[i - 1]]);

else

dp[i, w] = dp[i - 1, w];

return dp[n, W];

Explanation:

Build table where dp[i,w] = max value using first i items and capacity w.

Follow on:

ListNode ReverseBetween(ListNode head, int m, int n) {

if (head == null || m == n) return head;

ListNode dummy = new ListNode(0);

dummy.next = head;

ListNode prev = dummy;

// Move prev to one before m-th node

for (int i = 1; i < m; i++) prev = prev.next;

ListNode start = prev.next;

ListNode then = start.next;

// Reverse the sublist

for (int i = 0; i < n - m; i++) {

Follow on:

start.next = then.next;

then.next = prev.next;

prev.next = then;

then = start.next;

return dummy.next;

Explanation:

Use a dummy node to simplify edge cases. Reverse nodes between m and n by changing

pointers.

public int EvaluateInfix(string expression) {

Stack<int> operands = new Stack<int>();

Stack<char> operators = new Stack<char>();

int i = 0;

while (i < expression.Length) {

if (char.IsWhiteSpace(expression[i])) {

i++;

continue;

if (char.IsDigit(expression[i])) {

int val = 0;

while (i < expression.Length &&

char.IsDigit(expression[i])) {

val = val * 10 + (expression[i] - '0');

i++;

operands.Push(val);

continue;

if (expression[i] == '(') {

operators.Push(expression[i]);

else if (expression[i] == ')') {

while (operators.Peek() != '(') {

ApplyOp(operands, operators);

operators.Pop(); // remove '('

Follow on:

else if (IsOperator(expression[i])) {

while (operators.Count > 0 &&

Precedence(operators.Peek()) >= Precedence(expression[i])) {

ApplyOp(operands, operators);

operators.Push(expression[i]);

i++;

while (operators.Count > 0) {

ApplyOp(operands, operators);

return operands.Pop();

bool IsOperator(char c) {

return c == '+' || c == '-' || c == '*' || c == '/';

int Precedence(char op) {

if (op == '+' || op == '-') return 1;

if (op == '*' || op == '/') return 2;

return 0;

void ApplyOp(Stack<int> operands, Stack<char> operators) {

int b = operands.Pop();

int a = operands.Pop();

char op = operators.Pop();

int result = 0;

switch (op) {

case '+': result = a + b; break;

case '-': result = a - b; break;

case '*': result = a * b; break;

Follow on:

case '/': result = a / b; break;

operands.Push(result);

Explanation:

Standard two-stack algorithm for evaluating infix expressions considering operator

precedence and parentheses.

public int CountOnes(int n) {

int count = 0;

while (n != 0) {

n &= (n - 1); // Drops the lowest set bit

count++;

return count;

Explanation:

Brian Kernighan’s algorithm efficiently removes the lowest set bit each iteration until zero.

void VerticalSum(TreeNode root, int hd, Dictionary<int, int> map) {

if (root == null) return;

VerticalSum(root.left, hd - 1, map);

if (map.ContainsKey(hd))

map[hd] += root.val;

else

map[hd] = root.val;

VerticalSum(root.right, hd + 1, map);

Dictionary<int, int> GetVerticalSum(TreeNode root) {

var map = new Dictionary<int, int>();

VerticalSum(root, 0, map);

return map;

Explanation:

Use horizontal distance (hd) from root; sum values of nodes at each hd.

Follow on:

int CountConnectedComponents(Dictionary<int, List<int>> graph) {

var visited = new HashSet<int>();

int count = 0;

Follow on:

foreach (var node in graph.Keys) {

if (!visited.Contains(node)) {

DFS(node, graph, visited);

count++;

return count;

void DFS(int node, Dictionary<int, List<int>> graph, HashSet<int>

visited) {

visited.Add(node);

foreach (var neighbor in graph[node]) {

if (!visited.Contains(neighbor)) {

DFS(neighbor, graph, visited);

Explanation:

Run DFS on unvisited nodes, count how many times DFS starts.

(Repeated here for convenience)

string LongestPalindrome(string s)

if (string.IsNullOrEmpty(s)) return "";

int start = 0, maxLen = 1;

for (int i = 0; i < s.Length; i++)

ExpandAroundCenter(s, i, i, ref start, ref maxLen);

ExpandAroundCenter(s, i, i + 1, ref start, ref maxLen);

return s.Substring(start, maxLen);

void ExpandAroundCenter(string s, int left, int right, ref int

start, ref int maxLen)

while (left >= 0 && right < s.Length && s[left] == s[right])

Follow on:

if (right - left + 1 > maxLen)

start = left;

maxLen = right - left + 1;

left--;

right++;

int ClimbStairs(int n) {

if (n <= 2) return n;

int a = 1, b = 2;

for (int i = 3; i <= n; i++) {

int c = a + b;

a = b;

b = c;

return b;

integer

public int CountSetBits(int n) {

int count = 0;

while (n != 0) {

count += n & 1;

n >>= 1;

return count;

Explanation:

Shift through each bit; add 1 to count if least significant bit is set.

Alternative using Brian Kernighan’s algorithm:

public int CountSetBits(int n) {

int count = 0;

while (n != 0) {

n &= (n - 1); // Drops the lowest set bit

count++;

return count;

Follow on:

public int[] SortNearlySorted(int[] nums, int k) {

var result = new List<int>();

var minHeap = new SortedSet<(int val, int index)>();

for (int i = 0; i < nums.Length; i++) {

minHeap.Add((nums[i], i));

if (minHeap.Count > k) {

var min = minHeap.Min;

minHeap.Remove(min);

result.Add(min.val);

while (minHeap.Count > 0) {

var min = minHeap.Min;

minHeap.Remove(min);

result.Add(min.val);

return result.ToArray();

Explanation:

Use a min-heap of size k+1 to always extract the smallest element in the current window.

int LCM(int a, int b)

return a / GCD(a, b) * b;

Explanation:

LCM × GCD = product of the two numbers.

public string LongestCommonPrefix(string[] strs) {

if (strs == null || strs.Length == 0) return "";

for (int i = 0; i < strs[0].Length; i++) {

char c = strs[0][i];

for (int j = 1; j < strs.Length; j++) {

if (i == strs[j].Length || strs[j][i] != c)

return strs[0].Substring(0, i);

return strs[0];

Follow on:

Explanation:

Check character by character across all strings until mismatch.

public class Edge {

public int Source, Dest, Weight;

public Edge(int s, int d, int w) {

Source = s; Dest = d; Weight = w;

int[] BellmanFord(int vertices, List<Edge> edges, int source) {

int[] dist = new int[vertices];

for (int i = 0; i < vertices; i++) dist[i] = int.MaxValue;

dist[source] = 0;

Follow on:

for (int i = 1; i < vertices; i++) {

foreach (var edge in edges) {

if (dist[edge.Source] != int.MaxValue &&

dist[edge.Source] + edge.Weight < dist[edge.Dest]) {

dist[edge.Dest] = dist[edge.Source] + edge.Weight;

// Detect negative weight cycle (optional)

foreach (var edge in edges) {

if (dist[edge.Source] != int.MaxValue && dist[edge.Source] +

edge.Weight < dist[edge.Dest]) {

throw new Exception("Graph contains negative weight

cycle");

return dist;

Explanation:

Relax edges V-1 times, then check for negative weight cycles.

int[] NextGreaterElements(int[] nums) {

int n = nums.Length;

int[] result = new int[n];

Stack<int> stack = new Stack<int>();

for (int i = n - 1; i >= 0; i--) {

while (stack.Count > 0 && stack.Peek() <= nums[i]) {

stack.Pop();

result[i] = stack.Count == 0 ? -1 : stack.Peek();

stack.Push(nums[i]);

return result;

Explanation:

Traverse from right to left, use stack to keep track of next greater elements in O(n).

bool IsMatch(string s, string p)

return IsMatchHelper(s, p, 0, 0);

bool IsMatchHelper(string s, string p, int i, int j)

if (j == p.Length) return i == s.Length;

bool firstMatch = (i < s.Length) && (p[j] == s[i] || p[j] ==

'.');

if (j + 1 < p.Length && p[j + 1] == '*')

// Two cases:

// 1) Use zero occurrence of p[j] (skip)

// 2) If firstMatch, consume one char in s and keep pattern

at j

return IsMatchHelper(s, p, i, j + 2) ||

(firstMatch && IsMatchHelper(s, p, i + 1, j));

else

return firstMatch && IsMatchHelper(s, p, i + 1, j + 1);

Follow on:

Explanation:

Recursively matches strings supporting '.' (any char) and '*' (zero or more of preceding).

public bool HaveOppositeSigns(int x, int y) {

return (x ^ y) < 0;

Explanation:

XOR of two numbers with opposite signs has the sign bit set (negative number).

x * half * half : (half * half) / x;

Explanation:

Uses fast exponentiation (divide and conquer) to calculate x^n in O(log n).

TreeNode prev = null;

void Flatten(TreeNode root) {

if (root == null) return;

Flatten(root.right);

Flatten(root.left);

root.right = prev;

root.left = null;

prev = root;

Explanation:

Postorder traversal (right-left-root) to flatten tree in place.

public class MedianFinder {

private PriorityQueue<int, int> maxHeap; // lower half (max

heap)

Follow on:

private PriorityQueue<int, int> minHeap; // upper half (min

heap)

public MedianFinder() {

maxHeap = new PriorityQueue<int,

int>(Comparer<int>.Create((a, b) => b.CompareTo(a)));

minHeap = new PriorityQueue<int, int>();

public void AddNum(int num) {

maxHeap.Enqueue(num, num);

minHeap.Enqueue(maxHeap.Dequeue(), maxHeap.Peek());

if (maxHeap.Count < minHeap.Count)

maxHeap.Enqueue(minHeap.Dequeue(), minHeap.Peek());

public double FindMedian() {

if (maxHeap.Count > minHeap.Count)

return maxHeap.Peek();

return (maxHeap.Peek() + minHeap.Peek()) / 2.0;

Explanation:

Maintain two heaps: maxHeap for lower half, minHeap for upper half. Balance their sizes.

List<List<int>> Subsets(int[] nums)

List<List<int>> result = new List<List<int>>();

GenerateSubsets(nums, 0, new List<int>(), result);

return result;

void GenerateSubsets(int[] nums, int index, List<int> current,

List<List<int>> result)

if (index == nums.Length)

result.Add(new List<int>(current));

return;

// Exclude nums[index]

GenerateSubsets(nums, index + 1, current, result);

// Include nums[index]

current.Add(nums[index]);

GenerateSubsets(nums, index + 1, current, result);

current.RemoveAt(current.Count - 1);

Follow on:

Explanation:

Backtracking approach includes/excludes each element.

int LCS(string s1, string s2)

int m = s1.Length, n = s2.Length;

int[,] dp = new int[m + 1, n + 1];

for (int i = 1; i <= m; i++)

for (int j = 1; j <= n; j++)

if (s1[i - 1] == s2[j - 1])

dp[i, j] = dp[i - 1, j - 1] + 1;

else

dp[i, j] = Math.Max(dp[i - 1, j], dp[i, j - 1]);

return dp[m, n];

Explanation:

Build DP table comparing chars; find longest subsequence common to both strings.

Follow on:

public class NestedIterator {

private Queue<int> queue;

public NestedIterator(IList<NestedInteger> nestedList) {

queue = new Queue<int>();

Flatten(nestedList);

private void Flatten(IList<NestedInteger> nestedList) {

foreach (var ni in nestedList) {

if (ni.IsInteger()) queue.Enqueue(ni.GetInteger());

else Flatten(ni.GetList());

public bool HasNext() {

return queue.Count > 0;

public int Next() {

return queue.Dequeue();

Note:

NestedInteger is an interface with methods: IsInteger(), GetInteger(),

GetList().

Explanation:

Pre-flatten the nested list into a queue and iterate over it.

Follow on:

bool IsPrime(int n)

if (n <= 1) return false;

if (n <= 3) return true;

if (n % 2 == 0 || n % 3 == 0) return false;

for (int i = 5; i * i <= n; i += 6)

if (n % i == 0 || n % (i + 2) == 0)

return false;

Follow on:

return true;

Explanation:

Check divisibility by 2, 3, then test possible divisors of form 6k ± 1 up to √n.

int HouseRobber(int[] nums) {

if (nums.Length == 0) return 0;

if (nums.Length == 1) return nums[0];

int prev1 = 0, prev2 = 0;

foreach (var num in nums) {

int temp = prev1;

prev1 = Math.Max(prev2 + num, prev1);

prev2 = temp;

return prev1;

ListNode MergeKLists(ListNode[] lists) {

if (lists == null || lists.Length == 0) return null;

PriorityQueue<ListNode, int> pq = new PriorityQueue<ListNode,

int>();

foreach (var list in lists)

if (list != null)

pq.Enqueue(list, list.val);

ListNode dummy = new ListNode(0);

ListNode current = dummy;

while (pq.Count > 0) {

var node = pq.Dequeue();

current.next = node;

current = current.next;

if (node.next != null)

pq.Enqueue(node.next, node.next.val);

return dummy.next;

Follow on:

Explanation:

Use a min-heap (priority queue) to always pick the smallest head node among k lists.

int FirstOccurrence(int[] arr, int target)

int low = 0, high = arr.Length - 1, result = -1;

while (low <= high)

int mid = low + (high - low) / 2;

if (arr[mid] == target)

result = mid;

Follow on:

high = mid - 1; // search left side

else if (arr[mid] < target)

low = mid + 1;

else

high = mid - 1;

return result;

Explanation:

Binary search but continue left to find first occurrence.

bool IsIdentical(TreeNode p, TreeNode q) {

if (p == null && q == null) return true;

if (p == null || q == null) return false;

if (p.val != q.val) return false;

return IsIdentical(p.left, q.left) && IsIdentical(p.right,

q.right);

Explanation:

Recursive check values and structure for equality.

public bool IsPerfectSquare(int num) {

if (num < 0) return false;

int left = 0, right = num;

while (left <= right) {

int mid = left + (right - left) / 2;

long sq = (long)mid * mid;

if (sq == num) return true;

else if (sq < num) left = mid + 1;

else right = mid - 1;

return false;

Explanation:

Binary search for integer square root and check if square equals num.

public int FindMissingNumber(int[] nums, int n) {

int xor = 0;

for (int i = 1; i <= n; i++) {

xor ^= i;

foreach (int num in nums) {

xor ^= num;

return xor;

Follow on:

Explanation:

XOR all numbers from 1 to n and XOR all elements in array; duplicates cancel out, leaving

missing number.

public int CountPartitions(int[] nums) {

int sum = 0;

foreach (int num in nums) sum += num;

if (sum % 2 != 0) return 0;

int target = sum / 2;

int[] dp = new int[target + 1];

dp[0] = 1;

Follow on:

foreach (int num in nums) {

for (int j = target; j >= num; j--) {

dp[j] += dp[j - num];

return dp[target];

Explanation:

Classic subset sum DP. dp[j] = ways to get sum j. Counting subsets summing to half the

total.

public int MaximumTotal(IList<IList<int>> triangle) {

int n = triangle.Count;

int[] dp = new int[n];

for (int i = 0; i < n; i++) dp[i] = triangle[n - 1][i];

for (int layer = n - 2; layer >= 0; layer--) {

for (int i = 0; i <= layer; i++) {

dp[i] = triangle[layer][i] + Math.Max(dp[i], dp[i + 1]);

return dp[0];

Explanation:

Start from bottom row, keep updating max path sums up to the top.

List<List<int>> TarjanSCC(Dictionary<int, List<int>> graph) {

int time = 0;

var stack = new Stack<int>();

var onStack = new HashSet<int>();

var low = new Dictionary<int, int>();

var disc = new Dictionary<int, int>();

var visited = new HashSet<int>();

var sccList = new List<List<int>>();

void DFS(int u) {

disc[u] = time;

Follow on:

low[u] = time;

time++;

stack.Push(u);

onStack.Add(u);

visited.Add(u);

foreach (var v in graph[u]) {

if (!disc.ContainsKey(v)) {

DFS(v);

low[u] = Math.Min(low[u], low[v]);

} else if (onStack.Contains(v)) {

low[u] = Math.Min(low[u], disc[v]);

if (low[u] == disc[u]) {

var scc = new List<int>();

int w;

do {

w = stack.Pop();

onStack.Remove(w);

scc.Add(w);

} while (w != u);

sccList.Add(scc);

foreach (var node in graph.Keys) {

if (!disc.ContainsKey(node))

DFS(node);

return sccList;

Explanation:

Tarjan's algorithm finds SCCs using low-link values and DFS stack.

Follow on:

int LongestPalindromeSubseq(string s) {

int n = s.Length;

int[,] dp = new int[n, n];

for (int i = n - 1; i >= 0; i--) {

dp[i, i] = 1;

for (int j = i + 1; j < n; j++) {

if (s[i] == s[j])

Follow on:

dp[i, j] = dp[i + 1, j - 1] + 2;

else

dp[i, j] = Math.Max(dp[i + 1, j], dp[i, j - 1]);

return dp[0, n - 1];

element in constant time

public class MinStack {

private Stack<int> stack = new Stack<int>();

private Stack<int> minStack = new Stack<int>();

Follow on:

public void Push(int x) {

stack.Push(x);

if (minStack.Count == 0 || x <= minStack.Peek())

minStack.Push(x);

public void Pop() {

if (stack.Peek() == minStack.Peek())

minStack.Pop();

stack.Pop();

public int Top() {

return stack.Peek();

public int GetMin() {

return minStack.Peek();

Explanation:

Use two stacks: one normal stack, one for minimum values. When pushing a smaller or

equal value, push it on minStack; when popping, pop minStack if needed.

headB : a.next;

b = (b == null) ? headA : b.next;

return a; // either intersection or null

Explanation:

Two pointers traverse both lists; if no intersection, both will reach null simultaneously.

Eratosthenes)

List<int> GeneratePrimes(int n)

bool[] isPrime = new bool[n + 1];

for (int i = 2; i <= n; i++) isPrime[i] = true;

for (int i = 2; i * i <= n; i++)

if (isPrime[i])

for (int j = i * i; j <= n; j += i)

isPrime[j] = false;

List<int> primes = new List<int>();

for (int i = 2; i <= n; i++)

if (isPrime[i]) primes.Add(i);

return primes;

Explanation:

Mark multiples of each prime as non-prime starting from its square.

Follow on:

String

Dictionary<char, int> CountCharacters(string s)

var dict = new Dictionary<char, int>();

foreach (char c in s)

if (dict.ContainsKey(c))

dict[c]++;

else

dict[c] = 1;

return dict;

Explanation:

Simple frequency map using Dictionary.

int LIS(int[] nums)

int n = nums.Length;

int[] dp = new int[n];

Array.Fill(dp, 1);

int maxLen = 1;

for (int i = 1; i < n; i++)

for (int j = 0; j < i; j++)

if (nums[i] > nums[j])

dp[i] = Math.Max(dp[i], dp[j] + 1);

maxLen = Math.Max(maxLen, dp[i]);

return maxLen;

Explanation:

For each element, find LIS ending there by checking previous smaller elements.

Follow on:

List<List<string>> SolveNQueens(int n)

List<List<string>> results = new List<List<string>>();

int[] board = new int[n]; // board[i] = column position of queen

in row i

Solve(0, board, results, n);

return results;

void Solve(int row, int[] board, List<List<string>> results, int n)

if (row == n)

results.Add(GenerateBoard(board, n));

return;

for (int col = 0; col < n; col++)

if (IsSafe(row, col, board))

board[row] = col;

Solve(row + 1, board, results, n);

bool IsSafe(int row, int col, int[] board)

for (int i = 0; i < row; i++)

Follow on:

if (board[i] == col || Math.Abs(board[i] - col) ==

Math.Abs(i - row))

return false;

return true;

List<string> GenerateBoard(int[] board, int n)

List<string> res = new List<string>();

for (int i = 0; i < n; i++)

char[] row = new char[n];

for (int j = 0; j < n; j++)

row[j] = '.';

row[board[i]] = 'Q';

res.Add(new string(row));

return res;

Explanation:

Backtracking places queens row by row while checking columns and diagonals.

int FindPeakElement(int[] nums)

int left = 0, right = nums.Length - 1;

while (left < right)

int mid = (left + right) / 2;

if (nums[mid] > nums[mid + 1])

right = mid;

else

left = mid + 1;

return left;

Explanation:

Binary search comparing mid element with right neighbor to find peak.

public int FindCelebrity(int n, Func<int, int, bool> knows) {

int candidate = 0;

for (int i = 1; i < n; i++) {

if (knows(candidate, i)) candidate = i;

for (int i = 0; i < n; i++) {

if (i != candidate && (knows(candidate, i) || !knows(i,

candidate)))

return -1;

return candidate;

Explanation:

First find candidate by elimination, then verify candidate.

Follow on:

int MinPathSum(int[][] grid) {

int m = grid.Length, n = grid[0].Length;

for (int i = 1; i < m; i++) grid[i][0] += grid[i - 1][0];

for (int j = 1; j < n; j++) grid[0][j] += grid[0][j - 1];

for (int i = 1; i < m; i++) {

for (int j = 1; j < n; j++) {

grid[i][j] += Math.Min(grid[i - 1][j], grid[i][j - 1]);

return grid[m - 1][n - 1];

int SumOfDigits(int n)

int sum = 0;

n = Math.Abs(n);

while (n > 0)

sum += n % 10;

n /= 10;

return sum;

Explanation:

Extract digits using modulo 10 and add.

appears twice except one

public int SingleNumber(int[] nums) {

int result = 0;

foreach (var num in nums) {

result ^= num;

return result;

Explanation:

XOR of all elements cancels duplicates, leaving the single unique element.

int[] Dijkstra(Dictionary<int, List<(int neighbor, int weight)>>

graph, int source, int vertices) {

int[] dist = new int[vertices];

for (int i = 0; i < vertices; i++) dist[i] = int.MaxValue;

dist[source] = 0;

var pq = new SortedSet<(int dist, int node)>();

pq.Add((0, source));

while (pq.Count > 0) {

var current = pq.Min;

pq.Remove(current);

int u = current.node;

foreach (var (v, w) in graph[u]) {

if (dist[u] + w < dist[v]) {

if (dist[v] != int.MaxValue)

pq.Remove((dist[v], v));

dist[v] = dist[u] + w;

pq.Add((dist[v], v));

return dist;

Explanation:

Uses a priority queue to pick node with min dist; relax edges.

Follow on:

int BinarySearch(int[] arr, int target)

int low = 0, high = arr.Length - 1;

while (low <= high)

int mid = low + (high - low) / 2;

if (arr[mid] == target)

return mid;

else if (arr[mid] < target)

low = mid + 1;

else

high = mid - 1;

return -1;

Explanation:

Standard binary search to find target’s index or -1 if not found.

int NumIslands(char[][] grid)

if (grid == null || grid.Length == 0) return 0;

int count = 0;

for (int i = 0; i < grid.Length; i++)

for (int j = 0; j < grid[0].Length; j++)

if (grid[i][j] == '1')

Follow on:

DFS(grid, i, j);

count++;

return count;

void DFS(char[][] grid, int i, int j)

if (i < 0 || j < 0 || i >= grid.Length || j >= grid[0].Length ||

grid[i][j] == '0')

return;

grid[i][j] = '0'; // Mark visited

DFS(grid, i + 1, j);

DFS(grid, i - 1, j);

DFS(grid, i, j + 1);

DFS(grid, i, j - 1);

Explanation:

Use DFS to mark all connected land cells, count islands by visiting unvisited lands.

public int GCD(int a, int b) {

Follow on:

while (b != 0) {

int temp = b;

b = a % b;

a = temp;

return a;

public int LCM(int a, int b) {

return (a / GCD(a, b)) * b;

Explanation:

GCD uses Euclidean algorithm. LCM calculated via LCM(a,b) = (a*b)/GCD(a,b).

TreeNode LCA(TreeNode root, int n1, int n2) {

if (root == null) return null;

if (root.val == n1 || root.val == n2) return root;

TreeNode left = LCA(root.left, n1, n2);

Follow on:

TreeNode right = LCA(root.right, n1, n2);

if (left != null && right != null) return root;

return left ?? right;

int FindLevel(TreeNode root, int val, int level) {

if (root == null) return -1;

if (root.val == val) return level;

int left = FindLevel(root.left, val, level + 1);

if (left != -1) return left;

return FindLevel(root.right, val, level + 1);

int DistanceBetweenNodes(TreeNode root, int n1, int n2) {

TreeNode lca = LCA(root, n1, n2);

int d1 = FindLevel(lca, n1, 0);

int d2 = FindLevel(lca, n2, 0);

return d1 + d2;

Explanation:

Find Lowest Common Ancestor (LCA) then sum distances from LCA to each node.

Amount)

int CoinChange(int[] coins, int amount)

int[] dp = new int[amount + 1];

Array.Fill(dp, amount + 1);

dp[0] = 0;

for (int i = 1; i <= amount; i++)

foreach (int coin in coins)

if (coin <= i)

dp[i] = Math.Min(dp[i], 1 + dp[i - coin]);

return dp[amount] > amount ? -1 : dp[amount];

Explanation:

Bottom-up DP: min coins needed for all amounts up to target.

Follow on:

public void Swap(ref int a, ref int b) {

if (a != b) {

a ^= b;

b ^= a;

a ^= b;

Explanation:

XOR swap algorithm exchanges values without extra storage. (Check a != b to avoid

zeroing when both are same.)

public int[] MaxSlidingWindow(int[] nums, int k) {

if (nums == null || k <= 0) return new int[0];

int n = nums.Length;

int[] result = new int[n - k + 1];

LinkedList<int> deque = new LinkedList<int>(); // store indices

for (int i = 0; i < n; i++) {

// Remove indices out of window

if (deque.Count > 0 && deque.First.Value <= i - k)

Follow on:

deque.RemoveFirst();

// Remove smaller values from the back

while (deque.Count > 0 && nums[deque.Last.Value] < nums[i])

deque.RemoveLast();

deque.AddLast(i);

if (i >= k - 1)

result[i - k + 1] = nums[deque.First.Value];

return result;

Explanation:

Use a deque to keep indexes of useful elements in current window, ensuring the front is

always max.

bool IsRotation(string s1, string s2)

if (s1.Length != s2.Length) return false;

string doubled = s1 + s1;

return doubled.Contains(s2);

Follow on:

Explanation:

If s2 is rotation of s1, it must be substring of s1+s1.

l1.val : 0;

int y = (l2 != null) ? l2.val : 0;

int sum = x + y + carry;

carry = sum / 10;

curr.next = new ListNode(sum % 10);

curr = curr.next;

if (l1 != null) l1 = l1.next;

if (l2 != null) l2 = l2.next;

Follow on:

return dummy.next;

Explanation:

Add digit by digit with carry, creating new nodes for the result.

public class MultiLevelNode {

public int val;

public MultiLevelNode next;

public MultiLevelNode child;

public MultiLevelNode(int x) { val = x; next = null; child =

null; }

MultiLevelNode Flatten(MultiLevelNode head) {

if (head == null) return null;

MultiLevelNode dummy = new MultiLevelNode(0);

MultiLevelNode prev = dummy;

Stack<MultiLevelNode> stack = new Stack<MultiLevelNode>();

stack.Push(head);

while (stack.Count > 0) {

var curr = stack.Pop();

prev.next = curr;

curr.child = null; // remove child pointer after flattening

prev = curr;

if (curr.next != null) stack.Push(curr.next);

if (curr.child != null) stack.Push(curr.child);

return dummy.next;

Follow on:

Explanation:

Use a stack to perform DFS; attach nodes and remove child pointers.

public int TrailingZeroes(int n) {

int count = 0;

while (n > 0) {

n /= 5;

count += n;

return count;

Explanation:

Count factors of 5 in factorial since 2s are plentiful, trailing zeros depend on 5s.

public class TrieNode {

public TrieNode[] Children = new TrieNode[26];

public bool IsEnd = false;

public class Trie {

private TrieNode root;

public Trie() {

root = new TrieNode();

public void Insert(string word) {

TrieNode node = root;

foreach (char c in word) {

int idx = c - 'a';

if (node.Children[idx] == null)

node.Children[idx] = new TrieNode();

node = node.Children[idx];

node.IsEnd = true;

public bool Search(string word) {

TrieNode node = SearchNode(word);

return node != null && node.IsEnd;

public bool StartsWith(string prefix) {

return SearchNode(prefix) != null;

private TrieNode SearchNode(string word) {

TrieNode node = root;

foreach (char c in word) {

int idx = c - 'a';

if (node.Children[idx] == null) return null;

Follow on:

node = node.Children[idx];

return node;

Explanation:

Standard trie with insert, search, and prefix checking.

bool WordBreak(string s, HashSet<string> wordDict)

bool[] dp = new bool[s.Length + 1];

dp[0] = true;

for (int i = 1; i <= s.Length; i++)

for (int j = 0; j < i; j++)

if (dp[j] && wordDict.Contains(s.Substring(j, i - j)))

dp[i] = true;

break;

return dp[s.Length];

Explanation:

DP to check if substring can be segmented using dictionary words.

Follow on:

(Others Appear Twice)

int SingleNonDuplicate(int[] nums)

int low = 0, high = nums.Length - 1;

while (low < high)

int mid = low + (high - low) / 2;

if (mid % 2 == 1) mid--; // ensure mid is even

if (nums[mid] == nums[mid + 1])

low = mid + 2;

else

high = mid;

Follow on:

return nums[low];

Explanation:

Pairs appear consecutively; use binary search on even indices to find mismatch.

public int[] SearchRange(int[] nums, int target) {

int left = FindBoundary(nums, target, true);

int right = FindBoundary(nums, target, false);

return new int[] { left, right };

private int FindBoundary(int[] nums, int target, bool findFirst) {

Follow on:

int left = 0, right = nums.Length - 1;

int boundary = -1;

while (left <= right) {

int mid = left + (right - left) / 2;

if (nums[mid] == target) {

boundary = mid;

if (findFirst)

right = mid - 1;

else

left = mid + 1;

else if (nums[mid] < target) left = mid + 1;

else right = mid - 1;

return boundary;

Explanation:

Binary search twice — once to find first occurrence and once for last occurrence.

public uint ReverseBits(uint n) {

uint result = 0;

for (int i = 0; i < 32; i++) {

result <<= 1;

result |= (n & 1);

n >>= 1;

return result;

Explanation:

Iteratively take least significant bit of n, add it to result’s LSB, shift both accordingly.

Follow on:

Follow on:

int EditDistance(string word1, string word2) {

int m = word1.Length, n = word2.Length;

int[,] dp = new int[m + 1, n + 1];

for (int i = 0; i <= m; i++) dp[i, 0] = i;

for (int j = 0; j <= n; j++) dp[0, j] = j;

for (int i = 1; i <= m; i++) {

for (int j = 1; j <= n; j++) {

if (word1[i - 1] == word2[j - 1])

dp[i, j] = dp[i - 1, j - 1];

else

Follow on:

dp[i, j] = 1 + Math.Min(dp[i - 1, j - 1],

Math.Min(dp[i - 1, j], dp[i, j - 1]));

return dp[m, n];

All Characters of String t

string MinWindow(string s, string t)

if (string.IsNullOrEmpty(s) || string.IsNullOrEmpty(t)) return

"";

Dictionary<char, int> dictT = new Dictionary<char, int>();

foreach (char c in t)

dictT[c] = dictT.ContainsKey(c) ? dictT[c] + 1 : 1;

int required = dictT.Count;

int formed = 0;

Dictionary<char, int> windowCounts = new Dictionary<char,

int>();

int left = 0, right = 0;

int minLen = int.MaxValue, minLeft = 0;

while (right < s.Length)

char c = s[right];

windowCounts[c] = windowCounts.ContainsKey(c) ?

windowCounts[c] + 1 : 1;

if (dictT.ContainsKey(c) && windowCounts[c] == dictT[c])

formed++;

while (left <= right && formed == required)

if (right - left + 1 < minLen)

minLen = right - left + 1;

Follow on:

minLeft = left;

char leftChar = s[left];

windowCounts[leftChar]--;

if (dictT.ContainsKey(leftChar) &&

windowCounts[leftChar] < dictT[leftChar])

formed--;

left++;

right++;

return minLen == int.MaxValue ? "" : s.Substring(minLeft,

minLen);

Explanation:

Sliding window with two pointers keeps track of counts of chars matching the target.

int TrailingZeroes(int n)

int count = 0;

for (int i = 5; i <= n; i *= 5)

count += n / i;

return count;

Explanation:

Trailing zeros come from factors of 10 = 2 × 5, but 2s are plenty, count 5s.

List<List<int>> LevelOrderBottom(TreeNode root) {

var res = new List<List<int>>();

if (root == null) return res;

Queue<TreeNode> queue = new Queue<TreeNode>();

queue.Enqueue(root);

while (queue.Count > 0) {

int size = queue.Count;

var level = new List<int>();

Follow on:

for (int i = 0; i < size; i++) {

TreeNode node = queue.Dequeue();

level.Add(node.val);

if (node.left != null) queue.Enqueue(node.left);

if (node.right != null) queue.Enqueue(node.right);

res.Insert(0, level); // prepend to get reverse order

return res;

Explanation:

Perform normal BFS, insert each level at front of result list for reversed order.

Array

int LongestConsecutive(int[] nums)

HashSet<int> set = new HashSet<int>(nums);

int longest = 0;

foreach (int num in set)

if (!set.Contains(num - 1))

Follow on:

int currentNum = num;

int length = 1;

while (set.Contains(currentNum + 1))

currentNum++;

length++;

longest = Math.Max(longest, length);

return longest;

Explanation:

Check only starts of sequences, count consecutive numbers using HashSet for O(n).

void PrintNodesAtLevel(Dictionary<int, List<int>> graph, int start,

int targetLevel) {

var visited = new HashSet<int>();

var queue = new Queue<(int node, int level)>();

Follow on:

queue.Enqueue((start, 0));

visited.Add(start);

while (queue.Count > 0) {

var (node, level) = queue.Dequeue();

if (level == targetLevel) {

Console.WriteLine(node);

if (level > targetLevel) break;

foreach (var neighbor in graph[node]) {

if (!visited.Contains(neighbor)) {

visited.Add(neighbor);

queue.Enqueue((neighbor, level + 1));

Explanation:

BFS traversal with level tracking; print nodes at the requested level.

int MinCutPalindromePartition(string s) {

int n = s.Length;

bool[,] dp = new bool[n, n];

int[] cuts = new int[n];

for (int i = 0; i < n; i++) {

int minCuts = i;

for (int j = 0; j <= i; j++) {

if (s[j] == s[i] && (i - j < 2 || dp[j + 1, i - 1])) {

dp[j, i] = true;

minCuts = j == 0 ? 0 : Math.Min(minCuts, cuts[j - 1]

+ 1);

cuts[i] = minCuts;

return cuts[n - 1];

1 : -1;

return candidate;

Explanation:

Boyer-Moore Voting Algorithm tracks majority element by counting net votes.

Follow on:

int MaxSubArray(int[] nums)

int maxSoFar = nums[0];

int maxEndingHere = nums[0];

for (int i = 1; i < nums.Length; i++)

maxEndingHere = Math.Max(nums[i], maxEndingHere + nums[i]);

maxSoFar = Math.Max(maxSoFar, maxEndingHere);

return maxSoFar;

Explanation:

Track max subarray ending at current position; update global max.

public int FindDuplicate(int[] nums) {

int slow = nums[0], fast = nums[0];

do {

slow = nums[slow];

fast = nums[nums[fast]];

} while (slow != fast);

fast = nums[0];

while (slow != fast) {

slow = nums[slow];

fast = nums[fast];

return slow;

Explanation:

Floyd’s Tortoise and Hare cycle detection, treating values as pointers.

bool PrintAncestors(TreeNode root, int target) {

if (root == null) return false;

if (root.val == target) return true;

if (PrintAncestors(root.left, target) ||

PrintAncestors(root.right, target)) {

Console.Write(root.val + " ");

return true;

return false;

Explanation:

Recursive check if target is in subtree; print current node on path back if yes.

1 : -1;

return candidate;

Explanation:

Maintain a candidate and count; majority element survives this cancellation.

x * temp * temp : (temp * temp) / x;

Explanation:

Recursive fast power divides exponent by 2 to reduce complexity to O(log n).

int KthSmallest(int[][] matrix, int k)

int n = matrix.Length;

int low = matrix[0][0], high = matrix[n - 1][n - 1];

while (low < high)

Follow on:

int mid = low + (high - low) / 2;

int count = 0, j = n - 1;

for (int i = 0; i < n; i++)

while (j >= 0 && matrix[i][j] > mid)

j--;

count += (j + 1);

if (count < k)

low = mid + 1;

else

high = mid;

return low;

Explanation:

Binary search on values, count how many elements ≤ mid using matrix’s sorted rows/cols.

int MaxSumIncreasingSubsequence(int[] nums) {

int n = nums.Length;

int[] dp = new int[n];

Array.Copy(nums, dp, n);

for (int i = 1; i < n; i++) {

for (int j = 0; j < i; j++) {

Follow on:

if (nums[i] > nums[j])

dp[i] = Math.Max(dp[i], dp[j] + nums[i]);

return dp.Max();

void RotateMatrix(int[][] matrix)

int n = matrix.Length;

// Transpose

for (int i = 0; i < n; i++)

for (int j = i; j < n; j++)

(matrix[i][j], matrix[j][i]) = (matrix[j][i],

matrix[i][j]);

// Reverse each row

for (int i = 0; i < n; i++)

int left = 0, right = n - 1;

while (left < right)

(matrix[i][left], matrix[i][right]) = (matrix[i][right],

matrix[i][left]);

left++;

right--;

Explanation:

Transpose matrix and then reverse each row to rotate clockwise by 90°.

public int LengthOfLongestSubstringTwoDistinct(string s) {

int left = 0, right = 0, maxLen = 0;

Dictionary<char, int> map = new Dictionary<char, int>();

while (right < s.Length) {

Follow on:

char c = s[right];

map[c] = right;

if (map.Count > 2) {

int delIndex = map.Values.Min();

map.Remove(s[delIndex]);

left = delIndex + 1;

maxLen = Math.Max(maxLen, right - left + 1);

right++;

return maxLen;

Explanation:

Sliding window with hashmap to track indices of distinct chars, remove the leftmost when

>2.

int CountOnes(int n)

int count = 0;

while (n != 0)

n &= (n - 1); // drops the lowest set bit

count++;

return count;

Explanation:

Brian Kernighan’s algorithm removes one set bit per iteration.

int RodCutting(int[] prices, int n)

int[] dp = new int[n + 1];

dp[0] = 0;

Follow on:

for (int i = 1; i <= n; i++)

int maxVal = int.MinValue;

for (int j = 0; j < i; j++)

maxVal = Math.Max(maxVal, prices[j] + dp[i - j - 1]);

dp[i] = maxVal;

return dp[n];

Explanation:

Max revenue by cutting rod into pieces of various lengths.

min, int? max) {

if (node == null) return true;

if ((min != null && node.val <= min) || (max != null && node.val

>= max)) return false;

return Validate(node.left, min, node.val) &&

Validate(node.right, node.val, max);

Explanation:

Pass down min and max bounds for subtree values; node must be in (min, max) range.

int MatrixChainOrder(int[] dims)

int n = dims.Length - 1;

int[,] dp = new int[n, n];

for (int l = 2; l <= n; l++)

Follow on:

for (int i = 0; i < n - l + 1; i++)

int j = i + l - 1;

dp[i, j] = int.MaxValue;

for (int k = i; k < j; k++)

int cost = dp[i, k] + dp[k + 1, j] + dims[i] *

dims[k + 1] * dims[j + 1];

dp[i, j] = Math.Min(dp[i, j], cost);

return dp[0, n - 1];

Explanation:

DP calculates minimal cost to multiply chain of matrices by trying all partitions.

void SortColors(int[] nums)

int low = 0, mid = 0, high = nums.Length - 1;

while (mid <= high)

if (nums[mid] == 0)

(nums[low++], nums[mid++]) = (nums[mid], nums[low]);

else if (nums[mid] == 1)

mid++;

else

(nums[mid], nums[high--]) = (nums[high], nums[mid]);

Follow on:

Explanation:

Partition array into three parts in one pass using three pointers.

public bool IsBalanced(string s) {

Stack<char> stack = new Stack<char>();

Dictionary<char, char> pairs = new Dictionary<char, char> {

{')', '('}, {']', '['}, {'}', '{'}

foreach (char c in s) {

if ("([{".Contains(c))

stack.Push(c);

else if (")]}".Contains(c)) {

if (stack.Count == 0 || stack.Pop() != pairs[c])

return false;

return stack.Count == 0;

Follow on:

Explanation:

Use a stack to match opening and closing brackets properly.

int WidthOfBinaryTree(TreeNode root) {

if (root == null) return 0;

int maxWidth = 0;

Queue<(TreeNode node, int idx)> queue = new Queue<(TreeNode,

int)>();

queue.Enqueue((root, 0));

while (queue.Count > 0) {

int size = queue.Count;

int start = queue.Peek().idx;

int end = start;

for (int i = 0; i < size; i++) {

var (node, idx) = queue.Dequeue();

end = idx;

if (node.left != null)

queue.Enqueue((node.left, 2 * idx + 1));

if (node.right != null)

queue.Enqueue((node.right, 2 * idx + 2));

maxWidth = Math.Max(maxWidth, end - start + 1);

Follow on:

return maxWidth;

Explanation:

Assign index to each node as if in a complete tree; width is max difference of indices per

level.

List<string> WordBreak(string s, IList<string> wordDict) {

var wordSet = new HashSet<string>(wordDict);

var memo = new Dictionary<int, List<string>>();

return DFS(0);

List<string> DFS(int start) {

if (memo.ContainsKey(start)) return memo[start];

var res = new List<string>();

if (start == s.Length) {

res.Add("");

return res;

for (int end = start + 1; end <= s.Length; end++) {

string word = s.Substring(start, end - start);

if (wordSet.Contains(word)) {

foreach (var sub in DFS(end)) {

string space = sub.Length == 0 ? "" : " ";

res.Add(word + space + sub);

memo[start] = res;

return res;

Follow on:

a / b : throw new DivideByZeroException(),

_ => throw new ArgumentException("Invalid operator"),

Explanation:

Simple switch statement performing basic arithmetic, with divide-by-zero check.

bool SolveSudoku(char[][] board)

for (int i = 0; i < 9; i++)

for (int j = 0; j < 9; j++)

Follow on:

if (board[i][j] == '.')

for (char c = '1'; c <= '9'; c++)

if (IsValid(board, i, j, c))

board[i][j] = c;

if (SolveSudoku(board))

return true;

else

board[i][j] = '.';

return false;

return true;

bool IsValid(char[][] board, int row, int col, char c)

for (int i = 0; i < 9; i++)

if (board[row][i] == c) return false;

if (board[i][col] == c) return false;

if (board[3 * (row / 3) + i / 3][3 * (col / 3) + i % 3] ==

c) return false;

return true;

Explanation:

Backtracking tries digits 1-9 in empty cells, validating constraints.

bool CanPartition(int[] nums)

int sum = nums.Sum();

Follow on:

if (sum % 2 != 0) return false;

int target = sum / 2;

bool[] dp = new bool[target + 1];

dp[0] = true;

foreach (int num in nums)

for (int j = target; j >= num; j--)

dp[j] = dp[j] || dp[j - num];

return dp[target];

Explanation:

Subset sum to check if half the total sum is achievable.

Sorting and Searching

Integers

int FindMissingNumber(int[] nums)

int low = 0, high = nums.Length - 1;

while (low <= high)

int mid = low + (high - low) / 2;

if (nums[mid] == mid)

low = mid + 1;

else

high = mid - 1;

return low;

Explanation:

In perfect array nums[i] == i; missing number breaks this property, use binary search to find

breakpoint.

Mathematical Problems

public void RotateMatrix(int[][] matrix) {

int n = matrix.Length;

// Transpose

for (int i = 0; i < n; i++) {

for (int j = i + 1; j < n; j++) {

int temp = matrix[i][j];

matrix[i][j] = matrix[j][i];

matrix[j][i] = temp;

// Reverse each row

for (int i = 0; i < n; i++) {

Array.Reverse(matrix[i]);

Explanation:

Transpose matrix and then reverse each row to rotate 90° clockwise.

Follow on:

List<string> GenerateParenthesis(int n)

List<string> result = new List<string>();

Generate("", 0, 0, n, result);

return result;

void Generate(string current, int open, int close, int max,

List<string> result)

if (current.Length == max * 2)

result.Add(current);

return;

if (open < max)

Generate(current + "(", open + 1, close, max, result);

if (close < open)

Generate(current + ")", open, close + 1, max, result);

Explanation:

Use backtracking to add '(' and ')' only when valid.

long LargestPrimeFactor(long n)

long maxPrime = -1;

while (n % 2 == 0)

maxPrime = 2;

n /= 2;

for (long i = 3; i * i <= n; i += 2)

while (n % i == 0)

maxPrime = i;

n /= i;

if (n > 2) maxPrime = n;

return maxPrime;

Follow on:

Explanation:

Divide out factors of 2, then test odd factors; leftover > 2 is prime.

Miscellaneous Problems