Interview Q&A

Short answer: 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;…

Explain a bit more

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.

Example code

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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 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,…

Example code

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];
}

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: void PrintNodesAtLevel(Dictionary<int, List<int>> graph, int start, int targetLevel) {

Example code

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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: List<List<int>> LevelOrderBottom(TreeNode root) {

Example code

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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 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…

Explain a bit more

(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.

Example code

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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: All Characters of String t string MinWindow(string s, string t) return minLen == int.MaxValue ? "" : s.Substring(minLeft, minLen); } Explanation: Sliding window with two pointers keeps track of counts of chars matching the target.

Example code

{
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++; }

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: Array int LongestConsecutive(int[] nums)

Example code

{
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).

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 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.

Example code

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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: (Others Appear Twice) int SingleNonDuplicate(int[] nums)

Example code

{
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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 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:

Example code

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:

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: string input = "programming"; Dictionary<char, int> map = new Dictionary<char, int>(); foreach (char c in input) map[c] = map.ContainsKey(c) ? map[c] + 1 : 1; foreach (var item in map) { if (item.Value > 1) Console.WriteLine(item.Key); }

Example code

string input = "programming";
Dictionary<char, int> map = new Dictionary<char, int>();
foreach (char c in input)
map[c] = map.ContainsKey(c) ? map[c] + 1 : 1;
foreach (var item in map)
{
if (item.Value > 1) Console.WriteLine(item.Key); }

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: int MajorityElement(int[] nums) { int count = 0, candidate = 0; foreach (var num in nums) { if (count == 0) candidate = num; count += (num == candidate) ? 1 : -1; } return candidate; } Explanation: Boyer-Moore Voting Algorithm tracks majority element by counting net votes. Follow on:

Example code

int MajorityElement(int[] nums)
{
int count = 0, candidate = 0;
foreach (var num in nums)
{
if (count == 0)
candidate = num;
count += (num == candidate) ? 1 : -1;
}
return candidate;
} Explanation: Boyer-Moore Voting Algorithm tracks majority element by counting net votes. Follow on:

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: double Power(double x, int n) Follow on: {

Example code

if (n == 0) return 1;
double temp = Power(x, n / 2);
if (n % 2 == 0)
return temp * temp; else return (n > 0) ? x * temp * temp : (temp * temp) / x;
} Explanation: Recursive fast power divides exponent by 2 to reduce complexity to O(log n).

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: Algorithm) int MajorityElement(int[] nums)

Example code

{
int count = 0, candidate = 0;
foreach (var num in nums)
{
if (count == 0)
candidate = num;
count += (num == candidate) ? 1 : -1;
}
return candidate;
} Explanation: Maintain a candidate and count; majority element survives this cancellation.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 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.

Example code

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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 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]; }

Example code

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];
}

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 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.

Example code

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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 1 : -1; return candidate; Explanation: Boyer-Moore Voting Algorithm tracks majority element by counting net votes. Follow on:

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: x * temp * temp : (temp * temp) / x; Explanation: Recursive fast power divides exponent by 2 to reduce complexity to O(log n).

Explain a bit more

x * temp * temp : (temp * temp) / x; Explanation: Recursive fast power divides exponent by 2 to reduce complexity to O(log n). x * temp * temp : (temp * temp) / x; Explanation: Recursive fast power divides exponent by 2 to reduce complexity to O(log n). x * temp * temp : (temp * temp) / x; Explanation: Recursive fast power divides exponent by 2 to reduce complexity to O(log n).

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 1 : -1; return candidate; Explanation: Maintain a candidate and count; majority element survives this cancellation.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 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.

Example code

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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: public sealed class Singleton { private static readonly object lockObj = new object(); private static Singleton instance; private Singleton() { } public static Singleton Instance { get { if (instance == null) { lock (lockObj) { if (instance == null) instance = new Singleton(); } } return instance; } } }

Example code

public sealed class Singleton
{
private static readonly object lockObj = new object();
private static Singleton instance;
private Singleton() { }
public static Singleton Instance
{ get {
if (instance == null)
{ lock (lockObj) {
if (instance == null)
instance = new Singleton();
}
}
return instance;
}
}
}

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: rray.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(); } rray.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…… rray.Copy(nums, dp, n); for (int i = 1; i < n; i++) { for (int…

Explain a bit more

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(); } rray.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(); }

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: bool IsValidBST(TreeNode root) { return Validate(root, null, null); } Follow on: bool Validate(TreeNode node, int?

Explain a bit more

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.

Example code

bool IsValidBST(TreeNode root) { return Validate(root, null, null);
} Follow on: bool Validate(TreeNode node, int? 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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.

Short answer: 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:…

Explain a bit more

Sliding window with hashmap to track indices of distinct chars, remove the leftmost when >2.

Example code

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.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.