Rod Cutting Problem?

Short answer: 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.

Example code

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.

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.