Longest Increasing Subsequence (LIS) int LIS(int[] nums) { int n = nums.Length; int[] dp = new int[n];?

Short answer: rray.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.

Explain a bit more

Follow on: rray.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. for (int i = 1; i < n; i++)

Example code

{
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:

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.