Dynamic Programming: Knapsack - Perfect Squares
What is the output of the bottom-up DP function
numSquares when called with n = 1?What is the output of the bottom-up DP function numSquares when called with n = 1?
medium🧾 Code Trace Q4 of Q15
Step-by-Step Solution
Solution:
Step 1: Initialize dp array for n=1
dp = [0, inf]Step 2: Compute dp[1]
Check squares ≤1: only 1. dp[1] = min(inf, 1 + dp[0]) = 1.Final Answer:
Option D -> Option DQuick Check:
1 is a perfect square itself -> minimal count 1 [OK]
Quick Trick: dp[1] = 1 + dp[0] = 1 [OK]
Common Mistakes:
MISTAKES- Forgetting dp[0] = 0 base case
- Returning inf if base case missing
- Off-by-one errors in loop
Trap Explanation:
PITFALL- Without dp[0] = 0, dp[1] remains inf causing wrong output.
Interviewer Note:
CONTEXT- Tests handling of base cases and boundary conditions in DP
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