Dynamic Programming - Perfect Squares

Given an integer n, return the least number of perfect square numbers (1, 4, 9, 16, ...) that sum to n.

Input: 12 → Output: 3 (4+4+4)

Input: 13 → Output: 2 (4+9)

DP: dp[i] = minimum number of perfect squares summing to i. For each i, try all perfect squares <= i.

Initialize dp[] with i (worst case: all 1s).
For each i from 1 to n: for each j where j*j <= i: dp[i] = min(dp[i], dp[i-j*j]+1).
Return dp[n].

class Solution { public int numSquares(int n) { int[] dp = new int[n + 1]; java.util.Arrays.fill(dp, Integer.MAX_VALUE); dp[0] = 0; for (int i = 1; i <= n; i++) { for (int j = 1; j * j <= i; j++) { dp[i] = Math.min(dp[i], dp[i - j * j] + 1); } } return dp[n]; } }

Time Complexity: O(N * sqrt(N))
Space Complexity: O(N)