LeetCode 279.

Tags:

Complexity

  • Time complexity: O(nsqrt(n))*

  • Space complexity: O(n)

Code

class Solution {
    public int numSquares(int n) {
        int[] dp = new int[10001];
        Arrays.fill(dp, 10000);

        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];
    }
}

Check out the description of this problem at LC 279.