Perfect Squares

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

Analyse: dp[m] = min{ dp[m - i * i] + 1 } as long as i * i <= m. 

Runtime: 400ms.

class Solution {
public:
    int numSquares(int n) {
        if(n <= 1) return n;
        
        vector<int> dp(n + 1, INT_MAX);
        dp[0] = 0;
        dp[1] = 1;
        
        for(int i = 1; i <= n; i++){
            for(int j = 1; j * j <= i; j++)
                dp[i] = min(dp[i], dp[i - j * j] + 1);
        }
        return dp[n];
    }
};