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.

MinSquare.java

// dp[0] = 0
// dp[i] =  min(dp[n-i*i] + 1) where i > 0 && n-i*i >=0
public int numSquares(int n) {
    int dp[] = new int[n+1];

    dp[0] = 0;

    for(int i = 1; i <= n; i++) {
        int min = Integer.MAX_VALUE;
        int j = 1; 
        while(i - j*j >= 0) {
            min = Math.min(min, dp[i - j*j] + 1);
            j++;
        }
        dp[i] = min;
    }

    return dp[n];
}

For every i, it must be the sum of i-j^2 and a perfect square j^2

dp[] = [0 1 2 3 1 2 3 4 2 1 2 3 3 3...]

@shailrshah. I would really appreciate if can explain your thought process explicitly?