279. Perfect Squares

Problem:

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.

Analysis:

Any number i can be composted of j*j and i - j*j. So that for 1 <= j*j<= i, dp[i] = min(dp[1 -j*j] + 1).

Solution:

class Solution {
    public int numSquares(int n) {
        int[] dp = new int[n + 1];
        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];
    }
}