279. 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.

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

The idea using DP is simple, For a number N, if some int i*i = N, then dp[N] = 1;

else for every k from 1 -> n - 1 dp[N] = min(dp[k] + dp[n-k]);

Code:

public class Solution {    public int numSquares(int n) {        int[] dp = new int[n + 1];        int i = 0;        while(i*i <= n){            dp[i*i] = 1;            i++;        }        for(i = 2; i < dp.length; i++){            if(dp[i] == 0){                int temp = Integer.MAX_VALUE;                for(int k = 1; k < i; k++){                    temp = Math.min(temp,dp[k] + dp[i - k]);                }                dp[i] = temp;            }        }        return dp[n];    }}

However, this gets TLE, since it actually requires O(N2) time.

How to optimize, for example 12 = 4 + 4 + 4, we get 12 = 1 + 11, 2 + 10 3 + 9....., which is unnecessary.

for 2 + 10 , it actually the same with 1 + 11, because they all will get factor 1,

So what we could do is to pick k  that is the square of some int.

public class Solution {    public int numSquares(int n) {        int[] dp = new int[n + 1];        int i = 0;        while(i*i <= n){            dp[i*i] = 1;            i++;        }        for(i = 2; i < dp.length; i++){            if(dp[i] == 0){                int temp = Integer.MAX_VALUE;                for(int k = 1; k * k < i; k++){                    temp = Math.min(temp,1 + dp[i - k*k]);                }                dp[i] = temp;            }        }        return dp[n];    }}