# Leetcode 279 Perfect Square.

Leetcode 279 Perfect Square

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.

DP question:

```dp[0] = 0
dp[1] = dp[0]+1 = 1
dp[2] = dp[1]+1 = 2
dp[3] = dp[2]+1 = 3
dp[4] = Min{ dp[4-1*1]+1, dp[4-2*2]+1 }
= Min{ dp[3]+1, dp[0]+1 }
= 1
dp[5] = Min{ dp[5-1*1]+1, dp[5-2*2]+1 }
= Min{ dp[4]+1, dp[1]+1 }
= 2
.
.
.
dp[13] = Min{ dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1 }
= Min{ dp[12]+1, dp[9]+1, dp[4]+1 }
= 2
.
.
.
dp[n] = Min{ dp[n - i*i] + 1 },  n - i*i >=0 && i >= 1
```

So the code can be like this. For DP questions, the first thing to do is write down all possible situation, the we can find the solution that we want.

``````public 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=0){
min=Math.min(min,dp[i-j*j]+1);
j++;
}
dp[i]=min;
}
return dp[n];
}
}``````