Perfect Squares

This in No.279 in LeetCode.

The math theory is:

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
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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
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dp[n] = Min{ dp[n - i*i] + 1 },  n - i*i >=0 && i >= 1

Because 0 = 0; 1 = 0 + 1; 2 = 1+ 1; 3 = 1 + 1 + 1 = 2 + 1 ....

So we set a dp array and set the first one as 0 ( 0 = 0).
And then we use a for loop to go over the array, update the minimum value for each integer n.

The solution is post below:

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++ ){
            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];
    }

}

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