以下代码段的时间复杂度是否为O（n ^ 2）？

Question

I was wondering if the time complexity of the following code snippet is O(n^2) :

class Solution {
public:

    int numSquares(int n) {
        if(n<=0)
            return 0;

        vector<int> dp(n+1, INT_MAX);
        dp[0]=0;
        for(int i=1; i<=n; i++) {
            for(int j=1; j*j<=i; j++) {
                //+1 because you are adding the current `j`
                dp[i]=min(dp[i], dp[i-j*j]+1);
            }
        }

        return dp[n];
    }
};

I am not sure because in the inner loop, we are checking for perfect squares less than i , which would be very less in comparison to i (and I think so less, that they can be assumed to be constant). In this case then, the complexity would be just O(n) . So, can I say that the complexity is O(n) or is it O(n^2) ?

Note: The code snippet is a solution to a question from LeetCode.com which apparently has a collection of interview questions.

Answer 1

The outer loop is O(N) .
The inner loop is O(sqrt(i)) .

The sum will be:

1 + sqrt(2) + ... + sqrt(N)

It's greater than O(N) but is less than O(N^2) .

Without going into a very accurate computation of the above sum, I would say, it's close to O(N*sqrt(N)) .

Update

From http://ramanujan.sirinudi.org/Volumes/published/ram09.pdf , the above sum is:

C1 + (2.0/3)*N*SQRT(N) + (1.0/2)*SQRT(N) + ....