I came up to an algorithm to solve this problem. however i am not preety sure what excactly is the time complexity of this solution. Is it O(n^3) or O(n^2) .I want to know what is the time complexity of the given function subarraysDivByK? Please help

{
    public int subarraysDivByK(int[] A, int K) 
    {
        int n = A.length;
        int count = 0;
        for(int i=0;i<n;i  )
        {
            for(int j=i;j<n;j  )
            {
                int sumWindow = findSum(A,i,j);
                if(sumWindow%K==0)
                {
                    count  ;
                }
            }
        }
        return count;
    }
    
    public int findSum(int[] a, int start, int end)
    {
        int sum = 0;
        for(int i = start;i<=end;i  )
        {
            sum = sum   a[i];
        }
        return sum;
    }
}

CodePudding user response：

For some value of i you'll try all values of j from i to n. And for each j you'll calculate the sum of j-i elements. You'll need O(0 1 ... (n-i))=O((n-i)^2) operations for that. So for all possible values of i you'll need O(n^2 (n-1)^2 ... 2^2 1^2) = O(n^3) operations. Answer: O(n^3).