Maximum sum in a contiguous subarray of a given 2D array

I have solved this problem but all the explanation given are with DP solutions having complexity at least N^3. I have solved it in N^2 I am wondering if my solution is wrong.

We are given a 2D array of integers (+ve and -ve), we are two find the subarray of this array having the maximum possible sum.

Algorithm that I have used is to fix 4 indices. For 2 columns I span from 0 to N to find where the maximum sum occurs and then I span from N to 0. From these two results I deduce from which column to which one will the sum be maximum. S Same approach I use for rows. Once, I get the 4 indices, I am done. Complexity: O(N^2)

Here is my code:

#include <iostream>

using namespace std;

template<size_t n>
int find_col_sum(int A[][n], int m, int r1, int r2, int c) {
    int sum = 0;
    for (int i = r1; i <= r2; i++)
        sum += A[i][c];
    return sum;
}
template<size_t n>
int find_row_sum(int A[][n], int m, int c1, int c2, int r) {
    int sum = 0;
    for (int i = c1; i <= c2; i++)
        sum += A[r][i];
    return sum;
}

template<size_t n>
void max_sum_array(int A[][n], int m, int &i1, int &i2, int &i3, int &i4) {
    int mx1 = 0, mx2 = 0;
    int sum = 0, max1_sum = 0;
    int max2_sum = 0;

    //0 to col
    for (int i = 0; i < n; i++) {
        sum += find_col_sum(A,m,0,n-1,i);
        if (sum > max1_sum) {
            mx1 = i;
            max1_sum = sum;
        }
    }
    //n-1 to col
    sum = 0; mx2 = n-1;
    for (int i = n-1; i >= 0; i--) {
        sum += find_col_sum(A,m,0,n-1,i);
        if (sum > max2_sum) {
            mx2 = i;
            max2_sum = sum;
        }
    }
    if (mx1 <= mx2) {
        if (max1_sum >= max2_sum) {
            mx2 = mx1; mx1 = 0;
        }
        else {
            mx1 = mx2; mx2 = n-1;
        }
    }
    else
        swap(mx1,mx2);
    i1 = mx1; i2 = mx2;
    mx1 = 0; mx2 = m-1;
    max1_sum = 0; max2_sum = 0;
    sum = 0;
    //0 to row
    for (int i = 0; i < m; i++) {
        sum += find_row_sum(A,m,i1,i2,i);
        if (sum > max1_sum) {
            mx1 = i;
            max1_sum = sum;
        }
    }
    sum = 0;
    //m-1 to row
    for (int i = m-1; i >= 0; i--) {
        sum += find_row_sum(A,m,i1,i2,i);
        if (sum > max2_sum) {
            mx2 = i;
            max2_sum = sum;
        }
    }
    if (mx1 <= mx2) {
        if (max1_sum >= max2_sum) {
            mx2 = mx1; mx1 = 0;
        }
        else {
            mx1 = mx2; mx2 = m-1;
        }
    } else
        swap(mx1,mx2);
    i3 = mx1; i4 = mx2;
}

int main() {
    int A[][4] = {{0,-2,-7,0},{9,2,-6,2},{-4,1-4,1},{-1,8,0,-2}};
    int i1 = 0, i2 = 0, i3 = 0, i4 = 0;

    max_sum_array(A,4,i1,i2,i3,i4);
    cout<<i1<<" "<<i2<<" "<<i3<<" "<<i4<<endl;

    return 0;
}