11. Count pairs Sum in matrices

The problem can be found at the following link: Question Link

My Approach

To solve this problem, I have used a two-pointer approach.

• I initialize one pointer at the top-right corner of mat1 and another pointer at the bottom-left corner of mat2.

• Then, we move these pointers inward while adjusting them based on the sum of the elements at their respective positions.

• If the sum is equal to x, we increment the count and move both pointers inward. If the sum is greater than x, we decrement the right pointer, and if it's less than x, we increment the left pointer.

• We repeat this process until both pointers meet or cross each other.

Time and Auxiliary Space Complexity

• Time Complexity : O(n*n), where n is the size of the matrices. We traverse both matrices once using the two-pointer approach, resulting in linear time complexity.

• Auxiliary Space Complexity : O(1). We use only a constant amount of extra space for storing variables like l, r, cnt, etc.

Code (C++)

class Solution {
public:
    int countPairs(vector<vector<int>> &mat1, vector<vector<int>> &mat2, int n, int x)
    {
        int sz = n * n;
        int l = 0, r = sz - 1;
        int cnt = 0;
        
        while (l < sz && r >= 0)
        {
            int sum = mat1[l / n][l % n] + mat2[r / n][r % n];
            
            if (sum == x) {
                l++;
                r--;
                ++cnt;
            }
            else if (sum > x)
                --r;
            else
                ++l;
        }
        
        return cnt;
    }
};

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