Summed Matrix

A matrix is constructed of size n*n and given an integer ‘q’. The value at every cell of the matrix is given as, M(i,j) = i+j, where ‘M(i,j)‘ is the value of a cell, ‘i’ is the row number, and ‘j’ is the column number. Return the number of cells having value ‘q’.

Note: Assume, the array is in 1-based indexing.

Examples:

Input:n = 4, q = 7
Output: 2
Explanation: Matrix becomes
2 3 4 5 
3 4 5 6 
4 5 6 7
5 6 7 8
The count of 7 is 2.

Input:n = 5, q = 4
Output: 3
Explanation: Matrix becomes
2 3 4 5 6 
3 4 5 6 7 
4 5 6 7 8 
5 6 7 8 9 
6 7 8 9 10 
The count of 4 is 3.

Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)

Constraints:
1 ≤ n,q ≤ 10¹⁸

Approach 01:

#include <iostream>

class Solution {
public:
    long long sumMatrix(long long rows, long long query_position) {
        if (query_position < 2 || query_position > 2 * rows)
            return 0;
        else if (query_position >= 2 && query_position <= rows + 1)
            return query_position - 1;
        return 2 * rows - query_position + 1;
    }
};

class Solution:
    def sumMatrix(self, rows: int, query_position: int) -> int:
        if query_position < 2 or query_position > 2 * rows:
            return 0
        elif 2 <= query_position <= rows + 1:
            return query_position - 1
        return 2 * rows - query_position + 1