LeetCode //C - 2352. Equal Row and Column Pairs

2352. Equal Row and Column Pairs

Given a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal.

A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array).
 

Example 1:

Input: grid = [[3,2,1],[1,7,6],[2,7,7]]
Output: 1
Explanation: There is 1 equal row and column pair:

• (Row 2, Column 1): [2,7,7]

Example 2:

Input: grid = [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]
Output: 3
Explanation: There are 3 equal row and column pairs:

• (Row 0, Column 0): [3,1,2,2]
• (Row 2, Column 2): [2,4,2,2]
• (Row 3, Column 2): [2,4,2,2]

Constraints:

• n == grid.length == grid[i].length
• 1 <= n <= 200
• $1<=grid[i][j]<=10^5$

From: LeetCode
Link: 2352. Equal Row and Column Pairs

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Solution:

Ideas：

This function counts the number of equal pairs by comparing each row with each column. If the elements in a row and column match for the entire length, it increments the count. This solution has a time complexity of $O(n^3)$ which is acceptable given the constraints $1\le n\le 200$.

Code:

int equalPairs(int** grid, int gridSize, int* gridColSize) {
    
    
    int count = 0;
    for (int i = 0; i < gridSize; ++i) {
    
     // Iterate over rows
        for (int j = 0; j < gridSize; ++j) {
    
     // Iterate over columns
            int k;
            for (k = 0; k < gridSize; ++k) {
    
     // Check if row[i] equals column[j]
                if (grid[i][k] != grid[k][j]) {
    
    
                    break; // If any element does not match, break out of the loop
                }
            }
            if (k == gridSize) {
    
     // If we didn't break, the row and column are equal
                count++;
            }
        }
    }
    return count;
}