LeetCode 2352. Equal Row and Column Pairs Solution in Java, C++, Python & More | Explanation + Code

Description

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] <= 105

Solutions

Solution 1: Simulation

We directly compare each row and column of the matrix grid. If they are equal, then it is a pair of equal row-column pairs, and we increment the answer by one.

The time complexity is O(n³), where n is the number of rows or columns in the matrix grid. The space complexity is O(1).

PythonJavaC++GoTypeScript

class Solution: def equalPairs(self, grid: List[List[int]]) -> int: n = len(grid) ans = 0 for i in range(n): for j in range(n): ans += all(grid[i][k] == grid[k][j] for k in range(n)) return ans(code-box)

class Solution { public int equalPairs(int[][] grid) { int n = grid.length; int ans = 0; for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { int ok = 1; for (int k = 0; k < n; ++k) { if (grid[i][k] != grid[k][j]) { ok = 0; break; } } ans += ok; } } return ans; } }(code-box)

class Solution { public: int equalPairs(vector<vector<int>>& grid) { int n = grid.size(); int ans = 0; for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { int ok = 1; for (int k = 0; k < n; ++k) { if (grid[i][k] != grid[k][j]) { ok = 0; break; } } ans += ok; } } return ans; } };(code-box)

func equalPairs(grid [][]int) (ans int) { for i := range grid { for j := range grid { ok := 1 for k := range grid { if grid[i][k] != grid[k][j] { ok = 0 break } } ans += ok } } return }(code-box)

function equalPairs(grid: number[][]): number { const n = grid.length; let ans = 0; for (let i = 0; i < n; ++i) { for (let j = 0; j < n; ++j) { let ok = 1; for (let k = 0; k < n; ++k) { if (grid[i][k] !== grid[k][j]) { ok = 0; break; } } ans += ok; } } return ans; }(code-box)