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Formatted question description: https://leetcode.ca/all/2352.html

# 2352. Equal Row and Column Pairs

• Difficulty: Medium.
• Related Topics: Array, Hash Table, Matrix, Simulation.
• Similar Questions: .

## Problem

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

## Solution (Java, C++, Python)

• class Solution {
public int equalPairs(int[][] grid) {
int[] tmpCol = new int[grid.length];
Map<Integer, Integer> pairsMap = new HashMap<>();
int pairsCounter = 0;
for (int col = 0; col < grid[0].length; col++) {
for (int row = 0; row < grid.length; row++) {
tmpCol[row] = grid[row][col];
}
int hashCode = Arrays.hashCode(tmpCol);
pairsMap.put(hashCode, pairsMap.getOrDefault(hashCode, 0) + 1);
}
for (int[] row : grid) {
int hashCode = Arrays.hashCode(row);
if (pairsMap.containsKey(hashCode)) {
pairsCounter += pairsMap.get(hashCode);
}
}
return pairsCounter;
}
}

############

class Solution {
public int equalPairs(int[][] grid) {
int n = grid.length;
int[][] g = new int[n][n];
for (int j = 0; j < n; ++j) {
for (int i = 0; i < n; ++i) {
g[i][j] = grid[j][i];
}
}
int ans = 0;
for (var row : grid) {
for (var col : g) {
int ok = 1;
for (int i = 0; i < n; ++i) {
if (row[i] != col[i]) {
ok = 0;
break;
}
}
ans += ok;
}
}
return ans;
}
}

• class Solution:
def equalPairs(self, grid: List[List[int]]) -> int:
g = [list(col) for col in zip(*grid)]
return sum(row == col for row in grid for col in g)

############

# 2352. Equal Row and Column Pairs
# https://leetcode.com/problems/equal-row-and-column-pairs/

class Solution:
def equalPairs(self, grid: List[List[int]]) -> int:
rows, cols = len(grid), len(grid[0])
R = Counter()
C = Counter()

for row in grid:
R[tuple(row)] += 1

for col in zip(*grid):
C[tuple(col)] += 1

res = 0
for row, cnt in R.items():
res += cnt * C[row]

return res


• class Solution {
public:
int equalPairs(vector<vector<int>>& grid) {
int n = grid.size();
vector<vector<int>> g(n, vector<int>(n));
for (int j = 0; j < n; ++j) {
for (int i = 0; i < n; ++i) {
g[i][j] = grid[j][i];
}
}
int ans = 0;
for (auto& row : grid) {
for (auto& col : g) {
ans += row == col;
}
}
return ans;
}
};

• func equalPairs(grid [][]int) (ans int) {
n := len(grid)
g := make([][]int, n)
for i := range g {
g[i] = make([]int, n)
for j := 0; j < n; j++ {
g[i][j] = grid[j][i]
}
}
for _, row := range grid {
for _, col := range g {
ok := 1
for i, v := range row {
if v != col[i] {
ok = 0
break
}
}
ans += ok
}
}
return
}

• function equalPairs(grid: number[][]): number {
const n = grid.length;
const g = Array.from({ length: n }, () =>
Array.from({ length: n }, () => 0),
);
for (let j = 0; j < n; ++j) {
for (let i = 0; i < n; ++i) {
g[i][j] = grid[j][i];
}
}
let ans = 0;
for (const row of grid) {
for (const col of g) {
ans += Number(row.toString() === col.toString());
}
}
return ans;
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).