2352.Equal Row and Column Pairs
===

###### tags: `Medium`,`Array`,`Hash Table`,`Matrix`,`Simulation`

[2352. Equal Row and Column Pairs](https://leetcode.com/problems/equal-row-and-column-pairs/)

### 題目描述

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).

### 範例

**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

### 解答

#### Python

class Solution:
    def equalPairs(self, grid: List[List[int]]) -> int:
        d = defaultdict(int)
        ans = 0
        for row in grid:
            d[str(row)] += 1
        for col in map(list, zip(*grid)):
            ans += d[str(col)]
        return ans

> [name=Yen-Chi Chen][time=Wed, Jun 14, 2023]

#### TypeScript

function equalPairs(grid: number[][]): number {
  const n = grid.length;
  const rows = new Map<string, number>();
  const cols = new Map<string, number>();

  for (let i = 0; i < n; i++) {
    const row = grid[i].join(',');
    const col = grid.map((row) => row[i]).join(',');
    rows.set(row, (rows.get(row) || 0) + 1);
    cols.set(col, (cols.get(col) || 0) + 1);
  }

  let count = 0;
  for (const [key, value] of rows.entries()) {
    count += value * (cols.get(key) || 0);
  }

  return count;
}

> [name=Sheep][time=Wed, Jun 14, 2023]

#### C#

public class Solution {
    public int EqualPairs(int[][] grid) {
        int n = grid.Length;
        var dic = new Dictionary<string, int>();

        for (int i = 0; i < n; i++) {
            string key = string.Join('-', grid[i]);
            dic.TryGetValue(key, out int count);
            dic[key] = count + 1;
        }

        int total = 0;
        for (int j = 0; j < n; j++) {
            string key = string.Join('-', grid.Select(row => row[j]));
            if (dic.TryGetValue(key, out int count)) {
                total += count;
            }
        }

        return total;
    }
}

>[name=Jim][time=Jun 14, 2023]
### Reference

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