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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)

• Java
• Python
• C++
• Go
• TypeScript

• 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).

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