Question

2352. Equal Row and Column Pairs

中文文档

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 $O(n^3)$, where $n$ is the number of rows or columns in the matrix $grid$. The space complexity is $O(1)$.

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

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;
  }
}

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;
  }
};

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
}

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;
}