# [leetcode] 2352. Equal Row and Column Pairs _ Algorithm Problem Solve for python

## 1. Problem

2352. Equal Row and Column Pairs

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] <= 10^5

## 2. Solution

I solved this problem like this.

• Complexity
• Time complexity : O(N^2)
• Space complexity : O(N^2)
• Step
• I solve this problem using hash. Make key using row & column.
• Count each key and get result.
class Solution:
def equalPairs(self, grid: List[List[int]]) -> int:
dic_row = {}
dic_col = {}
for r_idx in range(len(grid)):
row = grid[r_idx]
key = str(row)
if key not in dic_row:
dic_row[key] = 0
dic_row[key] += 1

for c_idx in range(len(grid)):
col = [g[c_idx] for g in grid]
key = str(col)
if key not in dic_col:
dic_col[key] = 0
dic_col[key] += 1