[Solved] Equal Row and Column Pairs LeetCode Contest 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).

Equal Row and Column Pairs LeetCode Contest

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

class Solution {
public:
    int equalPairs(vector<vector<int>>& grid) 
    {
        // Number to store the count of equal pairs.
        int ans = 0;
        map<vector<int>, int> mp;
        // Storing each row int he map
        for (int i = 0; i < grid.size(); i++)
            mp[grid[i]]++;
        
        for (int i = 0; i < grid[0].size(); i++)
        {
            vector<int> v;
            // extracting column in a vector.
            for (int j = 0; j < grid.size(); j++)
                v.push_back(grid[j][i]);
            // Add the number of times that column appeared as a row.
            ans += mp[v];
        }
        // Return the number of count
        return ans;
    }
};

class Solution {
    public int equalPairs(int[][] g) {
		int n = g.length;
        int res = 0;
		//convert the numbers in each row to a single string
        Map<String, Integer> sToFreq = new HashMap<>();
        for (int i = 0; i < n; i++) {
            StringBuilder sb = new StringBuilder();
            for (int j = 0; j < n; j++) { 
                sb.append(g[i][j]);
                sb.append("+");
            }
            String s = sb.toString();
            sToFreq.put(s, sToFreq.getOrDefault(s, 0) + 1);
        }
        
		//convert the numbers in each col to string and find the matches
        for (int j = 0; j < n; j++) {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < n; i++) { 
                sb.append(g[i][j]);
                sb.append("+");
            }
            String s = sb.toString();
            if (sToFreq.containsKey(s)) {
                res += sToFreq.get(s);
            }
        }
        return res;
    }
}

    def equalPairs(self, grid: List[List[int]]) -> int:
        pairs = 0
        cnt = Counter(tuple(row) for row in grid)
        for tpl in zip(*grid):
            pairs += cnt[tpl]
        return pairs

    def equalPairs(self, A):
        count = Counter(zip(*A))
        return sum(count[tuple(r)] for r in A)

Happy Learning – If you require any further information, feel free to contact me.