Problem of The Day: Equal Row and Column Pairs

Problem Statement

leetcode problem link

My Solution [Accepted]

public class Solution {
    public Dictionary<int, int[]> ColDict {set;get;}
    public Dictionary<int, int[]> RowDict {set;get;}
    public bool IsValid(int col, int row, int[][] grid) {
        int[] colItems = ColDict[col];
        int[] rowItems = RowDict[row];
        return colItems.SequenceEqual(rowItems);
    }
    public int EqualPairs(int[][] grid) {
        var n = grid.Length;
        var res = 0;
        ColDict = new Dictionary<int, int[]>();
        RowDict = new Dictionary<int, int[]>();

        for(int i = 0; i < n; i++) {
            ColDict.Add(i, grid[i].Select(x => x).ToArray());
        }

        for(int j = 0; j < n; j++) {
            int[] temp = new int[n];
            for (int i = 0; i < n; i++) {
                temp[i] = grid[i][j];
            }
            RowDict.Add(j, temp.Select(x => x).ToArray());
        }

        for(int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (IsValid(i, j, grid)) {
                    res++;
                }
            }
        }
        return res;
    }
}

Improved Code

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

        // Convert each row to a string key and count occurrences
        for (int i = 0; i < n; i++) {
            string key = string.Join(",", grid[i]);
            if (!rowCount.ContainsKey(key))
                rowCount[key] = 0;
            rowCount[key]++;
        }

        int result = 0;

        // Build each column and check if it matches any row
        for (int col = 0; col < n; col++) {
            int[] colArr = new int[n];
            for (int row = 0; row < n; row++)
                colArr[row] = grid[row][col];

            string colKey = string.Join(",", colArr);

            if (rowCount.ContainsKey(colKey))
                result += rowCount[colKey];
        }

        return result;
    }
}

Editorial

Approach 2: Hash Map

class Solution:
    def equalPairs(self, grid: List[List[int]]) -> int:
        count = 0
        n = len(grid)

        # Keep track of the frequency of each row.
        row_counter = collections.Counter(tuple(row) for row in grid)

        # Add up the frequency of each column in map.
        for c in range(n):
            col = [grid[i][c] for i in range(n)]
            count += row_counter[tuple(col)]


        return count