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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
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;
}
};
public int equalPairs(int[][] grid) {
int res = 0, n = grid.length;
HashMap<String, Integer> x = new HashMap<>();
HashMap<String, Integer> y = new HashMap<>();
for(int i=0; i<n; i++){
StringBuilder sb1 = new StringBuilder(), sb2 = new StringBuilder();
for(int j=0; j<n; j++){
sb1.append(grid[i][j]);
sb2.append(grid[j][i]);
sb1.append(','); sb2.append(',');
}
String curr1 = sb1.toString(), curr2 = sb2.toString();
x.put(curr1, x.getOrDefault(curr1, 0)+1);
y.put(curr2, y.getOrDefault(curr2, 0)+1);
}
for(String str : x.keySet())
if(y.containsKey(str))
res += x.get(str)*y.get(str);
return res;
}
def equalPairs(self, A):
count = Counter(zip(*A))
return sum(count[tuple(r)] for r in A)
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