We are given a m*n matrix of positive integers and a row number. The task is to find all rows in given matrix which are permutations of given row elements. It is also given that values in every row are distinct.

Examples:

Input : mat[][] = {{3, 1, 4, 2}, , {1, 6, 9, 3}, {1, 2, 3, 4}, {4, 3, 2, 1}} row = 3 Output: 0, 2 Rows at indexes 0 and 2 are permutations of row at index 3.

A **simple solution** is to one by one sort all rows and check all rows. If any row is completely equal to given row that means current row is a permutation of given row. Time complexity for this approach will be O(m*n log n).

An **efficient approach** is to use a hashing. Simply create a hash set for given row. After hash set creation, traverse through remaining rows and for every row check if all of its elements are present in hash set or not.

## CPP

// C++ program to find all permutations of a given row #include<bits/stdc++.h> #define MAX 100 using namespace std; // Function to find all permuted rows of a given row r void permutatedRows(int mat[][MAX], int m, int n, int r) { // Creating an empty set unordered_set<int> s; // Count frequencies of elements in given row r for (int j=0; j<n; j++) s.insert(mat[r][j]); // Traverse through all remaining rows for (int i=0; i<m; i++) { // we do not need to check for given row r if (i==r) continue; // initialize hash i.e; count frequencies // of elements in row i int j; for (j=0; j<n; j++) if (s.find(mat[i][j]) == s.end()) break; if (j != n) continue; cout << i << ", "; } } // Driver program to run the case int main() { int m = 4, n = 4,r = 3; int mat[][MAX] = {{3, 1, 4, 2}, {1, 6, 9, 3}, {1, 2, 3, 4}, {4, 3, 2, 1}}; permutatedRows(mat, m, n, r); return 0; }

## Python3

# Python program to find all # permutations of a given row # Function to find all # permuted rows of a given row r def permutatedRows(mat, m, n, r): # Creating an empty set s=set() # Count frequencies of # elements in given row r for j in range(n): s.add(mat[r][j]) # Traverse through all remaining rows for i in range(m): # we do not need to check # for given row r if i == r: continue # initialize hash i.e # count frequencies # of elements in row i for j in range(n): if mat[i][j] not in s: # to avoid the case when last # element does not match j = j - 2 break; if j + 1 != n: continue print(i) # Driver program to run the case m = 4 n = 4 r = 3 mat = [[3, 1, 4, 2], [1, 6, 9, 3], [1, 2, 3, 4], [4, 3, 2, 1]] permutatedRows(mat, m, n, r) # This code is contributed # by Upendra Singh Bartwal.

Output:

0, 2

Time complexity : O(m*n)

Auxiliary space : O(n)

**Exercise : **

Extend the above solution to work for input matrix where all elements of a row don’t have be distinct. (Hit : We can use Hash Map instead of Hash Set)

This article is contributed by **Shashank Mishra ( Gullu )**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.