Find all permuted rows of a given row in a matrix

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

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// 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;
}

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Java

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// Java program to find all permutations of a given row
import java.util.*;
  
class GFG
{
      
static int MAX = 100;
  
// Function to find all permuted rows of a given row r
static void permutatedRows(int mat[][], int m, int n, int r)
{
    // Creating an empty set
    LinkedHashSet<Integer> s = new LinkedHashSet<>();
  
  
    // Count frequencies of elements in given row r
    for (int j = 0; j < n; j++)
        s.add(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.contains(mat[i][j]))
                break;
        if (j != n)
        continue;
  
        System.out.print(i+", ");
    }
}
  
// Driver program to run the case
public static void main(String[] args)
{
    int m = 4, n = 4,r = 3;
    int mat[][] = {{3, 1, 4, 2},
                    {1, 6, 9, 3},
                    {1, 2, 3, 4},
                    {4, 3, 2, 1}};
    permutatedRows(mat, m, n, r);
}
}
  
// This code has been contributed by 29AjayKumar

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Python3

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# 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.

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C#

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// C# program to find all permutations of a given row 
using System;
using System.Collections.Generic;
  
class GFG 
      
static int MAX = 100; 
  
// Function to find all permuted rows of a given row r 
static void permutatedRows(int [,]mat, int m, int n, int r) 
    // Creating an empty set 
    HashSet<int> s = new HashSet<int>(); 
  
  
    // Count frequencies of elements in given row r 
    for (int j = 0; j < n; j++) 
        s.Add(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.Contains(mat[i,j])) 
                break
        if (j != n) 
        continue
  
        Console.Write(i+", "); 
    
  
// Driver program to run the case 
public static void Main(String[] args) 
    int m = 4, n = 4,r = 3; 
    int [,]mat = {{3, 1, 4, 2}, 
                    {1, 6, 9, 3}, 
                    {1, 2, 3, 4}, 
                    {4, 3, 2, 1}}; 
    permutatedRows(mat, m, n, r); 
  
/* This code contributed by PrinciRaj1992 */

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Output:

0, 2

Time complexity : O(m*n)
Auxiliary space : O(n)

Another approach to the solution using Standard Template Library(STL):

CPP

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// 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)
{
   for (int i=0; i<m&&i!=r; i++){
        if(is_permutation(mat[i],mat[i]+n,mat[r])) 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;
}

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Output:

0, 2

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.

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