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Find distinct elements common to all rows of a matrix

  • Difficulty Level : Medium
  • Last Updated : 09 Jul, 2021

Given a n x n matrix. The problem is to find all the distinct elements common to all rows of the matrix. The elements can be printed in any order.

Examples: 

Input : mat[][] = {  {2, 1, 4, 3},
                     {1, 2, 3, 2},  
                     {3, 6, 2, 3},  
                     {5, 2, 5, 3}  }
Output : 2 3

Input : mat[][] = {  {12, 1, 14, 3, 16},
                     {14, 2, 1, 3, 35},  
                     {14, 1, 14, 3, 11},  
                     {14, 25, 3, 2, 1},
                     {1, 18, 3, 21, 14}  }
Output : 1 3 14

Method 1: Using three nested loops. Check if an element of 1st row is present in all the subsequent rows. Time Complexity of O(n3). Extra space could be required to handle the duplicate elements.
 
Method 2: Sort all the rows of the matrix individually in increasing order. Then apply a modified approach to the problem of finding common elements in 3 sorted arrays. Below an implementation for the same is given. 

C++




// C++ implementation to find distinct elements
// common to all rows of a matrix
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
 
// function to individually sort
// each row in increasing order
void sortRows(int mat[][MAX], int n)
{
    for (int i=0; i<n; i++)
        sort(mat[i], mat[i] + n);
}
 
// function to find all the common elements
void findAndPrintCommonElements(int mat[][MAX], int n)
{
    // sort rows individually
    sortRows(mat, n);
 
    // current column index of each row is stored
    // from where the element is being searched in
    // that row
    int curr_index[n];
    memset(curr_index, 0, sizeof(curr_index));
    int f = 0;
 
    for (; curr_index[0]<n; curr_index[0]++)
    {
        // value present at the current column index
        // of 1st row
        int value = mat[0][curr_index[0]];
 
        bool present = true;
 
        // 'value' is being searched in all the
        // subsequent rows
        for (int i=1; i<n; i++)
        {
            // iterate through all the elements of
            // the row from its current column index
            // till an element greater than the 'value'
            // is found or the end of the row is
            // encountered
            while (curr_index[i] < n &&
                   mat[i][curr_index[i]] <= value)
                curr_index[i]++;
 
            // if the element was not present at the column
            // before to the 'curr_index' of the row
            if (mat[i][curr_index[i]-1] != value)
                present = false;
 
            // if all elements of the row have
            // been traversed
            if (curr_index[i] == n)
            {
                f = 1;
                break;
            }
        }
 
        // if the 'value' is common to all the rows
        if (present)
            cout << value << " ";
 
        // if any row have been completely traversed
        // then no more common elements can be found
        if (f == 1)
            break;
    }
}
 
// Driver program to test above
int main()
{
    int mat[][MAX] = {  {12, 1, 14, 3, 16},
        {14, 2, 1, 3, 35},
        {14, 1, 14, 3, 11},
        {14, 25, 3, 2, 1},
        {1, 18, 3, 21, 14}
    };
 
    int n = 5;
    findAndPrintCommonElements(mat, n);
    return 0;
}

Java




// JAVA Code to find distinct elements
// common to all rows of a matrix
import java.util.*;
 
class GFG {
     
    // function to individually sort
    // each row in increasing order
    public static void sortRows(int mat[][], int n)
    {
        for (int i=0; i<n; i++)
            Arrays.sort(mat[i]);
    }
      
    // function to find all the common elements
    public static void findAndPrintCommonElements(int mat[][],
                                                     int n)
    {
        // sort rows individually
        sortRows(mat, n);
      
        // current column index of each row is stored
        // from where the element is being searched in
        // that row
        int curr_index[] = new int[n];
         
        int f = 0;
      
        for (; curr_index[0]<n; curr_index[0]++)
        {
            // value present at the current column index
            // of 1st row
            int value = mat[0][curr_index[0]];
      
            boolean present = true;
      
            // 'value' is being searched in all the
            // subsequent rows
            for (int i=1; i<n; i++)
            {
                // iterate through all the elements of
                // the row from its current column index
                // till an element greater than the 'value'
                // is found or the end of the row is
                // encountered
                while (curr_index[i] < n &&
                       mat[i][curr_index[i]] <= value)
                    curr_index[i]++;
      
                // if the element was not present at the
                // column before to the 'curr_index' of the
               // row
                if (mat[i][curr_index[i]-1] != value)
                    present = false;
      
                // if all elements of the row have
                // been traversed
                if (curr_index[i] == n)
                {
                    f = 1;
                    break;
                }
            }
      
            // if the 'value' is common to all the rows
            if (present)
               System.out.print(value+" ");
      
            // if any row have been completely traversed
            // then no more common elements can be found
            if (f == 1)
                break;
        }
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int mat[][] = {  {12, 1, 14, 3, 16},
                         {14, 2, 1, 3, 35},
                         {14, 1, 14, 3, 11},
                         {14, 25, 3, 2, 1},
                         {1, 18, 3, 21, 14}
                                            };
          
            int n = 5;
            findAndPrintCommonElements(mat, n);
    }
  }
// This code is contributed by Arnav Kr. Mandal.

Python




# Python3 implementation to find distinct
# elements common to all rows of a matrix
MAX = 100
 
# function to individually sort
# each row in increasing order
def sortRows(mat, n):
 
    for i in range(0, n):
        mat[i].sort();
 
# function to find all the common elements
def findAndPrintCommonElements(mat, n):
 
    # sort rows individually
    sortRows(mat, n)
 
    # current column index of each row is
    # stored from where the element is being
    # searched in that row
     
    curr_index = [0] * n
    for i in range (0, n):
        curr_index[i] = 0
         
    f = 0
 
    while(curr_index[0] < n):
     
        # value present at the current
        # column index of 1st row
        value = mat[0][curr_index[0]]
 
        present = True
 
        # 'value' is being searched in
        # all the subsequent rows
        for i in range (1, n):
         
            # iterate through all the elements
            # of the row from its current column
            # index till an element greater than
            # the 'value' is found or the end of
            # the row is encountered
            while (curr_index[i] < n and
                   mat[i][curr_index[i]] <= value):
                curr_index[i] = curr_index[i] + 1
                 
            # if the element was not present at
            # the column before to the 'curr_index'
            # of the row
            if (mat[i][curr_index[i] - 1] != value):
                present = False
 
            # if all elements of the row have
            # been traversed)
            if (curr_index[i] == n):
             
                f = 1
                break
             
        # if the 'value' is common to all the rows
        if (present):
            print(value, end = " ")
 
        # if any row have been completely traversed
        # then no more common elements can be found
        if (f == 1):
            break
     
        curr_index[0] = curr_index[0] + 1
 
# Driver Code
mat = [[12, 1, 14, 3, 16],
       [14, 2, 1, 3, 35],
       [14, 1, 14, 3, 11],
       [14, 25, 3, 2, 1],
       [1, 18, 3, 21, 14]]
 
n = 5
findAndPrintCommonElements(mat, n)
 
# This code is contributed by iAyushRaj

C#




// C# Code to find distinct elements
// common to all rows of a matrix
using System;
 
class GFG
{
 
// function to individually sort
// each row in increasing order
public static void sortRows(int[][] mat, int n)
{
    for (int i = 0; i < n; i++)
    {
        Array.Sort(mat[i]);
    }
}
 
// function to find all the common elements
public static void findAndPrintCommonElements(int[][] mat,
                                              int n)
{
    // sort rows individually
    sortRows(mat, n);
 
    // current column index of each row is stored
    // from where the element is being searched in
    // that row
    int[] curr_index = new int[n];
 
    int f = 0;
 
    for (; curr_index[0] < n; curr_index[0]++)
    {
        // value present at the current column index
        // of 1st row
        int value = mat[0][curr_index[0]];
 
        bool present = true;
 
        // 'value' is being searched in all the
        // subsequent rows
        for (int i = 1; i < n; i++)
        {
            // iterate through all the elements of
            // the row from its current column index
            // till an element greater than the 'value'
            // is found or the end of the row is
            // encountered
            while (curr_index[i] < n &&
                   mat[i][curr_index[i]] <= value)
            {
                curr_index[i]++;
            }
 
            // if the element was not present at the column
            // before to the 'curr_index' of the row
            if (mat[i][curr_index[i] - 1] != value)
            {
                present = false;
            }
 
            // if all elements of the row have
            // been traversed
            if (curr_index[i] == n)
            {
                f = 1;
                break;
            }
        }
 
        // if the 'value' is common to all the rows
        if (present)
        {
            Console.Write(value + " ");
        }
 
        // if any row have been completely traversed
        // then no more common elements can be found
        if (f == 1)
        {
            break;
        }
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    int[][] mat = new int[][]
    {
        new int[] {12, 1, 14, 3, 16},
        new int[] {14, 2, 1, 3, 35},
        new int[] {14, 1, 14, 3, 11},
        new int[] {14, 25, 3, 2, 1},
        new int[] {1, 18, 3, 21, 14}
    };
 
    int n = 5;
    findAndPrintCommonElements(mat, n);
}
}
 
// This code is contributed by Shrikant13

Javascript




<script>
 
// JavaScript Code to find distinct elements
// common to all rows of a matrix
 
    // function to individually sort
    // each row in increasing order
    function sortRows(mat,n)
    {
        for (let i=0; i<n; i++)
            mat[i].sort(function(a,b){return a-b;});
    }
     
    // function to find all the common elements
    function findAndPrintCommonElements(mat,n)
    {
        // sort rows individually
        sortRows(mat, n);
        // current column index of each row is stored
        // from where the element is being searched in
        // that row
        let curr_index = new Array(n);
        for(let i=0;i<n;i++)
        {
            curr_index[i]=0;
        }
        let f = 0;
       
        for (; curr_index[0]<n; curr_index[0]++)
        {
            // value present at the current column index
            // of 1st row
            let value = mat[0][curr_index[0]];
       
            let present = true;
       
            // 'value' is being searched in all the
            // subsequent rows
            for (let i=1; i<n; i++)
            {
                // iterate through all the elements of
                // the row from its current column index
                // till an element greater than the 'value'
                // is found or the end of the row is
                // encountered
                while (curr_index[i] < n &&
                       mat[i][curr_index[i]] <= value)
                    curr_index[i]++;
       
                // if the element was not present at the
                // column before to the 'curr_index' of the
               // row
                if (mat[i][curr_index[i]-1] != value)
                    present = false;
       
                // if all elements of the row have
                // been traversed
                if (curr_index[i] == n)
                {
                    f = 1;
                    break;
                }
            }
       
            // if the 'value' is common to all the rows
            if (present)
               document.write(value+" ");
       
            // if any row have been completely traversed
            // then no more common elements can be found
            if (f == 1)
                break;
        }
    }
     
    /* Driver program to test above function */
    let mat = [[12, 1, 14, 3, 16],
       [14, 2, 1, 3, 35],
       [14, 1, 14, 3, 11],
       [14, 25, 3, 2, 1],
       [1, 18, 3, 21, 14]]
     
    let n = 5;
    findAndPrintCommonElements(mat, n);
      
// This code is contributed by patel2127
 
</script>

Output:  

1 3 14

Time Complexity: O(n2log n), each row of size n requires O(nlogn) for sorting and there are total n rows. 
Auxiliary Space: O(n) to store current column indexes for each row.
 
Method 3: It uses the concept of hashing. The following steps are:  



  1. Map the element of the 1st row in a hash table. Let it be hash.
  2. Fow row = 2 to n
  3. Map each element of the current row into a temporary hash table. Let it be temp.
  4. Iterate through the elements of hash and check that the elements in hash are present in temp. If not present then delete those elements from hash.
  5. When all the rows are being processed in this manner, then the elements left in hash are the required common elements.

C++




// C++ program to find distinct elements
// common to all rows of a matrix
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 100;
 
// function to individually sort
// each row in increasing order
void findAndPrintCommonElements(int mat[][MAX], int n)
{
    unordered_set<int> us;
 
    // map elements of first row
    // into 'us'
    for (int i=0; i<n; i++)
        us.insert(mat[0][i]);
 
    for (int i=1; i<n; i++)
    {
        unordered_set<int> temp;
       
        // mapping elements of current row
        // in 'temp'
        for (int j=0; j<n; j++)
            temp.insert(mat[i][j]);
 
        unordered_set<int>:: iterator itr;
 
        // iterate through all the elements
        // of 'us'
        for (itr=us.begin(); itr!=us.end(); itr++)
 
            // if an element of 'us' is not present
            // into 'temp', then erase that element
            // from 'us'
            if (temp.find(*itr) == temp.end())
                us.erase(*itr);
 
        // if size of 'us' becomes 0,
        // then there are no common elements
        if (us.size() == 0)
            break;
    }
 
    // print the common elements
    unordered_set<int>:: iterator itr;
    for (itr=us.begin(); itr!=us.end(); itr++)
        cout << *itr << " ";
}
 
// Driver program to test above
int main()
{
    int mat[][MAX] = { {2, 1, 4, 3},
                       {1, 2, 3, 2},
                       {3, 6, 2, 3},
                       {5, 2, 5, 3}  };
    int n = 4;
    findAndPrintCommonElements(mat, n);
    return 0;
}

Python




# Python3 program to find distinct elements
# common to all rows of a matrix
MAX = 100
 
# function to individually sort
# each row in increasing order
def findAndPrintCommonElements(mat, n):
    us = dict()
 
    # map elements of first row
    # into 'us'
    for i in range(n):
        us[mat[0][i]] = 1
 
    for i in range(1, n):
        temp = dict()
         
        # mapping elements of current row
        # in 'temp'
        for j in range(n):
            temp[mat[i][j]] = 1
 
        # iterate through all the elements
        # of 'us'
        for itr in list(us):
 
            # if an element of 'us' is not present
            # into 'temp', then erase that element
            # from 'us'
            if itr not in temp:
                del us[itr]
 
        # if size of 'us' becomes 0,
        # then there are no common elements
        if (len(us) == 0):
            break
 
    # prthe common elements
    for itr in list(us)[::-1]:
        print(itr, end = " ")
 
# Driver Code
mat = [[2, 1, 4, 3],
       [1, 2, 3, 2],
       [3, 6, 2, 3],
       [5, 2, 5, 3]]
n = 4
findAndPrintCommonElements(mat, n)
 
# This code is contributed by Mohit Kumar

Javascript




<script>
 
// Javascript program to find distinct elements
// common to all rows of a matrix
var MAX = 100;
 
// function to individually sort
// each row in increasing order
function findAndPrintCommonElements(mat, n)
{
    var us = new Set();
 
    // map elements of first row
    // into 'us'
    for (var i = 0; i < n; i++)
        us.add(mat[0][i]);
 
    for(var i = 1; i < n; i++)
    {
        var temp = new Set();
       
        // mapping elements of current row
        // in 'temp'
        for (var j = 0; j < n; j++)
            temp.add(mat[i][j]);
 
        // iterate through all the elements
        // of 'us'
        for(var itr of us)
        {
 
            // if an element of 'us' is not present
            // into 'temp', then erase that element
            // from 'us'
            if(!temp.has(itr))
                us.delete(itr);
        }
 
        // if size of 'us' becomes 0,
        // then there are no common elements
        if (us.size == 0)
            break;
    }
 
    // print the common elements
    for(var itr of [...us].sort((a,b)=>b-a))
        document.write( itr + " ");
}
 
// Driver program to test above
var mat = [ [2, 1, 4, 3],
                   [1, 2, 3, 2],
                   [3, 6, 2, 3],
                   [5, 2, 5, 3]];
var n = 4;
findAndPrintCommonElements(mat, n);
 
// This code is contributed by noob2000.
</script>

 Output: 

3 2

Time Complexity: O(n2
Space Complexity: O(n)

Method 4: Using Map

  1. Insert all the elements of the 1st row in the map.
  2. Now we check that the elements present in the map are present in each row or not.
  3. If the element is present in the map and is not duplicated in the current row, then we increment the count of the element in the map by 1.
  4. If we reach the last row while traversing and if the element appears (N-1) times before then we print the element.

Java




// JAVA Code to find distinct elements
// common to all rows of a matrix
import java.io.*;
import java.util.*;
 
class GFG {
  static void distinct(int matrix[][], int N)
  {
    // make a empty map
    Map<Integer, Integer> ans = new HashMap<>();
 
    // Insert the elements of
    // first row in the map and
    // initialize with 1
    for (int j = 0; j < N; j++) {
      ans.put(matrix[0][j], 1);
    }
 
    // Traverse the matrix from 2nd row
    for (int i = 1; i < N; i++) {
      for (int j = 0; j < N; j++) {
 
        // If the element is present in the map
        // and is not duplicated in the current row
        if (ans.get(matrix[i][j]) != null
            && ans.get(matrix[i][j]) == i) {
 
          // Increment count of the element in
          // map by 1
          ans.put(matrix[i][j], i + 1);
 
          // If we have reached the last row
          if (i == N - 1) {
 
            // Print the element
            System.out.print(matrix[i][j]
                             + " ");
          }
        }
      }
    }
  }
 
  /* Driver program to test above function */
  public static void main(String[] args)
  {
    int matrix[][] = { { 2, 1, 4, 3 },
                      { 1, 2, 3, 2 },
                      { 3, 6, 2, 3 },
                      { 5, 2, 5, 3 } };
    int n = 4;
    distinct(matrix, n);
  }
}
// This code is Contributed by Darshit Shukla

C#




// C# code to find distinct elements
// common to all rows of a matrix
using System;
using System.Collections.Generic;
 
class GFG{
     
static void distinct(int[,] matrix, int N)
{
     
    // Make a empty map
    Dictionary<int,
               int> ans = new Dictionary<int,
                                         int>();
 
    // Insert the elements of
    // first row in the map and
    // initialize with 1
    for(int j = 0; j < N; j++)
    {
        ans[matrix[0, j]] = 1;
    }
 
    // Traverse the matrix from 2nd row
    for(int i = 1; i < N; i++)
    {
        for(int j = 0; j < N; j++)
        {
             
            // If the element is present in the map
            // and is not duplicated in the current row
            if (ans.ContainsKey(matrix[i, j]) &&
                            ans[matrix[i, j]] == i)
            {
                 
                // Increment count of the element in
                // map by 1
                ans[matrix[i, j]] = i + 1;
 
                // If we have reached the last row
                if (i == N - 1)
                {
                     
                    // Print the element
                    Console.Write(matrix[i, j] + " ");
                }
            }
        }
    }
}
 
// Driver code
public static void Main(string[] args)
{
    int[,] matrix = { { 2, 1, 4, 3 },
                      { 1, 2, 3, 2 },
                      { 3, 6, 2, 3 },
                      { 5, 2, 5, 3 } };
    int n = 4;
     
    distinct(matrix, n);
}
}
 
// This code is contributed by ukasp

Javascript




<script>
 
// Javascript code to find distinct elements
// common to all rows of a matrix
 
     
function distinct(matrix, N)
{
     
    // Make a empty map
    var ans = new Map()
 
    // Insert the elements of
    // first row in the map and
    // initialize with 1
    for(var j = 0; j < N; j++)
    {
        ans.set(matrix[0][j], 1);
    }
 
    // Traverse the matrix from 2nd row
    for(var i = 1; i < N; i++)
    {
        for(var j = 0; j < N; j++)
        {
             
            // If the element is present in the map
            // and is not duplicated in the current row
            if (ans.has(matrix[i][j]) &&
                            ans.get(matrix[i][j]) == i)
            {
                 
                // Increment count of the element in
                // map by 1
                ans.set(matrix[i][j], i + 1);
 
                // If we have reached the last row
                if (i == N - 1)
                {
                     
                    // Print the element
                    document.write(matrix[i][j] + " ");
                }
            }
        }
    }
}
 
// Driver code
var matrix = [ [ 2, 1, 4, 3 ],
                  [ 1, 2, 3, 2 ],
                  [ 3, 6, 2, 3 ],
                  [ 5, 2, 5, 3 ] ];
var n = 4;
 
distinct(matrix, n);
 
// This code is contributed by rutvik_56.
</script>

Output:

2 3

Time Complexity: O(n2

Space Complexity: O(n)

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