Find distinct elements common to all rows of a matrix

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 of 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. 

Python3

# 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 

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:

Map the element of 1st row in a hash table. Let it be hash.
Fow row = 2 to n
Map each element of the current row into a temporary hash table. Let it be temp.
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.
When all the rows are being processed in this manner, then the elements left in hash are the required common elements.

// 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; 
} 

Output:

3 2

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

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Find distinct elements common to all rows of a matrix

C++

Java

Python3

C#

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