Find distinct elements common to all rows of a matrix

• Difficulty Level : Medium
• Last Updated : 13 Jan, 2022

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 using namespace std;const int MAX = 100; // function to individually sort// each row in increasing ordervoid sortRows(int mat[][MAX], int n){    for (int i=0; i

Java

 // JAVA Code to find distinct elements// common to all rows of a matriximport 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

Python

 # Python3 implementation to find distinct# elements common to all rows of a matrixMAX = 100 # function to individually sort# each row in increasing orderdef sortRows(mat, n):     for i in range(0, n):        mat[i].sort(); # function to find all the common elementsdef 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 =  * n    for i in range (0, n):        curr_index[i] = 0             f = 0     while(curr_index < n):             # value present at the current        # column index of 1st row        value = mat[curr_index]         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 = curr_index + 1 # Driver Codemat = [[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 = 5findAndPrintCommonElements(mat, n) # This code is contributed by iAyushRaj

C#

 // C# Code to find distinct elements// common to all rows of a matrixusing System; class GFG{ // function to individually sort// each row in increasing orderpublic static void sortRows(int[][] mat, int n){    for (int i = 0; i < n; i++)    {        Array.Sort(mat[i]);    }} // function to find all the common elementspublic 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 < n; curr_index++)    {        // value present at the current column index        // of 1st row        int value = mat[curr_index];         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 Codepublic 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



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 using namespace std; const int MAX = 100; // function to individually sort// each row in increasing ordervoid findAndPrintCommonElements(int mat[][MAX], int n){    unordered_set us;     // map elements of first row    // into 'us'    for (int i=0; i temp;               // mapping elements of current row        // in 'temp'        for (int j=0; j:: 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++);              else               itr++;         // if size of 'us' becomes 0,        // then there are no common elements        if (us.size() == 0)            break;    }     // print the common elements    unordered_set:: iterator itr;    for (itr=us.begin(); itr!=us.end(); itr++)        cout << *itr << " ";} // Driver program to test aboveint 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 matrixMAX = 100 # function to individually sort# each row in increasing orderdef findAndPrintCommonElements(mat, n):    us = dict()     # map elements of first row    # into 'us'    for i in range(n):        us[mat[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     # print common elements    for itr in list(us)[::-1]:        print(itr, end = " ") # Driver Codemat = [[2, 1, 4, 3],       [1, 2, 3, 2],       [3, 6, 2, 3],       [5, 2, 5, 3]]n = 4findAndPrintCommonElements(mat, n) # This code is contributed by Mohit Kumar

Javascript



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 matriximport java.io.*;import java.util.*; class GFG {  static void distinct(int matrix[][], int N)  {    // make a empty map    Map 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[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 matrixusing System;using System.Collections.Generic; class GFG{     static void distinct(int[,] matrix, int N){         // Make a empty map    Dictionary ans = new Dictionary();     // 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 codepublic 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



Output:

2 3

Time Complexity: O(n2

Space Complexity: O(n)

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