# Find distinct elements common to all rows of a Matrix in C++

## Concept

With respect of a given an m x m matrix, the problem is to determine all the distinct elements common to all rows of the matrix. So the elements can be displayed in any order.

## Input

mat[][] = { {13, 2, 15, 4, 17},
{15, 3, 2, 4, 36},
{15, 2, 15, 4, 12},
{15, 26, 4, 3, 2},
{2, 19, 4, 22, 15}
}

## Output

2 4 15

## Methods

First Method: Implement three nested loops. Verify if an element of 1st row is present in all the subsequent rows. Here, time Complexity is O(m^3). Additional space could be required to control the duplicate elements.

Second Method: Arrange or sort all the rows of the matrix individually in increasing order. So we then implement a modified approach of the problem of determining common elements in 3 sorted arrays. Following an implementation for the same is given.

## Example

Live Demo

// C++ implementation to find distinct elements
// common to all rows of a matrix
#include <bits/stdc++.h>
using namespace std;
const int MAX1 = 100;
// Shows function to individually sort
// each row in increasing order
void sortRows1(int mat1[][MAX1], int m){
for (int i=0; i<m; i++)
sort(mat1[i], mat1[i] + m);
}
// Shows function to find all the common elements
void findAndPrintCommonElements1(int mat1[][MAX1], int m){
//Used to sort rows individually
sortRows1(mat1, m);
// Shows current column index of each row is stored
// from where the element is being searched in
// that row
int curr_index1[m];
memset(curr_index1, 0, sizeof(curr_index1));
int f = 0;
for (; curr_index1[0]<m; curr_index1[0]++){
//Indicates value present at the current column index
// of 1st row
int value1 = mat1[0][curr_index1[0]];
bool present1 = true;
//Indicates 'value' is being searched in all the
// subsequent rows
for (int i=1; i<m; i++){
// Used to 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_index1[i] < m &&
mat1[i][curr_index1[i]] <= value1)
curr_index1[i]++;
// Now if the element was not present at the column
// before to the 'curr_index' of the row
if (mat1[i][curr_index1[i]-1] != value1)
present1 = false;
// Now if all elements of the row have
// been traversed
if (curr_index1[i] == m){
f = 1;
break;
}
}
// Now if the 'value' is common to all the rows
if (present1)
cout << value1 << " ";
// Now 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 mat1[][MAX1] = { {13, 2, 15, 4, 17},{15, 3, 2, 4, 36},{15, 2, 15, 4, 12},
{15, 26, 4, 3, 2},{2, 19, 4, 22, 15}};
int m = 5;
findAndPrintCommonElements1(mat1, m);
return 0;
}

## Output

2 4 15