C++ program to print common elements in three arrays

Given three arrays sorted in non-decreasing order, print all common elements in these arrays.

Examples:

Input:
ar1[] = {1, 5, 10, 20, 40, 80}
ar2[] = {6, 7, 20, 80, 100}
ar3[] = {3, 4, 15, 20, 30, 70, 80, 120}
Output: 20, 80

Input:
ar1[] = {1, 5, 5}
ar2[] = {3, 4, 5, 5, 10}
ar3[] = {5, 5, 10, 20}
Output: 5, 5

A simple solution is to first find [intersection of two arrays and store the intersection in a temporary array, then find the intersection of third array and temporary array.
Time complexity of this solution is O(n1 + n2 + n3) where n1, n2 and n3 are sizes of ar1[], ar2[] and ar3[] respectively.
The above solution requires extra space and two loops, we can find the common elements using a single loop and without extra space. The idea is similar to [intersection of two arrays. Like two arrays loop, we run a loop and traverse three arrays.
Let the current element traversed in ar1[] be x, in ar2[] be y and in ar3[] be z. We can have following cases inside the loop.

  • If x, y and z are same, we can simply print any of them as common element and move ahead in all three arrays.
  • Else If x < y, we can move ahead in ar1[] as x cannot be a common element.
  • Else If x > z and y > z), we can simply move ahead in ar3[] as z cannot be a common element.

Below is the implementation of the above approach:

  • C++

// C++ program to print common elements in three arrays

#include <bits/stdc++.h>

using namespace std;

// This function prints common elements in ar1

void findCommon( int ar1[], int ar2[], int ar3[], int n1, int n2, int n3)

{

// Initialize starting indexes for ar1[], ar2[] and ar3[]

int i = 0, j = 0, k = 0;

// Iterate through three arrays while all arrays have elements

while (i < n1 && j < n2 && k < n3)

{

// If x = y and y = z, print any of them and move ahead

// in all arrays

if (ar1[i] == ar2[j] && ar2[j] == ar3[k])

{ cout << ar1[i] << " " ; i++; j++; k++; }

// x < y

else if (ar1[i] < ar2[j])

i++;

// y < z

else if (ar2[j] < ar3[k])

j++;

// We reach here when x > y and z < y, i.e., z is smallest

else

k++;

}

}

// Driver code

int main()

{

int ar1[] = {1, 5, 10, 20, 40, 80};

int ar2[] = {6, 7, 20, 80, 100};

int ar3[] = {3, 4, 15, 20, 30, 70, 80, 120};

int n1 = sizeof (ar1)/ sizeof (ar1[0]);

int n2 = sizeof (ar2)/ sizeof (ar2[0]);

int n3 = sizeof (ar3)/ sizeof (ar3[0]);

cout << "Common Elements are " ;

findCommon(ar1, ar2, ar3, n1, n2, n3);

return 0;

}

Output

Common Elements are 20 80

Time complexity of the above solution is O(n1 + n2 + n3). In the worst case, the largest sized array may have all small elements and middle-sized array has all middle elements.