# 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.