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.