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:

- Javascript

`<script>`

` `

`// JavaScript program to print`

` `

`// common elements in three arrays`

` `

`// This function prints common elements in ar1`

` `

`function`

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

` `

`{`

` `

` `

`// Initialize starting indexes`

` `

`// for ar1[], ar2[] and ar3[]`

` `

`var`

`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])`

` `

`{`

` `

`document.write(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`

` `

`var`

`ar1 = [1, 5, 10, 20, 40, 80];`

` `

`var`

`ar2 = [6, 7, 20, 80, 100];`

` `

`var`

`ar3 = [3, 4, 15, 20, 30, 70, 80, 120];`

` `

`var`

`n1 = ar1.length;`

` `

`var`

`n2 = ar2.length;`

` `

`var`

`n3 = ar3.length;`

` `

`document.write(`

`"Common Elements are "`

`);`

` `

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

` `

` `

`</script>`

**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.