# 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 find common elements in`

`// three arrays`

`using` `System;`

`class` `GFG {`

` `

` ` `// This function prints common element`

` ` `// s in ar1`

` ` `static` `void` `findCommon(` `int` `[]ar1, ` `int` `[]ar2,`

` ` `int` `[]ar3)`

` ` `{`

` `

` ` `// 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 < ar1.Length && j < ar2.Length`

` ` `&& k < ar3.Length)`

` ` `{`

` `

` ` `// 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])`

` ` `{`

` ` `Console.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 to test above`

` ` `public` `static` `void` `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};`

` ` `Console.Write(` `"Common elements are "` `);`

` `

` ` `findCommon(ar1, ar2, ar3);`

` ` `}`

`}`

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.