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

Examples:

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 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.

1) If x, y and z are same, we can simply print any of them as common element and move ahead in all three arrays.

2) Else If x < y, we can move ahead in ar1[] as x cannot be a common element
3) Else If y y and y > z), we can simply move ahead in ar3[] as z cannot be a common element.

Following are implementations of the above idea.

## C++

// C++ program to print common elements in three arrays #include <iostream> 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 program to test above function 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; }

## Java

// Java program to find common elements in three arrays class FindCommon { // This function prints common elements in ar1 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]) { System.out.print(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(String args[]) { FindCommon ob = new FindCommon(); 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}; System.out.print("Common elements are "); ob.findCommon(ar1, ar2, ar3); } } /*This code is contributed by Rajat Mishra */

## Python

# Python function to print common elements in three sorted arrays def findCommon(ar1, ar2, ar3, n1, n2, n3): # Initialize starting indexes for ar1[], ar2[] and ar3[] i, j, k = 0, 0, 0 # Iterate through three arrays while all arrays have elements while (i < n1 and j < n2 and k< n3): # If x = y and y = z, print any of them and move ahead # in all arrays if (ar1[i] == ar2[j] and ar2[j] == ar3[k]): print ar1[i], i += 1 j += 1 k += 1 # x < y elif ar1[i] < ar2[j]: i += 1 # y < z elif ar2[j] < ar3[k]: j += 1 # We reach here when x > y and z < y, i.e., z is smallest else: k += 1 # Driver program to check above function ar1 = [1, 5, 10, 20, 40, 80] ar2 = [6, 7, 20, 80, 100] ar3 = [3, 4, 15, 20, 30, 70, 80, 120] n1 = len(ar1) n2 = len(ar2) n3 = len(ar3) print "Common elements are", findCommon(ar1, ar2, ar3, n1, n2, n3) # This code is contributed by __Devesh Agrawal__

## 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); } } // This code is contributed by Sam007.

Output:

Common Elements are 20 80

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

This article is compiled by **Rahul Gupta** Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.