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Given three Sorted arrays 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

Recommended Practice

Common elements in three sorted arrays using two pointer:

A simple solution is to first find the intersection of two arrays and store the intersection in a temporary array, then find the intersection of the third array and temporary array. 

  • Initialize both pointers i and j to 0, and an empty list common.
  • While both pointers i and j are within the bounds of the two arrays:
    • If arr1[i] is less than arr2[j], increment i by 1.
    • If arr2[j] is less than arr1[i], increment j by 1.
    • If arr1[i] is equal to arr2[j]:
    • Add arr1[i] to the common list.
    • Increment both i and j by 1.
  • Return the common list containing the common elements of the two arrays.

Below is the implementation of the above approach:

C++

#include <iostream>
using namespace std;
 
// Function to find the intersection of two arrays
void FindIntersection(int arr1[], int arr2[], int temp[],
                      int m, int n, int& k)
{
    int i = 0, j = 0;
    // vector to store the intersection of the arr1[] and
    // arr2[]
    while (i < m && j < n) {
        // ith element can not be common element
        if (arr1[i] < arr2[j]) {
            i++;
        }
 
        // jth element can not be common element
        else if (arr2[j] < arr1[i]) {
            j++;
        }
 
        // if arr1[i] == arr2[j]
        else {
            temp[k] = arr1[i];
            i++;
            j++;
            k++;
        }
    }
}
 
int main()
{
 
    int arr1[] = { 1, 5, 10, 20, 40, 80 };
    int arr2[] = { 6, 7, 20, 80, 100 };
    int arr3[] = { 3, 4, 15, 20, 30, 70, 80, 120 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    int n3 = sizeof(arr3) / sizeof(arr3[0]);
 
    // temp array to store the common elements of arr1 and
    // arr2 arrays
    int temp[200000];
 
    // ans array to store the common elements of temp and
    // arr3 arrays (i.e common elements of all 3 arrays)
    int ans[200000];
 
    int k = 0;
 
    // function call to find the temp array
    FindIntersection(arr1, arr2, temp, n1, n2, k);
    int tempSize = k;
    k = 0;
 
    // function call to find the ans array.
    FindIntersection(arr3, temp, ans, n3, tempSize, k);
    cout << "Common elements present in arrays are : \n";
 
    for (int i = 0; i < k; i++) {
        cout << ans[i] << " ";
    }
    cout << endl;
 
    return 0;
}
 
// This code is contributed by Hem Kishan

                    

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
public class GFG {
 
    // Function to find the intersection of two arrays
    static void findIntersection(int[] arr1, int[] arr2,
                                 int[] temp, int m, int n,
                                 int[] k)
    {
        int i = 0, j = 0;
        // Loop to find the intersection of arr1[] and
        // arr2[]
        while (i < m && j < n) {
            // ith element can't be a common element
            if (arr1[i] < arr2[j]) {
                i++;
            }
            // jth element can't be a common element
            else if (arr2[j] < arr1[i]) {
                j++;
            }
            // if arr1[i] == arr2[j]
            else {
                temp[k[0]] = arr1[i];
                i++;
                j++;
                k[0]++;
            }
        }
    }
 
    public static void main(String[] args)
    {
 
        int[] arr1 = { 1, 5, 10, 20, 40, 80 };
        int[] arr2 = { 6, 7, 20, 80, 100 };
        int[] arr3 = { 3, 4, 15, 20, 30, 70, 80, 120 };
        int n1 = arr1.length;
        int n2 = arr2.length;
        int n3 = arr3.length;
 
        // temp array to store the common elements of arr1
        // and arr2 arrays
        int[] temp = new int[200000];
 
        // ans array to store the common elements of temp
        // and arr3 arrays
        int[] ans = new int[200000];
 
        int[] k = {
            0
        }; // Mutable integer to simulate pass-by-reference
 
        // function call to find the temp array
        findIntersection(arr1, arr2, temp, n1, n2, k);
        int tempSize = k[0];
        k[0] = 0;
 
        // function call to find the ans array
        findIntersection(arr3, temp, ans, n3, tempSize, k);
 
        System.out.println(
            "Common elements present in arrays are :");
 
        // Printing the common elements
        for (int i = 0; i < k[0]; i++) {
            System.out.print(ans[i] + " ");
        }
        System.out.println();
    }
}

                    

Python3

# Function to find the intersection of two arrays
def find_intersection(arr1, arr2, temp, m, n, k):
    i = 0
    j = 0
    while i < m and j < n:
        if arr1[i] < arr2[j]:
            i += 1
        elif arr2[j] < arr1[i]:
            j += 1
        else:
            temp[k[0]] = arr1[i]
            i += 1
            j += 1
            k[0] += 1
 
# Main function
 
 
def main():
    arr1 = [1, 5, 10, 20, 40, 80]
    arr2 = [6, 7, 20, 80, 100]
    arr3 = [3, 4, 15, 20, 30, 70, 80, 120]
    n1 = len(arr1)
    n2 = len(arr2)
    n3 = len(arr3)
 
    temp = [0] * 200000
    ans = [0] * 200000
 
    k = [0# Mutable list to simulate pass-by-reference
 
    # Function call to find the temp array
    find_intersection(arr1, arr2, temp, n1, n2, k)
    temp_size = k[0]
    k[0] = 0
 
    # Function call to find the ans array
    find_intersection(arr3, temp, ans, n3, temp_size, k)
 
    print("Common elements present in arrays are:")
 
    # Printing the common elements
    for i in range(k[0]):
        print(ans[i], end=" ")
    print()
 
 
if __name__ == "__main__":
    main()

                    

C#

using System;
 
public class GFG {
    // Function to find the intersection of two arrays
    static void FindIntersection(int[] arr1, int[] arr2,
                                 int[] temp, int m, int n,
                                 int[] k)
    {
        int i = 0, j = 0;
        // Loop to find the intersection of arr1[] and
        // arr2[]
        while (i < m && j < n) {
            // ith element can't be a common element
            if (arr1[i] < arr2[j]) {
                i++;
            }
            // jth element can't be a common element
            else if (arr2[j] < arr1[i]) {
                j++;
            }
            // if arr1[i] == arr2[j]
            else {
                temp[k[0]] = arr1[i];
                i++;
                j++;
                k[0]++;
            }
        }
    }
 
    public static void Main(string[] args)
    {
        int[] arr1 = { 1, 5, 10, 20, 40, 80 };
        int[] arr2 = { 6, 7, 20, 80, 100 };
        int[] arr3 = { 3, 4, 15, 20, 30, 70, 80, 120 };
        int n1 = arr1.Length;
        int n2 = arr2.Length;
        int n3 = arr3.Length;
 
        // temp array to store the common elements of arr1
        // and arr2 arrays
        int[] temp = new int[200000];
 
        // ans array to store the common elements of temp
        // and arr3 arrays
        int[] ans = new int[200000];
 
        int[] k = {
            0
        }; // Mutable integer to simulate pass-by-reference
 
        // Function call to find the temp array
        FindIntersection(arr1, arr2, temp, n1, n2, k);
        int tempSize = k[0];
        k[0] = 0;
 
        // Function call to find the ans array
        FindIntersection(arr3, temp, ans, n3, tempSize, k);
 
        Console.WriteLine(
            "Common elements present in arrays are :");
 
        // Printing the common elements
        for (int i = 0; i < k[0]; i++) {
            Console.Write(ans[i] + " ");
        }
        Console.WriteLine();
    }
}

                    

Javascript

// Function to find the intersection of two arrays
function findIntersection(arr1, arr2) {
    let i = 0;
    let j = 0;
    const temp = [];
 
    while (i < arr1.length && j < arr2.length) {
        if (arr1[i] < arr2[j]) {
            i++;
        } else if (arr2[j] < arr1[i]) {
            j++;
        } else {
            temp.push(arr1[i]);
            i++;
            j++;
        }
    }
 
    return temp;
}
 
function main() {
    const arr1 = [1, 5, 10, 20, 40, 80];
    const arr2 = [6, 7, 20, 80, 100];
    const arr3 = [3, 4, 15, 20, 30, 70, 80, 120];
 
    // Find the intersection of arr1 and arr2
    const temp = findIntersection(arr1, arr2);
 
    // Find the intersection of temp and arr3 to get common elements
    const ans = findIntersection(temp, arr3);
 
    console.log("Common elements present in arrays are:");
    for (const element of ans) {
        console.log(element);
    }
}
 
// Call the main function to start the program
main();

                    

Output
Common elements present in arrays are : 
20 80 









Time complexity: O(n1 + n2 + n3), where n1, n2 and n3 are sizes of ar1[], ar2[] and ar3[] respectively.
Auxiliary Space: O(max(n1,n2,n3))

Common elements in three sorted arrays using three pointer:

It is known that the arrays are sorted in a non-decreasing order. When a common integer has been found, we want to move forward in each array in search of another common integer. Otherwise, the smaller integer among the three must not be common.

The reason for this is that at least one of the other integers is a larger integer, and as we move forward in the array, we only encounter larger integers. In this case, we want to proceed with only the array that contains the smaller integer.

  • Create and initialize three variables i, j, and k with 0, it will point to the indices of the arrays.
  • Repeat the following steps until we reach the end of any one of the arrays:
    • Check whether the integers appointed by i, j, and k are equal or not.
    • If they are equal, print any of the integers and increase i, j, and k by 1.
    • Otherwise, increase the index that points to the smaller integer by 1.

Illustration:

common elements in three sorted arrays

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;
}
 
// This code is contributed by Sania Kumari Gupta
// (kriSania804)

                    

C

// C program to print common elements in three arrays
#include <stdio.h>
 
// 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]) {
            printf("%d ", 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]);
 
    printf("Common Elements are ");
    findCommon(ar1, ar2, ar3, n1, n2, n3);
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta
// (kriSania804)

                    

Java

// Java program to find common elements in three arrays
import java.io.*;
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.

                    

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);
       
      // This code is contributed by rdtank.
       
</script>

                    

PHP

<?php
// PHP 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[]
    $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])
        {
            echo $ar1[$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
    $ar1 = array(1, 5, 10, 20, 40, 80);
    $ar2 = array(6, 7, 20, 80, 100);
    $ar3 = array(3, 4, 15, 20, 30, 70,
                                80, 120);
    $n1 = count($ar1);
    $n2 = count($ar2);
    $n3 = count($ar3);
 
    echo "Common Elements are ";
     
    findCommon($ar1, $ar2, $ar3,$n1, $n2, $n3);
     
// This code is contributed by anuj_67.
?>

                    

Output
Common Elements are 20 80 








Time complexity: O(n1 + n2 + n3), In the worst case, the largest-sized array may have all small elements and the middle-sized array has all middle elements.
Auxiliary Space:  O(1)

This article is compiled by Rahul Gupta



Last Updated : 02 Oct, 2023
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