Union and Intersection of two sorted arrays

Difficulty Level : Easy
Last Updated : 21 Sep, 2021

Given two sorted arrays, find their union and intersection.

Example:

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Input : arr1[] = {1, 3, 4, 5, 7}
        arr2[] = {2, 3, 5, 6} 
Output : Union : {1, 2, 3, 4, 5, 6, 7} 
         Intersection : {3, 5}

Input : arr1[] = {2, 5, 6}
        arr2[] = {4, 6, 8, 10} 
Output : Union : {2, 4, 5, 6, 8, 10} 
         Intersection : {6}

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Union of arrays arr1[] and arr2[]

To find union of two sorted arrays, follow the following merge procedure :

1) Use two index variables i and j, initial values i = 0, j = 0
2) If arr1[i] is smaller than arr2[j] then print arr1[i] and increment i.
3) If arr1[i] is greater than arr2[j] then print arr2[j] and increment j.
4) If both are same then print any of them and increment both i and j.
5) Print remaining elements of the larger array.

Below is the implementation of the above approach :

C++

// C++ program to find union of
// two sorted arrays
#include <bits/stdc++.h>
using namespace std;
  
/* Function prints union of arr1[] and arr2[]
   m is the number of elements in arr1[]
   n is the number of elements in arr2[] */
void printUnion(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
    while (i < m && j < n) {
        if (arr1[i] < arr2[j])
            cout << arr1[i++] << " ";
  
        else if (arr2[j] < arr1[i])
            cout << arr2[j++] << " ";
  
        else {
            cout << arr2[j++] << " ";
            i++;
        }
    }
  
    /* Print remaining elements of the larger array */
    while (i < m)
        cout << arr1[i++] << " ";
  
    while (j < n)
        cout << arr2[j++] << " ";
}
  
/* Driver program to test above function */
int main()
{
    int arr1[] = { 1, 2, 4, 5, 6 };
    int arr2[] = { 2, 3, 5, 7 };
  
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
  
    // Function calling
    printUnion(arr1, arr2, m, n);
  
    return 0;
}

C

// C program to find union of
// two sorted arrays
#include <stdio.h>
  
/* Function prints union of arr1[] and arr2[]
   m is the number of elements in arr1[]
   n is the number of elements in arr2[] */
void printUnion(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
    while (i < m && j < n) {
        if (arr1[i] < arr2[j])
            printf(" %d ", arr1[i++]);
        else if (arr2[j] < arr1[i])
            printf(" %d ", arr2[j++]);
        else {
            printf(" %d ", arr2[j++]);
            i++;
        }
    }
  
    /* Print remaining elements of the larger array */
    while (i < m)
        printf(" %d ", arr1[i++]);
    while (j < n)
        printf(" %d ", arr2[j++]);
}
  
/* Driver program to test above function */
int main()
{
    int arr1[] = { 1, 2, 4, 5, 6 };
    int arr2[] = { 2, 3, 5, 7 };
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    printUnion(arr1, arr2, m, n);
    getchar();
    return 0;
}

Java

// Java program to find union of
// two sorted arrays
  
class FindUnion {
    /* Function prints union of arr1[] and arr2[]
    m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    static int printUnion(int arr1[], int arr2[], int m, int n)
    {
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                System.out.print(arr1[i++] + " ");
            else if (arr2[j] < arr1[i])
                System.out.print(arr2[j++] + " ");
            else {
                System.out.print(arr2[j++] + " ");
                i++;
            }
        }
  
        /* Print remaining elements of 
         the larger array */
        while (i < m)
            System.out.print(arr1[i++] + " ");
        while (j < n)
            System.out.print(arr2[j++] + " ");
  
        return 0;
    }
  
    public static void main(String args[])
    {
        int arr1[] = { 1, 2, 4, 5, 6 };
        int arr2[] = { 2, 3, 5, 7 };
        int m = arr1.length;
        int n = arr2.length;
        printUnion(arr1, arr2, m, n);
    }
}

Python

# Python program to find union of
# two sorted arrays
# Function prints union of arr1[] and arr2[]
# m is the number of elements in arr1[]
# n is the number of elements in arr2[]
def printUnion(arr1, arr2, m, n):
    i, j = 0, 0
    while i < m and j < n:
        if arr1[i] < arr2[j]:
            print(arr1[i])
            i += 1
        elif arr2[j] < arr1[i]:
            print(arr2[j])
            j+= 1
        else:
            print(arr2[j])
            j += 1
            i += 1
  
    # Print remaining elements of the larger array
    while i < m:
        print(arr1[i])
        i += 1
  
    while j < n:
        print(arr2[j])
        j += 1
  
# Driver program to test above function
arr1 = [1, 2, 4, 5, 6]
arr2 = [2, 3, 5, 7]
m = len(arr1)
n = len(arr2)
printUnion(arr1, arr2, m, n)
  
# This code is contributed by Pratik Chhajer

C#

// C# program to find union of
// two sorted arrays
  
using System;
  
class GFG {
    /* Function prints union of arr1[] and arr2[]
    m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    static int printUnion(int[] arr1,
                          int[] arr2, int m, int n)
    {
        int i = 0, j = 0;
  
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                Console.Write(arr1[i++] + " ");
            else if (arr2[j] < arr1[i])
                Console.Write(arr2[j++] + " ");
            else {
                Console.Write(arr2[j++] + " ");
                i++;
            }
        }
  
        /* Print remaining elements of 
        the larger array */
        while (i < m)
            Console.Write(arr1[i++] + " ");
        while (j < n)
            Console.Write(arr2[j++] + " ");
  
        return 0;
    }
  
    public static void Main()
    {
        int[] arr1 = { 1, 2, 4, 5, 6 };
        int[] arr2 = { 2, 3, 5, 7 };
        int m = arr1.Length;
        int n = arr2.Length;
  
        printUnion(arr1, arr2, m, n);
    }
}
  
// This code is contributed by Sam007

PHP

<?php
// PHP program to find union of
// two sorted arrays
  
/* Function prints union of 
  arr1[] and arr2[] m is the
  number of elements in arr1[]
  n is the number of elements
  in arr2[] */
function printUnion($arr1, $arr2,
                         $m, $n)
{
    $i = 0; $j = 0;
    while ($i < $m && $j < $n)
    {
        if ($arr1[$i] < $arr2[$j])
            echo($arr1[$i++] . " ");
          
        else if ($arr2[$j] < $arr1[$i])
            echo($arr2[$j++] . " ");
          
        else
        {
            echo($arr2[$j++] . " ");
            $i++;
        }
    }
      
    // Print remaining elements
    // of the larger array
    while($i < $m)
        echo($arr1[$i++] . " ");
      
    while($j < $n)
        echo($arr2[$j++] . " ");
}
  
// Driver Code
$arr1 = array(1, 2, 4, 5, 6);
$arr2 = array(2, 3, 5, 7);
  
$m = sizeof($arr1);
$n = sizeof($arr2);
  
// Function calling
printUnion($arr1, $arr2, $m, $n);
  
// This code is contributed by Ajit.
?>

Javascript

<script>
// JavaScript program to find union of
// two sorted arrays
  
    /* Function prints union of arr1[] and arr2[]
    m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    function printUnion( arr1,  arr2,  m,  n)
    {
        var i = 0, j = 0;
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                document.write(arr1[i++] + " ");
            else if (arr2[j] < arr1[i])
                document.write(arr2[j++] + " ");
            else {
                document.write(arr2[j++] + " ");
                i++;
            }
        }
  
        /* Print remaining elements of
        the larger array */
        while (i < m)
            document.write(arr1[i++] + " ");
        while (j < n)
            document.write(arr2[j++] + " ");
  
        return 0;
    }
  
        var arr1 = [ 1, 2, 4, 5, 6 ];
        var arr2 = [ 2, 3, 5, 7 ];
        var m = arr1.length;
        var n = arr2.length;
        printUnion(arr1, arr2, m, n);
          
// this code is contributed by shivanisinghss2110
</script>

Output:

1 2 3 4 5 6 7

Time Complexity : O(m + n)

Handling duplicates in any of the array : Above code does not handle duplicates in any of the array. To handle the duplicates, just check for every element whether adjacent elements are equal.

Below is the implementation of this approach.

Java

// Java program to find union of two
// sorted arrays (Handling Duplicates)
class FindUnion {
  
    static void UnionArray(int arr1[],
                           int arr2[])
    {
        // Taking max element present in either array
        int m = arr1[arr1.length - 1];
        int n = arr2[arr2.length - 1];
  
        int ans = 0;
  
        if (m > n) {
            ans = m;
        }
        else
            ans = n;
  
        // Finding elements from 1st array
        // (non duplicates only). Using
        // another array for storing union
        // elements of both arrays
        // Assuming max element present
        // in array is not more than 10^7
        int newtable[] = new int[ans + 1];
  
        // First element is always
        // present in final answer
        System.out.print(arr1[0] + " ");
  
        // Incrementing the First element's count
        // in it's corresponding index in newtable
        ++newtable[arr1[0]];
  
        // Starting traversing the first
        // array from 1st index till last
        for (int i = 1; i < arr1.length; i++) {
            // Checking whether current element
            // is not equal to it's previous element
            if (arr1[i] != arr1[i - 1]) {
                System.out.print(arr1[i] + " ");
                ++newtable[arr1[i]];
            }
        }
  
        // Finding only non common
        // elements from 2nd array
        for (int j = 0; j < arr2.length; j++) {
            // By checking whether it's already
            // present in newtable or not
            if (newtable[arr2[j]] == 0) {
                System.out.print(arr2[j] + " ");
                ++newtable[arr2[j]];
            }
        }
    }
  
    // Driver Code
    public static void main(String args[])
    {
        int arr1[] = { 1, 2, 2, 2, 3 };
        int arr2[] = { 2, 3, 4, 5 };
  
        UnionArray(arr1, arr2);
    }
}

Python3

# Python3 program to find union of two 
# sorted arrays (Handling Duplicates) 
def UnionArray(arr1, arr2): 
      
    # Taking max element present in either array 
    m = arr1[-1] 
    n = arr2[-1] 
    ans = 0
          
    if m > n: 
        ans = m 
    else:
        ans = n 
          
    # Finding elements from 1st array 
    # (non duplicates only). Using 
    # another array for storing union 
    # elements of both arrays 
    # Assuming max element present 
    # in array is not more than 10 ^ 7 
    newtable = [0] * (ans + 1) 
          
    # First element is always 
    # present in final answer 
    print(arr1[0], end = " ") 
          
    # Incrementing the First element's count 
    # in it's corresponding index in newtable 
    newtable[arr1[0]] += 1
          
    # Starting traversing the first 
    # array from 1st index till last 
    for i in range(1, len(arr1)): 
          
        # Checking whether current element 
        # is not equal to it's previous element 
        if arr1[i] != arr1[i - 1]: 
              
            print(arr1[i], end = " ") 
            newtable[arr1[i]] += 1
              
    # Finding only non common 
    # elements from 2nd array         
    for j in range(0, len(arr2)): 
          
        # By checking whether it's already 
        # present in newtable or not 
        if newtable[arr2[j]] == 0: 
              
            print(arr2[j], end = " ") 
            newtable[arr2[j]] += 1
      
# Driver Code 
if __name__ == "__main__": 
      
    arr1 = [1, 2, 2, 2, 3]
    arr2 = [2, 3, 4, 5]
          
    UnionArray(arr1, arr2) 
  
# This code is contributed by Rituraj Jain

C#

// C# program to find union of two
// sorted arrays (Handling Duplicates)
using System;
  
class GFG {
  
    static void UnionArray(int[] arr1,
                           int[] arr2)
    {
  
        // Taking max element present
        // in either array
        int m = arr1[arr1.Length - 1];
        int n = arr2[arr2.Length - 1];
  
        int ans = 0;
  
        if (m > n)
            ans = m;
        else
            ans = n;
  
        // Finding elements from 1st array
        // (non duplicates only). Using
        // another array for storing union
        // elements of both arrays
        // Assuming max element present
        // in array is not more than 10^7
        int[] newtable = new int[ans + 1];
  
        // First element is always
        // present in final answer
        Console.Write(arr1[0] + " ");
  
        // Incrementing the First element's
        // count in it's corresponding
        // index in newtable
        ++newtable[arr1[0]];
  
        // Starting traversing the first
        // array from 1st index till last
        for (int i = 1; i < arr1.Length; i++) {
            // Checking whether current
            // element is not equal to
            // it's previous element
            if (arr1[i] != arr1[i - 1]) {
                Console.Write(arr1[i] + " ");
                ++newtable[arr1[i]];
            }
        }
  
        // Finding only non common
        // elements from 2nd array
        for (int j = 0; j < arr2.Length; j++) {
            // By checking whether it's already
            // present in newtable or not
            if (newtable[arr2[j]] == 0) {
                Console.Write(arr2[j] + " ");
                ++newtable[arr2[j]];
            }
        }
    }
  
    // Driver Code
    public static void Main()
    {
        int[] arr1 = { 1, 2, 2, 2, 3 };
        int[] arr2 = { 2, 3, 4, 5 };
  
        UnionArray(arr1, arr2);
    }
}
  
// This code is contributed by anuj_67.

Thanks to Rajat Rawat for suggesting this solution.

Intersection of arrays arr1[] and arr2[]

To find intersection of 2 sorted arrays, follow the below approach :

1) Use two index variables i and j, initial values i = 0, j = 0
2) If arr1[i] is smaller than arr2[j] then increment i.
3) If arr1[i] is greater than arr2[j] then increment j.
4) If both are same then print any of them and increment both i and j.

Below is the implementation of the above approach :

C++

// C++ program to find intersection of
// two sorted arrays
#include <bits/stdc++.h>
using namespace std;
  
/* Function prints Intersection of arr1[] and arr2[]
m is the number of elements in arr1[]
n is the number of elements in arr2[] */
void printIntersection(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
    while (i < m && j < n) {
        if (arr1[i] < arr2[j])
            i++;
        else if (arr2[j] < arr1[i])
            j++;
        else /* if arr1[i] == arr2[j] */
        {
            cout << arr2[j] << " ";
            i++;
            j++;
        }
    }
}
  
/* Driver program to test above function */
int main()
{
    int arr1[] = { 1, 2, 4, 5, 6 };
    int arr2[] = { 2, 3, 5, 7 };
  
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
  
    // Function calling
    printIntersection(arr1, arr2, m, n);
  
    return 0;
}

C

// C program to find intersection of
// two sorted arrays
#include <stdio.h>
  
/* Function prints Intersection of arr1[] and arr2[]
   m is the number of elements in arr1[]
   n is the number of elements in arr2[] */
void printIntersection(int arr1[], int arr2[], int m, int n)
{
    int i = 0, j = 0;
    while (i < m && j < n) {
        if (arr1[i] < arr2[j])
            i++;
        else if (arr2[j] < arr1[i])
            j++;
        else /* if arr1[i] == arr2[j] */
        {
            printf(" %d ", arr2[j++]);
            i++;
        }
    }
}
  
/* Driver program to test above function */
int main()
{
    int arr1[] = { 1, 2, 4, 5, 6 };
    int arr2[] = { 2, 3, 5, 7 };
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    printIntersection(arr1, arr2, m, n);
    getchar();
    return 0;
}

Java

// Java program to find intersection of
// two sorted arrays
  
class FindIntersection {
    /* Function prints Intersection of arr1[] and arr2[]
       m is the number of elements in arr1[]
       n is the number of elements in arr2[] */
    static void printIntersection(int arr1[], int arr2[], int m, int n)
    {
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                i++;
            else if (arr2[j] < arr1[i])
                j++;
            else {
                System.out.print(arr2[j++] + " ");
                i++;
            }
        }
    }
  
    public static void main(String args[])
    {
        int arr1[] = { 1, 2, 4, 5, 6 };
        int arr2[] = { 2, 3, 5, 7 };
        int m = arr1.length;
        int n = arr2.length;
        printIntersection(arr1, arr2, m, n);
    }
}

Python

# Python program to find intersection of
# two sorted arrays
# Function prints Intersection of arr1[] and arr2[]
# m is the number of elements in arr1[]
# n is the number of elements in arr2[]
def printIntersection(arr1, arr2, m, n):
    i, j = 0, 0
    while i < m and j < n:
        if arr1[i] < arr2[j]:
            i += 1
        elif arr2[j] < arr1[i]:
            j+= 1
        else:
            print(arr2[j])
            j += 1
            i += 1
  
# Driver program to test above function
arr1 = [1, 2, 4, 5, 6]
arr2 = [2, 3, 5, 7]
m = len(arr1)
n = len(arr2)
printIntersection(arr1, arr2, m, n)
  
# This code is contributed by Pratik Chhajer

C#

// C# program to find Intersection of
// two sorted arrays
  
using System;
  
class GFG {
  
    /* Function prints Intersection of arr1[]
    and arr2[] m is the number of elements in arr1[]
    n is the number of elements in arr2[] */
    static void printIntersection(int[] arr1,
                                  int[] arr2, int m, int n)
    {
        int i = 0, j = 0;
  
        while (i < m && j < n) {
            if (arr1[i] < arr2[j])
                i++;
            else if (arr2[j] < arr1[i])
                j++;
            else {
                Console.Write(arr2[j++] + " ");
                i++;
            }
        }
    }
  
    // driver code
    public static void Main()
    {
        int[] arr1 = { 1, 2, 4, 5, 6 };
        int[] arr2 = { 2, 3, 5, 7 };
        int m = arr1.Length;
        int n = arr2.Length;
  
        printIntersection(arr1, arr2, m, n);
    }
}
  
// This code is contributed by Sam007

PHP

<?php
// PHP program to find intersection of
// two sorted arrays
  
/* Function prints Intersection 
   of arr1[] and arr2[] m is the
   number of elements in arr1[] 
   n is the number of elements
   in arr2[] */
function printIntersection($arr1, $arr2,
                                $m, $n)
{
    $i = 0 ;
    $j = 0;
    while ($i < $m && $j < $n)
    {
        if ($arr1[$i] < $arr2[$j])
            $i++;
        else if ($arr2[$j] < $arr1[$i])
            $j++;
              
        /* if arr1[i] == arr2[j] */
        else 
        {
            echo $arr2[$j], " ";
            $i++;
            $j++;
        }
    }
}
  
// Driver Code
$arr1 = array(1, 2, 4, 5, 6);
$arr2 = array(2, 3, 5, 7);
  
$m = count($arr1);
$n = count($arr2);
  
// Function calling
printIntersection($arr1, $arr2, $m, $n);
  
// This code is contributed by anuj_67.
?>

Javascript

<script>
  
// JavaScript program to find intersection of
// two sorted arrays
  
// Function prints Intersection of arr1[] and arr2[]
// m is the number of elements in arr1[]
// n is the number of elements in arr2[] 
function printIntersection(arr1, arr2, m, n)
{
    var i = 0, j = 0;
    while (i < m && j < n)
    {
        if (arr1[i] < arr2[j])
            i++;
        else if (arr2[j] < arr1[i])
            j++;
        else 
        {
            document.write(arr2[j++] + " ");
            i++;
        }
    }
}
  
// Driver code
var arr1 = [ 1, 2, 4, 5, 6 ];
var arr2 = [ 2, 3, 5, 7 ];
  
var m = arr1.length;
var n = arr2.length;
  
printIntersection(arr1, arr2, m, n);
  
// This code is contributed by shivanisinghss2110
  
</script>

Output:

2 5

Time Complexity : O(m + n)

Handling duplicate in Arrays :
Above code does not handle duplicate elements in arrays. The intersection should not count duplicate elements. To handle duplicates just check whether current element is already present in intersection list. Below is the implementation of this approach.

Python3

# Python3 program to find Intersection of two
# Sorted Arrays (Handling Duplicates)
def IntersectionArray(a, b, n, m):
    '''
    :param a: given sorted array a
    :param n: size of sorted array a
    :param b: given sorted array b
    :param m: size of sorted array b
    :return: array of intersection of two array or -1
    '''
  
    Intersection = []
    i = j = 0
      
    while i < n and j < m:
        if a[i] == b[j]:
  
            # If duplicate already present in Intersection list
            if len(Intersection) > 0 and Intersection[-1] == a[i]:
                i+= 1
                j+= 1
  
            # If no duplicate is present in Intersection list
            else:
                Intersection.append(a[i])
                i+= 1
                j+= 1
        elif a[i] < b[j]:
            i+= 1
        else:
            j+= 1
              
    if not len(Intersection):
        return [-1]
    return Intersection
  
# Driver Code
if __name__ == "__main__":
  
    arr1 = [1, 2, 2, 3, 4]
    arr2 = [2, 2, 4, 6, 7, 8]
      
    l = IntersectionArray(arr1, arr2, len(arr1), len(arr2))
    print(*l)
  
# This code is contributed by Abhishek Kumar

Output:

2 4

Time Complexity : O(m + n)
Auxiliary Space : O(min(m, n))

Another approach that is useful when difference between sizes of two given arrays is significant.
The idea is to iterate through the shorter array and do a binary search for every element of short array in big array (note that arrays are sorted). Time complexity of this solution is O(min(mLogn, nLogm)). This solution works better than the above approach when ratio of larger length to smaller is more than logarithmic order.

See following post for unsorted arrays.
Find Union and Intersection of two unsorted arrays
Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem.

My Personal Notes

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