Merge 3 Sorted Arrays

Merge two sorted arrays

Last Updated : 28 Feb, 2025

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Given two sorted arrays, the task is to merge them in a sorted manner.
Examples:

Input: arr1[] = { 1, 3, 4, 5}, arr2[] = {2, 4, 6, 8}
Output: arr3[] = {1, 2, 3, 4, 4, 5, 6, 8}

Input: arr1[] = { 5, 8, 9}, arr2[] = {4, 7, 8}
Output: arr3[] = {4, 5, 7, 8, 8, 9}

Table of Content

[Naive Approach] Concatenate and Sort – O((n1 + n2) log(n1 + n2)) time and O(1) space
[Expected Approach] Using Merge of Merge Sort O(n1 + n2) time and O(n1 + n2) space
Related Articles :

[Naive Approach] Concatenate and Sort – O((n1 + n2) log(n1 + n2)) time and O(1) space

This approach involves two key steps: first, combining elements from two separate arrays into a third result array, and then sorting the result array.

Since it requires sorting the entire merged array, resulting in a time complexity of O((n1 + n2) log(n1 + n2)), where n is the total number of elements.

C++

#include <bits/stdc++.h>
using namespace std;

void mergeArrays(vector<int>& arr1, vector<int>& arr2, 
                                    vector<int>& arr3) {
    int n1 = arr1.size();
    int n2 = arr2.size();
    int i = 0, j = 0, k = 0;

    // Traverse arr1 and insert its elements into arr3
    while (i < n1) {
        arr3[k++] = arr1[i++];
    }

    // Traverse arr2 and insert its elements into arr3
    while (j < n2) {
        arr3[k++] = arr2[j++];
    }

    // Sort the entire arr3
    sort(arr3.begin(), arr3.end());
}

// Driver code
int main() {
    vector<int> arr1 = {1, 3, 5, 7};
    vector<int> arr2 = {2, 4, 6, 8};
    vector<int> arr3(arr1.size() + arr2.size());

    mergeArrays(arr1, arr2, arr3);

    for (int i = 0; i < arr3.size(); i++)
        cout << arr3[i] << " ";

    return 0;
}

C

#include <stdio.h>
#include <stdlib.h>

// Function to merge two arrays and sort the result
void mergeArrays(int* arr1, int n1, int* arr2, int n2, int* arr3) {
    int i = 0, j = 0, k = 0;

    // Traverse arr1 and insert its elements into arr3
    while (i < n1) {
        arr3[k++] = arr1[i++];
    }

    // Traverse arr2 and insert its elements into arr3
    while (j < n2) {
        arr3[k++] = arr2[j++];
    }

    // Sort the entire arr3
    qsort(arr3, n1 + n2, sizeof(int), [](const void* a, const void* b) {
        return (*(int*)a - *(int*)b);
    });
}

// Driver code
int main() {
    int arr1[] = {1, 3, 5, 7};
    int arr2[] = {2, 4, 6, 8};
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    int arr3[n1 + n2];

    mergeArrays(arr1, n1, arr2, n2, arr3);

    for (int i = 0; i < n1 + n2; i++)
        printf("%d ", arr3[i]);

    return 0;
}

Java

import java.util.Arrays;

public class GfG {

    // Function to merge two arrays and sort the result
    public static void mergeArrays(int[] arr1, int[] arr2, int[] arr3) {
        int n1 = arr1.length;
        int n2 = arr2.length;
        int i = 0, j = 0, k = 0;

        // Traverse arr1 and insert its elements into arr3
        while (i < n1) {
            arr3[k++] = arr1[i++];
        }

        // Traverse arr2 and insert its elements into arr3
        while (j < n2) {
            arr3[k++] = arr2[j++];
        }

        // Sort the entire arr3
        Arrays.sort(arr3);
    }

    // Driver code
    public static void main(String[] args) {
        int[] arr1 = {1, 3, 5, 7};
        int[] arr2 = {2, 4, 6, 8};
        int[] arr3 = new int[arr1.length + arr2.length];

        mergeArrays(arr1, arr2, arr3);

        for (int value : arr3)
            System.out.print(value + " ");
    }
}

Python

# Function to merge two arrays and sort the result
def merge_arrays(arr1, arr2):
    arr3 = arr1 + arr2
    arr3.sort()
    return arr3

# Driver code
arr1 = [1, 3, 5, 7]
arr2 = [2, 4, 6, 8]
arr3 = merge_arrays(arr1, arr2)

for value in arr3:
    print(value, end=" ")

C#

using System;
using System.Linq;

public class GfG {

    // Function to merge two arrays and sort the result
    public static int[] MergeArrays(int[] arr1, int[] arr2) {
        int[] arr3 = arr1.Concat(arr2).ToArray();
        Array.Sort(arr3);
        return arr3;
    }

    // Driver code
    public static void Main() {
        int[] arr1 = {1, 3, 5, 7};
        int[] arr2 = {2, 4, 6, 8};
        int[] arr3 = MergeArrays(arr1, arr2);

        foreach (int value in arr3)
            Console.Write(value + " ");
    }
}

JavaScript

// Function to merge two arrays and sort the result
function mergeArrays(arr1, arr2) {
    let arr3 = arr1.concat(arr2);
    arr3.sort((a, b) => a - b);
    return arr3;
}

// Driver code
let arr1 = [1, 3, 5, 7];
let arr2 = [2, 4, 6, 8];
let arr3 = mergeArrays(arr1, arr2);

console.log(arr3.join(" "));

Output

1 2 3 4 5 6 7 8

[Expected Approach] Using Merge of Merge Sort O(n1 + n2) time and O(n1 + n2) space

The idea is to use Merge function of Merge sort.

Create an array arr3[] of size n1 + n2.
Simultaneously traverse arr1[] and arr2[].
Pick smaller of current elements in arr1[] and arr2[], copy this smaller element to next position in arr3[] and move ahead in arr3[] and the array whose element is picked.
If there are remaining elements in arr1[] or arr2[], copy them also in arr3[].

C++

#include <iostream>
#include <vector>
using namespace std;

// Merge ar1[0..n1-1] and ar2[0..n2-1] into ar3
void mergeArrays(vector<int>& ar1, vector<int>& ar2, 
                               vector<int>& ar3) {
    int i = 0, j = 0, k = 0;
    int n1 = ar1.size();
    int n2 = ar2.size();

    while (i < n1 && j < n2) {
      
        // Pick smaller of the two current
        // elements and move ahead in the
        // array of the picked element
        if (ar1[i] < ar2[j])
            ar3[k++] = ar1[i++];
        else
            ar3[k++] = ar2[j++];
    }

    // if there are remaining elements of
    // the first array, move them
    while (i < n1)
        ar3[k++] = ar1[i++];

    // Else if there are remaining elements of
    // the second array, move them
    while (j < n2)
        ar3[k++] = ar2[j++];
}

// Driver code
int main() {
    vector<int> ar1 = {1, 3, 5, 7};
    vector<int> ar2 = {2, 4, 6, 8};
    vector<int> ar3(ar1.size() + ar2.size());

    mergeArrays(ar1, ar2, ar3);
  
    for (int i = 0; i < ar3.size(); i++)
        cout << ar3[i] << " ";

    return 0;
}

C

#include <stdio.h>

// Merge ar1[0..n1-1] and ar2[0..n2-1] into ar3
void mergeArrays(int* ar1, int n1, int* ar2, int n2, int* ar3) {
    int i = 0, j = 0, k = 0;

    while (i < n1 && j < n2) {
      
        // Pick smaller of the two current elements and move ahead in the array of the picked element
        if (ar1[i] < ar2[j])
            ar3[k++] = ar1[i++];
        else
            ar3[k++] = ar2[j++];
    }

    // if there are remaining elements of the first array, move them
    while (i < n1)
        ar3[k++] = ar1[i++];

    // Else if there are remaining elements of the second array, move them
    while (j < n2)
        ar3[k++] = ar2[j++];
}

// Driver code
int main() {
    int ar1[] = {1, 3, 5, 7};
    int ar2[] = {2, 4, 6, 8};
    int n1 = sizeof(ar1) / sizeof(ar1[0]);
    int n2 = sizeof(ar2) / sizeof(ar2[0]);
    int ar3[n1 + n2];

    mergeArrays(ar1, n1, ar2, n2, ar3);

    for (int i = 0; i < n1 + n2; i++)
        printf("%d ", ar3[i]);

    return 0;
}

Java

import java.util.*;

public class GfG {

    // Merge ar1[0..n1-1] and ar2[0..n2-1] into ar3
    public static void mergeArrays(int[] ar1, int[] ar2, int[] ar3) {
        int i = 0, j = 0, k = 0;
        int n1 = ar1.length;
        int n2 = ar2.length;

        while (i < n1 && j < n2) {
          
            // Pick smaller of the two current elements and move ahead in the array of the picked element
            if (ar1[i] < ar2[j])
                ar3[k++] = ar1[i++];
            else
                ar3[k++] = ar2[j++];
        }

        // if there are remaining elements of the first array, move them
        while (i < n1)
            ar3[k++] = ar1[i++];

        // Else if there are remaining elements of the second array, move them
        while (j < n2)
            ar3[k++] = ar2[j++];
    }

    // Driver code
    public static void main(String[] args) {
        int[] ar1 = {1, 3, 5, 7};
        int[] ar2 = {2, 4, 6, 8};
        int[] ar3 = new int[ar1.length + ar2.length];

        mergeArrays(ar1, ar2, ar3);

        for (int value : ar3)
            System.out.print(value + " ");
    }
}

Python

# Merge ar1[0..n1-1] and ar2[0..n2-1] into ar3
def merge_arrays(ar1, ar2):
    i = 0
    j = 0
    k = 0
    n1 = len(ar1)
    n2 = len(ar2)
    ar3 = []

    while i < n1 and j < n2:
      
        # Pick smaller of the two current elements and move ahead in the array of the picked element
        if ar1[i] < ar2[j]:
            ar3.append(ar1[i])
            i += 1
        else:
            ar3.append(ar2[j])
            j += 1

    # if there are remaining elements of the first array, move them
    while i < n1:
        ar3.append(ar1[i])
        i += 1

    # Else if there are remaining elements of the second array, move them
    while j < n2:
        ar3.append(ar2[j])
        j += 1

    return ar3

# Driver code
ar1 = [1, 3, 5, 7]
ar2 = [2, 4, 6, 8]
ar3 = merge_arrays(ar1, ar2)

print(" ".join(map(str, ar3)))

C#

using System;

public class GfG {

    // Merge ar1[0..n1-1] and ar2[0..n2-1] into ar3
    public static void MergeArrays(int[] ar1, int[] ar2, int[] ar3) {
        int i = 0, j = 0, k = 0;
        int n1 = ar1.Length;
        int n2 = ar2.Length;

        while (i < n1 && j < n2) {
          
            // Pick smaller of the two current elements and move ahead in the array of the picked element
            if (ar1[i] < ar2[j])
                ar3[k++] = ar1[i++];
            else
                ar3[k++] = ar2[j++];
        }

        // if there are remaining elements of the first array, move them
        while (i < n1)
            ar3[k++] = ar1[i++];

        // Else if there are remaining elements of the second array, move them
        while (j < n2)
            ar3[k++] = ar2[j++];
    }

    // Driver code
    public static void Main() {
        int[] ar1 = {1, 3, 5, 7};
        int[] ar2 = {2, 4, 6, 8};
        int[] ar3 = new int[ar1.Length + ar2.Length];

        MergeArrays(ar1, ar2, ar3);

        foreach (int value in ar3)
            Console.Write(value + " ");
    }
}

JavaScript

// Merge ar1[0..n1-1] and ar2[0..n2-1] into ar3
function mergeArrays(ar1, ar2) {
    let i = 0, j = 0, k = 0;
    const n1 = ar1.length;
    const n2 = ar2.length;
    const ar3 = [];

    while (i < n1 && j < n2) {
    
        // Pick smaller of the two current elements and move ahead in the array of the picked element
        if (ar1[i] < ar2[j])
            ar3[k++] = ar1[i++];
        else
            ar3[k++] = ar2[j++];
    }

    // if there are remaining elements of the first array, move them
    while (i < n1)
        ar3[k++] = ar1[i++];

    // Else if there are remaining elements of the second array, move them
    while (j < n2)
        ar3[k++] = ar2[j++];

    return ar3;
}

// Driver code
const ar1 = [1, 3, 5, 7];
const ar2 = [2, 4, 6, 8];
const ar3 = mergeArrays(ar1, ar2);

console.log(ar3.join(" "));

Output

1 2 3 4 5 6 7 8

Problems related to Merging Two Sorted Arrays

Merge two sorted arrays with O(1) extra space
Merge k sorted arrays
Union of Two Sorted Arrays
Intersection of Two Sorted Arrays
Merge 2 sorted linked list in reverse order
Sort last M elements
Merge 2 sorted linked list in reverse order
Merge Sort for Doubly Linked List

Merge 3 Sorted Arrays

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