Merge two sorted arrays (original) (raw)

Last Updated : 28 Feb, 2025

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& arr1, vector& arr2, vector& 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 arr1 = {1, 3, 5, 7}; vector arr2 = {2, 4, 6, 8}; vector 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(" "));

[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 #include using namespace std;

// Merge ar1[0..n1-1] and ar2[0..n2-1] into ar3 void mergeArrays(vector& ar1, vector& ar2, vector& 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 ar1 = {1, 3, 5, 7}; vector ar2 = {2, 4, 6, 8}; vector 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(" "));

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Merge two sorted arrays (original) (raw)

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

Function to merge two arrays and sort the result

Driver code

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

Merge ar1[0..n1-1] and ar2[0..n2-1] into ar3

Driver code

**Problems related to Merging Two Sorted Arrays