Merge 3 Sorted Arrays

Given 3 arrays A[], B[], and C[] that are sorted in ascending order, the task is to merge them together in ascending order and output the array D[]. 

Examples: 

Input: A = [1, 2, 3, 4, 5], B = [2, 3, 4], C = [4, 5, 6, 7]
Output : D = [1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 7]

Input: A = [1, 2, 3, 5], B = [6, 7, 8, 9 ], C = [10, 11, 12]
Output: D = [1, 2, 3, 5, 6, 7, 8, 9. 10, 11, 12]

Naive Approach:

The naive approach to merge three sorted arrays would be to first concatenate all three arrays into a single array and then sort it in ascending order. This approach has a time complexity of O((n+m+p) log(n+m+p)), where n, m, and p are the lengths of the three arrays.

Algorithm:

1.    Take the size of the arrays A, B, and C as input from the user.
2.    Create arrays A, B, and C of the input size.
3.    Take the elements of arrays A, B, and C as input from the user.
4.    Merge arrays A, B, and C into a single array D.
5.    Sort array D in ascending order.
6.    Print the elements of array D.

Below is the implementation of the approach:

C++

// C++ code for the approach

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

// Function to merge three sorted arrays
vector<int> mergeThreeSortedArrays(vector<int>& A, vector<int>& B, vector<int>& C) {
    vector<int> D;

    // Insert all elements from A into D
    for (int i = 0; i < A.size(); i++) {
        D.push_back(A[i]);
    }

    // Insert all elements from B into D
    for (int i = 0; i < B.size(); i++) {
        D.push_back(B[i]);
    }

    // Insert all elements from C into D
    for (int i = 0; i < C.size(); i++) {
        D.push_back(C[i]);
    }

    // Sort the merged array in ascending order
    sort(D.begin(), D.end());

    return D;
}

// Driver code
int main() {
    vector<int> A = { 1, 2, 3, 5 };
    vector<int> B = { 6, 7, 8, 9 };
    vector<int> C = { 10, 11, 12 };

    // Merge three sorted arrays
    vector<int> D = mergeThreeSortedArrays(A, B, C);

    // Print the merged and sorted array
    for (int i = 0; i < D.size(); i++) {
        cout << D[i] << " ";
    }
    cout << endl;

    return 0;
}

Output

1 2 3 5 6 7 8 9 10 11 12 

Time Complexity: O( (n+m+p) * log(n+m+p) ), where n, m, and p are the lengths of the three arrays.
Space Complexity: O(n+m+p), where n, m, and p are the lengths of the three arrays. This is because vector D has been created of size (m+n+p). 

Merge 3 Sorted Arrays by merging Two Arrays at a time:

The idea is to first merge two arrays and then merge the resultant with the third array.

Follow the steps below to solve the problem:

• First, merge the arr1 and arr2 using the idea mentioned in Merge Two Sorted Arrays
• Now merge the resultant array (from arr1 and arr2) with arr3.
• Print the answer array.

Pseudocode:

function merge(A, B)
    Let m and n be the sizes of A and B
    Let D be the array to store result
    // Merge by taking smaller element from A and B
    while i < m and j < n
        if A[i] <= B[j]
            Add A[i] to D and increment i by 1
        else Add B[j] to D and increment j by 1
    // If array A has exhausted, put elements from B
    while j < n
        Add B[j] to D and increment j by 1
    // If array B has exhausted, put elements from A
    while i < n
        Add A[j] to D and increment i by 1
    Return D

function merge_three(A, B, C)
    T = merge(A, B)
    return merge(T, C)

Below is the implementation of the above approach:

C++

// C++ program to merge three sorted arrays
// by merging two at a time.
#include <iostream>
#include <vector>
using namespace std;

using Vector = vector<int>;

void printVector(const Vector& a)
{
    for (auto e : a)
        cout << e << " ";
    cout << endl;
}

Vector mergeTwo(Vector& A, Vector& B)
{
    // Get sizes of vectors
    int m = A.size();
    int n = B.size();

    // Vector for storing Result
    Vector D;
    D.reserve(m + n);

    int i = 0, j = 0;
    while (i < m && j < n) {

        if (A[i] <= B[j])
            D.push_back(A[i++]);
        else
            D.push_back(B[j++]);
    }

    // B has exhausted
    while (i < m)
        D.push_back(A[i++]);

    // A has exhausted
    while (j < n)
        D.push_back(B[j++]);

    return D;
}

// Driver Code
int main()
{
    Vector A = { 1, 2, 3, 5 };
    Vector B = { 6, 7, 8, 9 };
    Vector C = { 10, 11, 12 };

    // First Merge A and B
    Vector T = mergeTwo(A, B);

    // Print Result after merging T with C
    printVector(mergeTwo(T, C));
    return 0;
}

Java

import java.util.*;
// Java program to merge three sorted arrays
// by merging two at a time.
class GFG {

    static ArrayList<Integer> mergeTwo(List<Integer> A,
                                       List<Integer> B)
    {
        // Get sizes of vectors
        int m = A.size();
        int n = B.size();

        // ArrayList for storing Result
        ArrayList<Integer> D
            = new ArrayList<Integer>(m + n);

        int i = 0, j = 0;
        while (i < m && j < n) {

            if (A.get(i) <= B.get(j))
                D.add(A.get(i++));
            else
                D.add(B.get(j++));
        }

        // B has exhausted
        while (i < m)
            D.add(A.get(i++));

        // A has exhausted
        while (j < n)
            D.add(B.get(j++));

        return D;
    }

    // Driver code
    public static void main(String[] args)
    {
        Integer[] a = { 1, 2, 3, 5 };
        Integer[] b = { 6, 7, 8, 9 };
        Integer[] c = { 10, 11, 12 };
        List<Integer> A = Arrays.asList(a);
        List<Integer> B = Arrays.asList(b);
        List<Integer> C = Arrays.asList(c);

        // First Merge A and B
        ArrayList<Integer> T = mergeTwo(A, B);

        // Print Result after merging T with C
        ArrayList<Integer> ans = mergeTwo(T, C);
        for (int i = 0; i < ans.size(); i++)
            System.out.print(ans.get(i) + " ");
    }
}

/* This code contributed by PrinciRaj1992 */

Python3

# Python program to merge three sorted arrays
# by merging two at a time.


def merge_two(a, b):
    (m, n) = (len(a), len(b))
    i = j = 0

    # Destination Array
    d = []

    # Merge from a and b together
    while i < m and j < n:
        if a[i] <= b[j]:
            d.append(a[i])
            i += 1
        else:
            d.append(b[j])
            j += 1

    # Merge from a if b has run out
    while i < m:
        d.append(a[i])
        i += 1

    # Merge from b if a has run out
    while j < n:
        d.append(b[j])
        j += 1

    return d


def merge(a, b, c):
    t = merge_two(a, b)
    return merge_two(t, c)


if __name__ == "__main__":
    A = [1, 2, 3, 5]
    B = [6, 7, 8, 9]
    C = [10, 11, 12]
    D = merge(A, B, C)
    for x in D:
        print(x, end=" ")

C#

// C# program to merge three sorted arrays
// by merging two at a time.

using System;
using System.Collections.Generic;

public static class GFG {

    static List<int> mergeTwo(List<int> A, List<int> B)
    {
        // Get sizes of vectors
        int m = A.Count;
        int n = B.Count;

        // Vector for storing Result
        List<int> D = new List<int>();
        D.Capacity = m + n;

        int i = 0;
        int j = 0;
        while (i < m && j < n) {

            if (A[i] <= B[j]) {
                D.Add(A[i++]);
            }
            else {
                D.Add(B[j++]);
            }
        }

        // B has exhausted
        while (i < m) {
            D.Add(A[i++]);
        }

        // A has exhausted
        while (j < n) {
            D.Add(B[j++]);
        }

        return D;
    }

    // Driver Code
    public static void Main()
    {
        List<int> A = new List<int>() { 1, 2, 3, 5 };
        List<int> B = new List<int>() { 6, 7, 8, 9 };
        List<int> C = new List<int>() { 10, 11, 12 };

        // First Merge A and B
        List<int> T = mergeTwo(A, B);

        // Print Result after merging T with C
        List<int> ans = mergeTwo(T, C);
        for (int i = 0; i < ans.Count; i++)
            Console.Write(ans[i] + " ");
    }
}

// This code is contributed by Aarti_Rathi

Javascript

<script>
// Javascript program to merge three sorted arrays
// by merging two at a time.




function mergeTwo(A, B)
{
    // Get sizes of vectors
    let m = A.length;
    let n = B.length;

    // Vector for storing Result
    let D = [];

    let i = 0, j = 0;
    while (i < m && j < n) {

        if (A[i] <= B[j])
            D.push(A[i++]);
        else
            D.push(B[j++]);
    }

    // B has exhausted
    while (i < m)
        D.push(A[i++]);

    // A has exhausted
    while (j < n)
        D.push(B[j++]);

    return D;
}

// Driver Code

    let A = [ 1, 2, 3, 5 ];
    let B = [ 6, 7, 8, 9 ];
    let C = [ 10, 11, 12 ];

    // First Merge A and B
    let T = mergeTwo(A, B);

    // Print Result after merging T with C
    document.write(mergeTwo(T, C));

</script>

Output

1 2 3 5 6 7 8 9 10 11 12 

Time Complexity: O(M+N+O) where m, n, o are the lengths of the 1st, 2nd, 3rd Array.
Auxiliary Space: O(M+N+O).

Merge 3 Sorted Arrays by merging Three arrays at a time:

The idea is to merge three arrays at the same time just like two arrays.

Follow the steps below to solve the problem:

• Initialize three pointers i, j, and k which are associated with arr1, arr2, and arr3 respectively
• Now check the smallest number from the i, j, and k indexes.
  • Put that minimum value in the output array 
  • Increment the pointer which has minimum value by 1
  • Perform these steps till any of the indexes reach the end of the array.
• After that check which two arrays don’t reach the end and perform the same with those two arrays.
• At last, check which of the three arrays doesn’t reach the end and put all the remaining elements of that array into the output array.

Pseudocode:

function merge-three(A, B, C)
    Let m, n, o be size of A, B, and C
    Let D be the array to store the result
    
    // Merge three arrays at the same time
    while i < m and j < n and k < o
        Get minimum of A[i], B[j], C[i]
        if the minimum is from A, add it to 
           D and advance i
        else if the minimum is from B add it 
                to D and advance j
        else if the minimum is from C add it 
                to D and advance k
    
   // After above step at least 1 array has 
   // exhausted. Only C has exhausted
   while i < m and j < n
       put minimum of A[i] and B[j] into D
       Advance i if minimum is from A else advance j 
   
   // Only B has exhausted
   while i < m and k < o
       Put minimum of A[i] and C[k] into D
       Advance i if minimum is from A else advance k
 
   // Only A has exhausted
   while j < n and k < o
       Put minimum of B[j] and C[k] into D
       Advance j if minimum is from B else advance k

   // After above steps at least 2 arrays have 
   // exhausted
   if A and B have exhausted take elements from C
   if B and C have exhausted take elements from A
   if A and C have exhausted take elements from B
   
   return D

Below is the implementation of the above approach:

C++

// C++ program to merger three sorted arrays
// by merging three simultaneously.

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

vector<int> mergeThree(vector<int>& A, vector<int>& B,
                       vector<int>& C)
{
    int m, n, o, i, j, k;
    // Get Sizes of three vectors
    m = A.size();
    n = B.size();
    o = C.size();

    // Vector for storing output
    vector<int> D;
    D.reserve(m + n + o);

    i = j = k = 0;

    while (i < m && j < n && k < o) {

        // Get minimum of a, b, c
        int m = min(min(A[i], B[j]), C[k]);

        // Put m in D
        D.push_back(m);

        // Increment i, j, k
        if (m == A[i])
            i++;
        else if (m == B[j])
            j++;
        else
            k++;
    }

    // C has exhausted
    while (i < m && j < n) {
        if (A[i] <= B[j]) {
            D.push_back(A[i]);
            i++;
        }
        else {
            D.push_back(B[j]);
            j++;
        }
    }

    // B has exhausted
    while (i < m && k < o) {
        if (A[i] <= C[k]) {
            D.push_back(A[i]);
            i++;
        }
        else {
            D.push_back(C[k]);
            k++;
        }
    }

    // A has exhausted
    while (j < n && k < o) {
        if (B[j] <= C[k]) {
            D.push_back(B[j]);
            j++;
        }
        else {
            D.push_back(C[k]);
            k++;
        }
    }

    // A and B have exhausted
    while (k < o)
        D.push_back(C[k++]);

    // B and C have exhausted
    while (i < m)
        D.push_back(A[i++]);

    // A and C have exhausted
    while (j < n)
        D.push_back(B[j++]);

    return D;
}

// Driver Code
int main()
{
    vector<int> A = { 1, 2, 3, 5 };
    vector<int> B = { 6, 7, 8, 9 };
    vector<int> C = { 10, 11, 12 };

    // Print Result
    vector<int> ans = mergeThree(A, B, C);
    for (auto x : ans)
        cout << x << " ";
    return 0;
}

Java

import java.io.*;
import java.lang.*;
import java.util.*;
class Sorting {
    public static void main(String[] args)
    {
        int A[] = { 1, 2, 3, 5 };
        int B[] = { 6, 7, 8, 9 };
        int C[] = { 10, 11, 12 };

        // call the function to sort and print the sorted
        // numbers
        merge3sorted(A, B, C);
    }

    // Function to merge three sorted arrays
    // A[], B[], C[]: input arrays
    static void merge3sorted(int A[], int B[], int C[])
    {
        // creating an empty list to store sorted numbers
        ArrayList<Integer> list = new ArrayList<Integer>();
        int i = 0, j = 0, k = 0;

        // using merge concept and trying to find
        // smallest of three while all three arrays
        // contains at least one element
        while (i < A.length && j < B.length
               && k < C.length) {
            int a = A[i];
            int b = B[j];
            int c = C[k];
            if (a <= b && a <= c) {
                list.add(a);
                i++;
            }
            else if (b <= a && b <= c) {
                list.add(b);
                j++;
            }
            else {
                list.add(c);
                k++;
            }
        }
        // next three while loop is to sort two
        // of arrays if one of the three gets exhausted
        while (i < A.length && j < B.length) {
            if (A[i] < B[j]) {
                list.add(A[i]);
                i++;
            }
            else {
                list.add(B[j]);
                j++;
            }
        }
        while (j < B.length && k < C.length) {
            if (B[j] < C[k]) {
                list.add(B[j]);
                j++;
            }
            else {
                list.add(C[k]);
                k++;
            }
        }
        while (i < A.length && k < C.length) {
            if (A[i] < C[k]) {
                list.add(A[i]);
                i++;
            }
            else {
                list.add(C[k]);
                k++;
            }
        }

        // if one of the array are left then
        // simply appending them as there will
        // be only largest element left
        while (i < A.length) {
            list.add(A[i]);
            i++;
        }
        while (j < B.length) {
            list.add(B[j]);
            j++;
        }
        while (k < C.length) {
            list.add(C[k]);
            k++;
        }

        // finally print the list
        for (Integer x : list)
            System.out.print(x + " ");
    } // merge3sorted closing braces
}

Python3

# Python program to merge three sorted arrays
# simultaneously.


def merge_three(a, b, c):
    (m, n, o) = (len(a), len(b), len(c))
    i = j = k = 0

    # Destination array
    d = []

    while i < m and j < n and k < o:

        # Get Minimum element
        mini = min(a[i], b[j], c[k])

        # Add m to D
        d.append(mini)

        # Increment the source pointer which
        # gives m
        if a[i] == mini:
            i += 1
        elif b[j] == mini:
            j += 1
        elif c[k] == mini:
            k += 1

    # Merge a and b if c has exhausted
    while i < m and j < n:
        if a[i] <= b[j]:
            d.append(a[i])
            i += 1
        else:
            d.append(b[j])
            j += 1

    # Merge b and c if a has exhausted
    while j < n and k < o:
        if b[j] <= c[k]:
            d.append(b[j])
            j += 1
        else:
            d.append(c[k])
            k += 1

    # Merge a and c if b has exhausted
    while i < m and k < o:
        if a[i] <= c[k]:
            d.append(a[i])
            i += 1
        else:
            d.append(c[k])
            k += 1

    # Take elements from a if b and c
    # have exhausted
    while i < m:
        d.append(a[i])
        i += 1

    # Take elements from b if a and c
    # have exhausted
    while j < n:
        d.append(b[j])
        j += 1

    # Take elements from c if a and
    # b have exhausted
    while k < o:
        d.append(c[k])
        k += 1

    return d


if __name__ == "__main__":
    a = [1, 2, 3, 5]
    b = [6, 7, 8, 9]
    c = [10, 11, 12]

    ans = merge_three(a, b, c)
    for x in ans:
        print(x, end=" ")

C#

// Online C# Editor for free
// Write, Edit and Run your C# code using C# Online Compiler
using System;
using System.Collections;

class Sorting {

    // Function to merge three sorted arrays
    // A[], B[], C[]: input arrays
    static void merge3sorted(int[] A, int[] B, int[] C)
    {
        // creating an empty list to store sorted numbers
        ArrayList list = new ArrayList();
        int i = 0, j = 0, k = 0;

        // using merge concept and trying to find
        // smallest of three while all three arrays
        // contains at least one element
        while (i < A.Length && j < B.Length
               && k < C.Length) {
            int a = A[i];
            int b = B[j];
            int c = C[k];
            if (a <= b && a <= c) {
                list.Add(a);
                i++;
            }
            else if (b <= a && b <= c) {
                list.Add(b);
                j++;
            }
            else {
                list.Add(c);
                k++;
            }
        }
        // next three while loop is to sort two
        // of arrays if one of the three gets exhausted
        while (i < A.Length && j < B.Length) {
            if (A[i] < B[j]) {
                list.Add(A[i]);
                i++;
            }
            else {
                list.Add(B[j]);
                j++;
            }
        }
        while (j < B.Length && k < C.Length) {
            if (B[j] < C[k]) {
                list.Add(B[j]);
                j++;
            }
            else {
                list.Add(C[k]);
                k++;
            }
        }
        while (i < A.Length && k < C.Length) {
            if (A[i] < C[k]) {
                list.Add(A[i]);
                i++;
            }
            else {
                list.Add(C[k]);
                k++;
            }
        }

        // if one of the array are left then
        // simply appending them as there will
        // be only largest element left
        while (i < A.Length) {
            list.Add(A[i]);
            i++;
        }
        while (j < B.Length) {
            list.Add(B[j]);
            j++;
        }
        while (k < C.Length) {
            list.Add(C[k]);
            k++;
        }

        // Finally print the list
        for (int x = 0; x < list.Count; x++)
            Console.Write(list[x] + " ");
    }

    public static void Main(string[] args)
    {

        int[] A = { 1, 2, 3, 5 };
        int[] B = { 6, 7, 8, 9 };
        int[] C = { 10, 11, 12 };

        // call the function to sort and print the sorted
        // numbers
        merge3sorted(A, B, C);
    }
}

// This code is contributed by Aarti_Rathi

Javascript

<script>
// Javascript program to merger three sorted arrays
// by merging three simultaneously.
function printVector(a) {
    document.write("[");
    for (let e of a) {
        document.write(e + " ");
    }
    document.write("]" + "<br>");
}

function mergeThree(A, B, C)
{
    let m, n, o, i, j, k;

    // Get Sizes of three vectors
    m = A.length;
    n = B.length;
    o = C.length;

    // Vector for storing output
    let D = [];
    i = j = k = 0;
    while (i < m && j < n && k < o)
    {

        // Get minimum of a, b, c
        let m = Math.min(Math.min(A[i], B[j]), C[k]);

        // Put m in D
        D.push(m);

        // Increment i, j, k
        if (m == A[i])
            i++;
        else if (m == B[j])
            j++;
        else
            k++;
    }

    // C has exhausted
    while (i < m && j < n) {
        if (A[i] <= B[j]) {
            D.push(A[i]);
            i++;
        }
        else {
            D.push(B[j]);
            j++;
        }
    }

    // B has exhausted
    while (i < m && k < o) {
        if (A[i] <= C[k]) {
            D.push(A[i]);
            i++;
        }
        else {
            D.push(C[k]);
            k++;
        }
    }

    // A has exhausted
    while (j < n && k < o) {
        if (B[j] <= C[k]) {
            D.push(B[j]);
            j++;
        }
        else {
            D.push(C[k]);
            k++;
        }
    }

    // A and B have exhausted
    while (k < o)
        D.push(C[k++]);

    // B and C have exhausted
    while (i < m)
        D.push(A[i++]);

    // A and C have exhausted
    while (j < n)
        D.push(B[j++]);

    return D;
}

// Driver Code

let A = [1, 2, 3, 5];
let B = [6, 7, 8, 9];
let C = [10, 11, 12];

// Print Result
printVector(mergeThree(A, B, C));

// This code is contributed by gfgking.
</script>

Output

1 2 3 5 6 7 8 9 10 11 12 

Time Complexity: O(M+N+O), Traversing over all the three arrays of size M, N, and O
Auxiliary Space: O(M+N+O), Space used for the output array

Note: While it is relatively easy to implement direct procedures to merge two or three arrays, the process becomes cumbersome if we want to merge 4 or more arrays. In such cases, we should follow the procedure shown in Merge K Sorted Arrays .

Optimization of the above approach:

The code written above can be shortened by the below code. Here we do not need to write code if any array gets exhausted. If any of the arrays get exhausted then store the large value(INT_MAX) in the variable which represents the value of that array. When all the pointers reach the end of their respective array return the resultant array.

Below is the implementation of the above approach:

C++

// C++ program to merge three sorted arrays
// by merging two at a time.
#include <bits/stdc++.h>
using namespace std;

// A[], B[], C[]: input arrays
// Function to merge three sorted lists into a single
// list.
vector<int> merge3sorted(vector<int>& A, vector<int>& B,
                         vector<int>& C)
{
    vector<int> ans;
    int l1 = A.size();
    int l2 = B.size();
    int l3 = C.size();
    int i = 0, j = 0, k = 0;
    while (i < l1 || j < l2 || k < l3) {
        // Assigning a, b, c with max values so that if
        // any value is not present then also we can sort
        // the array.
        int a = INT_MAX, b = INT_MAX, c = INT_MAX;

        // a, b, c variables are assigned only if the
        // value exist in the array.
        if (i < l1)
            a = A[i];
        if (j < l2)
            b = B[j];
        if (k < l3)
            c = C[k];

        // Checking if 'a' is the minimum
        if (a <= b && a <= c) {
            ans.push_back(a);
            i++;
        }
        // Checking if 'b' is the minimum
        else if (b <= a && b <= c) {
            ans.push_back(b);
            j++;
        }
        // Checking if 'c' is the minimum
        else {
            if (c <= a && c <= b) {
                ans.push_back(c);
                k++;
            }
        }
    }
    return ans;
}

// A utility function to print array list
void printeSorted(vector<int> list)
{
    for (auto x : list)
        cout << x << " ";
}
// Driver program to test above functions
int main()
{
    vector<int> A = { 1, 2, 3, 5 };
    vector<int> B = { 6, 7, 8, 9 };
    vector<int> C = { 10, 11, 12 };

    vector<int> final_ans = merge3sorted(A, B, C);
    printeSorted(final_ans);
    return 0;
}

// This code is contributed by Pushpesh raj

Java

/*package whatever //do not write package name here */

import java.io.*;
import java.lang.*;
import java.util.*;

// Java program to merge three sorted arrays
// by merging two at a time.

// This code is contributed by Animesh Nag

class Solution {
    // A[], B[], C[]: input arrays
    // Function to merge three sorted lists into a single
    // list.
    static ArrayList<Integer> merge3sorted(int A[], int B[],
                                           int C[])
    {
        ArrayList<Integer> ans = new ArrayList<Integer>();
        int l1 = A.length;
        int l2 = B.length;
        int l3 = C.length;
        int i = 0, j = 0, k = 0;
        while (i < l1 || j < l2 || k < l3) {
            // Assigning a, b, c with max values so that if
            // any value is not present then also we can
            // sort the array.
            int a = Integer.MAX_VALUE,
                b = Integer.MAX_VALUE,
                c = Integer.MAX_VALUE;

            // a, b, c variables are assigned only if the
            // value exist in the array.
            if (i < l1)
                a = A[i];
            if (j < l2)
                b = B[j];
            if (k < l3)
                c = C[k];

            // Checking if 'a' is the minimum
            if (a <= b && a <= c) {
                ans.add(a);
                i++;
            }
            // Checking if 'b' is the minimum
            else if (b <= a && b <= c) {
                ans.add(b);
                j++;
            }
            // Checking if 'c' is the minimum
            else {
                if (c <= a && c <= b) {
                    ans.add(c);
                    k++;
                }
            }
        }
        return ans;
    }
}

class GFG {
    // Driver program to test above functions
    public static void main(String[] args)
    {
        int[] A = { 1, 2, 3, 5 };
        int[] B = { 6, 7, 8, 9 };
        int[] C = { 10, 11, 12 };

        Solution sol = new Solution();
        ArrayList<Integer> final_ans
            = sol.merge3sorted(A, B, C);
        printeSorted(final_ans);
    }
    // A utility function to print array list
    static void printeSorted(ArrayList<Integer> list)
    {
        for (Integer x : list)
            System.out.print(x + " ");
    }
}

Python3

# Python program to merge three sorted arrays
# simultaneously.


def merge3sorted(A, B, C):
    (l1, l2, l3) = (len(A), len(B), len(C))
    i = j = k = 0

    # Destination array
    ans = []

    while (i < l1 or j < l2 or k < l3):

        # Assigning a, b, c with max values so that if
        # any value is not present then also we can sort
        # the array
        a = 9999
        b = 9999
        c = 9999

        # a, b, c variables are assigned only if the
        # value exist in the array.
        if (i < l1):
            a = A[i]
        if (j < l2):
            b = B[j]
        if (k < l3):
            c = C[k]

        # Checking if 'a' is the minimum
        if (a <= b and a <= c):
            ans.append(a)
            i += 1

        # Checking if 'b' is the minimum
        elif (b <= a and b <= c):
            ans.append(b)
            j += 1

        # Checking if 'c' is the minimum
        elif (c <= a and c <= b):
            ans.append(c)
            k += 1

    return ans


if __name__ == "__main__":
    A = [1, 2, 3, 5]
    B = [6, 7, 8, 9]
    C = [10, 11, 12]

    ans = merge3sorted(A, B, C)
    for x in ans:
        print(x, end=" ")

# This code is contributed by Aarti_Rathi

C#

using System;
using System.Collections.Generic;

public static class GFG {
    // C# program to merge three sorted arrays
    // by merging two at a time.

    // A[], B[], C[]: input arrays
    // Function to merge three sorted lists into a single
    // list.
    public static List<int>
    merge3sorted(List<int> A, List<int> B, List<int> C)
    {
        List<int> ans = new List<int>();
        int l1 = A.Count;
        int l2 = B.Count;
        int l3 = C.Count;
        int i = 0;
        int j = 0;
        int k = 0;
        while (i < l1 || j < l2 || k < l3) {
            // Assigning a, b, c with max values so that if
            // any value is not present then also we can
            // sort the array.
            int a = int.MaxValue;
            int b = int.MaxValue;
            int c = int.MaxValue;

            // a, b, c variables are assigned only if the
            // value exist in the array.
            if (i < l1) {
                a = A[i];
            }
            if (j < l2) {
                b = B[j];
            }
            if (k < l3) {
                c = C[k];
            }

            // Checking if 'a' is the minimum
            if (a <= b && a <= c) {
                ans.Add(a);
                i++;
            }
            // Checking if 'b' is the minimum
            else if (b <= a && b <= c) {
                ans.Add(b);
                j++;
            }
            // Checking if 'c' is the minimum
            else {
                if (c <= a && c <= b) {
                    ans.Add(c);
                    k++;
                }
            }
        }
        return new List<int>(ans);
    }

    // A utility function to print array list
    public static void printeSorted(List<int> list)
    {
        foreach(var x in list)
        {
            Console.Write(x);
            Console.Write(" ");
        }
    }
    // Driver program to test above functions
    public static void Main()
    {
        List<int> A = new List<int>() { 1, 2, 3, 5 };
        List<int> B = new List<int>() { 6, 7, 8, 9 };
        List<int> C = new List<int>() { 10, 11, 12 };

        List<int> final_ans = merge3sorted(A, B, C);
        printeSorted(new List<int>(final_ans));
    }
}

// This code is contributed by Aarti_Rathi

Javascript

// javascript program to merge three sorted arrays
// by merging two at a time.

// A[], B[], C[]: input arrays
// Function to merge three sorted lists into a single
// list.
function merge3sorted(A, B, C)
{
    var ans = new Array();
    var l1 = A.length;
    var l2 = B.length;
    var l3 = C.length;
    var i = 0;
    var j = 0;
    var k = 0;
    while (i < l1 || j < l2 || k < l3)
    {
        // Assigning a, b, c with max values so that if
        // any value is not present then also we can sort
        // the array.
        var a = Number.MAX_VALUE;
        var b = Number.MAX_VALUE;
        var c = Number.MAX_VALUE;
        // a, b, c variables are assigned only if the
        // value exist in the array.
        if (i < l1)
        {
            a = A[i];
        }
        if (j < l2)
        {
            b = B[j];
        }
        if (k < l3)
        {
            c = C[k];
        }
        // Checking if 'a' is the minimum
        if (a <= b && a <= c)
        {
            (ans.push(a) > 0);
            i++;
        }
        else if (b <= a && b <= c)
        {
            (ans.push(b) > 0);
            j++;
        }
        else
        {
            if (c <= a && c <= b)
            {
                (ans.push(c) > 0);
                k++;
            }
        }
    }
    return ans;
}




// A utility function to print array list
function printeSorted(list)
{
    console.log("[ ");
    for ( const  x of list) {console.log(x + " ");}
    console.log(" ]");
}

// Driver program to test above functions
var A = [1, 2, 3, 5];
var B = [6, 7, 8, 9];
var C = [10, 11, 12];
var final_ans = merge3sorted(A, B, C);
printeSorted(final_ans);

// This code is contributed by Aarti_Rathi

Output

1 2 3 5 6 7 8 9 10 11 12 

Time Complexity: O(l1+l2+l3), Traversing over all the three arrays of size l1, l2,and l3
Auxiliary Space:  O(l1+l2+l3),  Space used for the output array