Merge 3 Sorted Arrays
Last Updated :
31 Dec, 2023
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:
- Take the size of the arrays A, B, and C as input from the user.
- Create arrays A, B, and C of the input size.
- Take the elements of arrays A, B, and C as input from the user.
- Merge arrays A, B, and C into a single array D.
- Sort array D in ascending order.
- Print the elements of array D.
Below is the implementation of the approach:
C++
#include <bits/stdc++.h>
using namespace std;
vector< int > mergeThreeSortedArrays(vector< int >& A, vector< int >& B, vector< int >& C) {
vector< int > D;
for ( int i = 0; i < A.size(); i++) {
D.push_back(A[i]);
}
for ( int i = 0; i < B.size(); i++) {
D.push_back(B[i]);
}
for ( int i = 0; i < C.size(); i++) {
D.push_back(C[i]);
}
sort(D.begin(), D.end());
return D;
}
int main() {
vector< int > A = { 1, 2, 3, 5 };
vector< int > B = { 6, 7, 8, 9 };
vector< int > C = { 10, 11, 12 };
vector< int > D = mergeThreeSortedArrays(A, B, C);
for ( int i = 0; i < D.size(); i++) {
cout << D[i] << " " ;
}
cout << endl;
return 0;
}
|
Java
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class Main {
static List<Integer>
mergeThreeSortedArrays(List<Integer> A, List<Integer> B,
List<Integer> C)
{
List<Integer> D = new ArrayList<>();
D.addAll(A);
D.addAll(B);
D.addAll(C);
Collections.sort(D);
return D;
}
public static void main(String[] args)
{
List<Integer> A = Arrays.asList( 1 , 2 , 3 , 5 );
List<Integer> B = Arrays.asList( 6 , 7 , 8 , 9 );
List<Integer> C = Arrays.asList( 10 , 11 , 12 );
List<Integer> D = mergeThreeSortedArrays(A, B, C);
for ( int i = 0 ; i < D.size(); i++) {
System.out.print(D.get(i) + " " );
}
System.out.println();
}
}
|
Python3
def mergeThreeSortedArrays(A, B, C):
D = []
for element in A:
D.append(element)
for element in B:
D.append(element)
for element in C:
D.append(element)
D.sort()
return D
A = [ 1 , 2 , 3 , 5 ]
B = [ 6 , 7 , 8 , 9 ]
C = [ 10 , 11 , 12 ]
D = mergeThreeSortedArrays(A, B, C)
for element in D:
print (element, end = " " )
print ()
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
class Program {
static List< int > MergeThreeSortedArrays(List< int > A,
List< int > B,
List< int > C)
{
List< int > D = new List< int >();
D.AddRange(A);
D.AddRange(B);
D.AddRange(C);
D.Sort();
return D;
}
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 > D = MergeThreeSortedArrays(A, B, C);
foreach ( int num in D) { Console.Write(num + " " ); }
Console.WriteLine();
}
}
|
Javascript
function mergeThreeSortedArrays(A, B, C) {
let D = [];
for (let i = 0; i < A.length; i++) {
D.push(A[i]);
}
for (let i = 0; i < B.length; i++) {
D.push(B[i]);
}
for (let i = 0; i < C.length; i++) {
D.push(C[i]);
}
D.sort((a, b) => a - b);
return D;
}
function main() {
const A = [1, 2, 3, 5];
const B = [6, 7, 8, 9];
const C = [10, 11, 12];
const D = mergeThreeSortedArrays(A, B, C);
console.log(D.join( " " ));
}
main();
|
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++
#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)
{
int m = A.size();
int n = B.size();
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++]);
}
while (i < m)
D.push_back(A[i++]);
while (j < n)
D.push_back(B[j++]);
return D;
}
int main()
{
Vector A = { 1, 2, 3, 5 };
Vector B = { 6, 7, 8, 9 };
Vector C = { 10, 11, 12 };
Vector T = mergeTwo(A, B);
printVector(mergeTwo(T, C));
return 0;
}
|
Java
import java.util.*;
class GFG {
static ArrayList<Integer> mergeTwo(List<Integer> A,
List<Integer> B)
{
int m = A.size();
int n = B.size();
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++));
}
while (i < m)
D.add(A.get(i++));
while (j < n)
D.add(B.get(j++));
return D;
}
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);
ArrayList<Integer> T = mergeTwo(A, B);
ArrayList<Integer> ans = mergeTwo(T, C);
for ( int i = 0 ; i < ans.size(); i++)
System.out.print(ans.get(i) + " " );
}
}
|
Python3
def merge_two(a, b):
(m, n) = ( len (a), len (b))
i = j = 0
d = []
while i < m and j < n:
if a[i] < = b[j]:
d.append(a[i])
i + = 1
else :
d.append(b[j])
j + = 1
while i < m:
d.append(a[i])
i + = 1
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#
using System;
using System.Collections.Generic;
public static class GFG {
static List< int > mergeTwo(List< int > A, List< int > B)
{
int m = A.Count;
int n = B.Count;
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++]);
}
}
while (i < m) {
D.Add(A[i++]);
}
while (j < n) {
D.Add(B[j++]);
}
return D;
}
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 > T = mergeTwo(A, B);
List< int > ans = mergeTwo(T, C);
for ( int i = 0; i < ans.Count; i++)
Console.Write(ans[i] + " " );
}
}
|
Javascript
<script>
function mergeTwo(A, B)
{
let m = A.length;
let n = B.length;
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++]);
}
while (i < m)
D.push(A[i++]);
while (j < n)
D.push(B[j++]);
return D;
}
let A = [ 1, 2, 3, 5 ];
let B = [ 6, 7, 8, 9 ];
let C = [ 10, 11, 12 ];
let T = mergeTwo(A, B);
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++
#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;
m = A.size();
n = B.size();
o = C.size();
vector< int > D;
D.reserve(m + n + o);
i = j = k = 0;
while (i < m && j < n && k < o) {
int m = min(min(A[i], B[j]), C[k]);
D.push_back(m);
if (m == A[i])
i++;
else if (m == B[j])
j++;
else
k++;
}
while (i < m && j < n) {
if (A[i] <= B[j]) {
D.push_back(A[i]);
i++;
}
else {
D.push_back(B[j]);
j++;
}
}
while (i < m && k < o) {
if (A[i] <= C[k]) {
D.push_back(A[i]);
i++;
}
else {
D.push_back(C[k]);
k++;
}
}
while (j < n && k < o) {
if (B[j] <= C[k]) {
D.push_back(B[j]);
j++;
}
else {
D.push_back(C[k]);
k++;
}
}
while (k < o)
D.push_back(C[k++]);
while (i < m)
D.push_back(A[i++]);
while (j < n)
D.push_back(B[j++]);
return D;
}
int main()
{
vector< int > A = { 1, 2, 3, 5 };
vector< int > B = { 6, 7, 8, 9 };
vector< int > C = { 10, 11, 12 };
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 };
merge3sorted(A, B, C);
}
static void merge3sorted( int A[], int B[], int C[])
{
ArrayList<Integer> list = new ArrayList<Integer>();
int i = 0 , j = 0 , k = 0 ;
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++;
}
}
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++;
}
}
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++;
}
for (Integer x : list)
System.out.print(x + " " );
}
}
|
Python3
def merge_three(a, b, c):
(m, n, o) = ( len (a), len (b), len (c))
i = j = k = 0
d = []
while i < m and j < n and k < o:
mini = min (a[i], b[j], c[k])
d.append(mini)
if a[i] = = mini:
i + = 1
elif b[j] = = mini:
j + = 1
elif c[k] = = mini:
k + = 1
while i < m and j < n:
if a[i] < = b[j]:
d.append(a[i])
i + = 1
else :
d.append(b[j])
j + = 1
while j < n and k < o:
if b[j] < = c[k]:
d.append(b[j])
j + = 1
else :
d.append(c[k])
k + = 1
while i < m and k < o:
if a[i] < = c[k]:
d.append(a[i])
i + = 1
else :
d.append(c[k])
k + = 1
while i < m:
d.append(a[i])
i + = 1
while j < n:
d.append(b[j])
j + = 1
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#
using System;
using System.Collections;
class Sorting {
static void merge3sorted( int [] A, int [] B, int [] C)
{
ArrayList list = new ArrayList();
int i = 0, j = 0, k = 0;
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++;
}
}
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++;
}
}
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++;
}
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 };
merge3sorted(A, B, C);
}
}
|
Javascript
<script>
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;
m = A.length;
n = B.length;
o = C.length;
let D = [];
i = j = k = 0;
while (i < m && j < n && k < o)
{
let m = Math.min(Math.min(A[i], B[j]), C[k]);
D.push(m);
if (m == A[i])
i++;
else if (m == B[j])
j++;
else
k++;
}
while (i < m && j < n) {
if (A[i] <= B[j]) {
D.push(A[i]);
i++;
}
else {
D.push(B[j]);
j++;
}
}
while (i < m && k < o) {
if (A[i] <= C[k]) {
D.push(A[i]);
i++;
}
else {
D.push(C[k]);
k++;
}
}
while (j < n && k < o) {
if (B[j] <= C[k]) {
D.push(B[j]);
j++;
}
else {
D.push(C[k]);
k++;
}
}
while (k < o)
D.push(C[k++]);
while (i < m)
D.push(A[i++]);
while (j < n)
D.push(B[j++]);
return D;
}
let A = [1, 2, 3, 5];
let B = [6, 7, 8, 9];
let C = [10, 11, 12];
printVector(mergeThree(A, B, C));
</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++
#include <bits/stdc++.h>
using namespace std;
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) {
int a = INT_MAX, b = INT_MAX, c = INT_MAX;
if (i < l1)
a = A[i];
if (j < l2)
b = B[j];
if (k < l3)
c = C[k];
if (a <= b && a <= c) {
ans.push_back(a);
i++;
}
else if (b <= a && b <= c) {
ans.push_back(b);
j++;
}
else {
if (c <= a && c <= b) {
ans.push_back(c);
k++;
}
}
}
return ans;
}
void printeSorted(vector< int > list)
{
for ( auto x : list)
cout << x << " " ;
}
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;
}
|
Java
import java.io.*;
import java.lang.*;
import java.util.*;
class Solution {
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) {
int a = Integer.MAX_VALUE,
b = Integer.MAX_VALUE,
c = Integer.MAX_VALUE;
if (i < l1)
a = A[i];
if (j < l2)
b = B[j];
if (k < l3)
c = C[k];
if (a <= b && a <= c) {
ans.add(a);
i++;
}
else if (b <= a && b <= c) {
ans.add(b);
j++;
}
else {
if (c <= a && c <= b) {
ans.add(c);
k++;
}
}
}
return ans;
}
}
class GFG {
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);
}
static void printeSorted(ArrayList<Integer> list)
{
for (Integer x : list)
System.out.print(x + " " );
}
}
|
Python3
def merge3sorted(A, B, C):
(l1, l2, l3) = ( len (A), len (B), len (C))
i = j = k = 0
ans = []
while (i < l1 or j < l2 or k < l3):
a = 9999
b = 9999
c = 9999
if (i < l1):
a = A[i]
if (j < l2):
b = B[j]
if (k < l3):
c = C[k]
if (a < = b and a < = c):
ans.append(a)
i + = 1
elif (b < = a and b < = c):
ans.append(b)
j + = 1
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 = " " )
|
C#
using System;
using System.Collections.Generic;
public static class GFG {
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) {
int a = int .MaxValue;
int b = int .MaxValue;
int c = int .MaxValue;
if (i < l1) {
a = A[i];
}
if (j < l2) {
b = B[j];
}
if (k < l3) {
c = C[k];
}
if (a <= b && a <= c) {
ans.Add(a);
i++;
}
else if (b <= a && b <= c) {
ans.Add(b);
j++;
}
else {
if (c <= a && c <= b) {
ans.Add(c);
k++;
}
}
}
return new List< int >(ans);
}
public static void printeSorted(List< int > list)
{
foreach ( var x in list)
{
Console.Write(x);
Console.Write( " " );
}
}
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));
}
}
|
Javascript
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)
{
var a = Number.MAX_VALUE;
var b = Number.MAX_VALUE;
var c = Number.MAX_VALUE;
if (i < l1)
{
a = A[i];
}
if (j < l2)
{
b = B[j];
}
if (k < l3)
{
c = C[k];
}
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;
}
function printeSorted(list)
{
console.log( "[ " );
for ( const x of list) {console.log(x + " " );}
console.log( " ]" );
}
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);
|
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
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