Merging two unsorted arrays in sorted order
Last Updated :
19 Sep, 2023
Write a SortedMerge() function that takes two lists, each of which is unsorted, and merges the two together into one new list which is in sorted (increasing) order. SortedMerge() should return the new list.
Examples :
Input : a[] = {10, 5, 15}
b[] = {20, 3, 2}
Output : Merge List :
{2, 3, 5, 10, 15, 20}
Input : a[] = {1, 10, 5, 15}
b[] = {20, 0, 2}
Output : Merge List :
{0, 1, 2, 5, 10, 15, 20}
There are many cases to deal with: either ‘a’ or ‘b’ may be empty, during processing either ‘a’ or ‘b’ may run out first, and finally, there’s the problem of starting the result list empty and building it up while going through ‘a’ and ‘b’.
Method 1 (first Concatenate then Sort): In this case, we first append the two unsorted lists. Then we simply sort the concatenated list.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void sortedMerge( int a[], int b[], int res[],
int n, int m)
{
int i = 0, j = 0, k = 0;
while (i < n) {
res[k] = a[i];
i += 1;
k += 1;
}
while (j < m) {
res[k] = b[j];
j += 1;
k += 1;
}
sort(res, res + n + m);
}
int main()
{
int a[] = { 10, 5, 15 };
int b[] = { 20, 3, 2, 12 };
int n = sizeof (a) / sizeof (a[0]);
int m = sizeof (b) / sizeof (b[0]);
int res[n + m];
sortedMerge(a, b, res, n, m);
cout << "Sorted merged list :" ;
for ( int i = 0; i < n + m; i++)
cout << " " << res[i];
cout << "n" ;
return 0;
}
|
Java
import java.util.*;
class GFG {
public static void sortedMerge( int a[], int b[],
int res[], int n,
int m)
{
int i = 0 , j = 0 , k = 0 ;
while (i < n) {
res[k] = a[i];
i++;
k++;
}
while (j < m) {
res[k] = b[j];
j++;
k++;
}
Arrays.sort(res);
}
public static void main(String[] args)
{
int a[] = { 10 , 5 , 15 };
int b[] = { 20 , 3 , 2 , 12 };
int n = a.length;
int m = b.length;
int res[]= new int [n + m];
sortedMerge(a, b, res, n, m);
System.out.print( "Sorted merged list :" );
for ( int i = 0 ; i < n + m; i++)
System.out.print( " " + res[i]);
}
}
|
Python3
def sortedMerge(a, b, res, n, m):
i, j, k = 0 , 0 , 0
while (i < n):
res[k] = a[i]
i + = 1
k + = 1
while (j < m):
res[k] = b[j]
j + = 1
k + = 1
res.sort()
a = [ 10 , 5 , 15 ]
b = [ 20 , 3 , 2 , 12 ]
n = len (a)
m = len (b)
res = [ 0 for i in range (n + m)]
sortedMerge(a, b, res, n, m)
print ( "Sorted merged list :" )
for i in range (n + m):
print (res[i],end = " " )
|
C#
using System;
class GFG {
public static void sortedMerge( int []a, int []b,
int []res, int n,
int m)
{
int i = 0, j = 0, k = 0;
while (i < n) {
res[k] = a[i];
i++;
k++;
}
while (j < m) {
res[k] = b[j];
j++;
k++;
}
Array.Sort(res);
}
public static void Main()
{
int []a = {10, 5, 15};
int []b = {20, 3, 2, 12};
int n = a.Length;
int m = b.Length;
int []res= new int [n + m];
sortedMerge(a, b, res, n, m);
Console.Write( "Sorted merged list :" );
for ( int i = 0; i < n + m; i++)
Console.Write( " " + res[i]);
}
}
|
PHP
<?php
function sortedMerge( $a , $b , $n , $m )
{
$res = array ();
$i = 0; $j = 0; $k = 0;
while ( $i < $n )
{
$res [ $k ] = $a [ $i ];
$i += 1;
$k += 1;
}
while ( $j < $m )
{
$res [ $k ] = $b [ $j ];
$j += 1;
$k += 1;
}
sort( $res );
echo "Sorted merged list :" ;
for ( $i = 0; $i < count ( $res ); $i ++)
echo $res [ $i ] . " " ;
}
$a = array ( 10, 5, 15 );
$b = array ( 20, 3, 2, 12 );
$n = count ( $a );
$m = count ( $b );
sortedMerge( $a , $b , $n , $m );
?>
|
Javascript
<script>
function sortedMerge(a, b, res,
n, m)
{
a.sort((a,b) => a-b);
b.sort((a,b) => a-b);
let i = 0, j = 0, k = 0;
while (i < n && j < m) {
if (a[i] <= b[j]) {
res[k] = a[i];
i += 1;
k += 1;
} else {
res[k] = b[j];
j += 1;
k += 1;
}
}
while (i < n) {
res[k] = a[i];
i += 1;
k += 1;
}
while (j < m) {
res[k] = b[j];
j += 1;
k += 1;
}
}
let a = [ 10, 5, 15 ];
let b = [ 20, 3, 2, 12 ];
let n = a.length;
let m = b.length;
let res = new Array(n + m);
sortedMerge(a, b, res, n, m);
document.write( "Sorted merge list :" );
for (let i = 0; i < n + m; i++)
document.write( " " + res[i]);
</script>
|
Output
Sorted merged list : 2 3 5 10 12 15 20n
Time Complexity: O ( (n + m) (log(n + m)) )
Auxiliary Space: O ( (n + m) )
Method 2 (First Sort then Merge): We first sort both the given arrays separately. Then we simply merge two sorted arrays.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void sortedMerge( int a[], int b[], int res[],
int n, int m)
{
sort(a, a + n);
sort(b, b + m);
int i = 0, j = 0, k = 0;
while (i < n && j < m) {
if (a[i] <= b[j]) {
res[k] = a[i];
i += 1;
k += 1;
} else {
res[k] = b[j];
j += 1;
k += 1;
}
}
while (i < n) {
res[k] = a[i];
i += 1;
k += 1;
}
while (j < m) {
res[k] = b[j];
j += 1;
k += 1;
}
}
int main()
{
int a[] = { 10, 5, 15 };
int b[] = { 20, 3, 2, 12 };
int n = sizeof (a) / sizeof (a[0]);
int m = sizeof (b) / sizeof (b[0]);
int res[n + m];
sortedMerge(a, b, res, n, m);
cout << "Sorted merge list :" ;
for ( int i = 0; i < n + m; i++)
cout << " " << res[i];
cout << "n" ;
return 0;
}
|
Java
import java.util.*;
class GFG {
public static void sortedMerge( int a[], int b[],
int res[], int n,
int m)
{
Arrays.sort(a);
Arrays.sort(b);
int i = 0 , j = 0 , k = 0 ;
while (i < n && j < m) {
if (a[i] <= b[j]) {
res[k] = a[i];
i += 1 ;
k += 1 ;
} else {
res[k] = b[j];
j += 1 ;
k += 1 ;
}
}
while (i < n) {
res[k] = a[i];
i += 1 ;
k += 1 ;
}
while (j < m) {
res[k] = b[j];
j += 1 ;
k += 1 ;
}
}
public static void main(String[] args)
{
int a[] = { 10 , 5 , 15 };
int b[] = { 20 , 3 , 2 , 12 };
int n = a.length;
int m = b.length;
int res[] = new int [n + m];
sortedMerge(a, b, res, n, m);
System.out.print( "Sorted merged list :" );
for ( int i = 0 ; i < n + m; i++)
System.out.print( " " + res[i]);
}
}
|
Python3
def sortedMerge(a, b, res, n, m):
a.sort()
b.sort()
i, j, k = 0 , 0 , 0
while (i < n and j < m):
if (a[i] < = b[j]):
res[k] = a[i]
i + = 1
k + = 1
else :
res[k] = b[j]
j + = 1
k + = 1
while (i < n):
res[k] = a[i]
i + = 1
k + = 1
while (j < m):
res[k] = b[j]
j + = 1
k + = 1
a = [ 10 , 5 , 15 ]
b = [ 20 , 3 , 2 , 12 ]
n = len (a)
m = len (b)
res = [ 0 for i in range (n + m)]
sortedMerge(a, b, res, n, m)
print ( "Sorted merged list :" )
for i in range (n + m):
print (res[i],end = " " )
|
C#
using System;
class GFG {
static void sortedMerge( int []a, int []b,
int []res, int n, int m)
{
Array.Sort(a);
Array.Sort(b);
int i = 0, j = 0, k = 0;
while (i < n && j < m)
{
if (a[i] <= b[j])
{
res[k] = a[i];
i += 1;
k += 1;
}
else
{
res[k] = b[j];
j += 1;
k += 1;
}
}
while (i < n)
{
res[k] = a[i];
i += 1;
k += 1;
}
while (j < m)
{
res[k] = b[j];
j += 1;
k += 1;
}
}
public static void Main()
{
int []a = { 10, 5, 15 };
int []b = { 20, 3, 2, 12 };
int n = a.Length;
int m = b.Length;
int []res = new int [n + m];
sortedMerge(a, b, res, n, m);
Console.Write( "Sorted merged list :" );
for ( int i = 0; i < n + m; i++)
Console.Write( " " + res[i]);
}
}
|
Javascript
<script>
function sortedMerge(a, b, res, n, m)
{
a.sort((a, b) => a - b);
b.sort((a, b) => a - b);
let i = 0, j = 0, k = 0;
while (i < n && j < m)
{
if (a[i] <= b[j])
{
res[k] = a[i];
i += 1;
k += 1;
}
else
{
res[k] = b[j];
j += 1;
k += 1;
}
}
while (i < n)
{
res[k] = a[i];
i += 1;
k += 1;
}
while (j < m)
{
res[k] = b[j];
j += 1;
k += 1;
}
}
let a = [ 10, 5, 15 ];
let b = [ 20, 3, 2, 12 ];
let n = a.length;
let m = b.length;
let res = new Array(n + m);
sortedMerge(a, b, res, n, m);
document.write( "Sorted merge list :" );
for (let i = 0; i < n + m; i++)
document.write( " " + res[i]);
</script>
|
Output
Sorted merge list : 2 3 5 10 12 15 20n
Time Complexity: O (nlogn + mlogm + (n + m))
Space Complexity: O ( (n + m) )
It is obvious from above time complexities that method 2 is better than method 1.
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