# Merging two unsorted arrays in sorted order

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}
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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.

## C++

```// CPP program to merge two unsorted lists
// in sorted order
#include <bits/stdc++.h>
using namespace std;

// Function to merge array in sorted order
void sortedMerge(int a[], int b[], int res[],
int n, int m)
{
// Concatenate two arrays
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;
}

// sorting the res array
sort(res, res + n + m);
}

// Driver code
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]);

// Final merge list
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

```// JAVA Code for Merging two unsorted
// arrays in sorted order
import java.util.*;

class GFG {

// Function to merge array in sorted order
public static void sortedMerge(int a[], int b[],
int res[], int n,
int m)
{
// Concatenate two arrays
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 ++;
}

// sorting the res array
Arrays.sort(res);
}

/* Driver program to test above function */
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;

// Final merge list
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]);
}
}
// This code is contributed by Arnav Kr. Mandal.
```

## Python

```# Python program to merge two unsorted lists
# in sorted order

# Function to merge array in sorted order
def sortedMerge(a, b, res, n, m):
# Concatenate two arrays
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

# sorting the res array
res.sort()

# Driver code
a = [ 10, 5, 15 ]
b = [ 20, 3, 2, 12 ]
n = len(a)
m = len(b)

# Final merge list
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],

# This code is contributed by Sachin Bisht
```

Output :

```Sorted merged list : 2 3 5 10 12 15 20
```

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.

## C++

```// CPP program to merge two unsorted lists
// in sorted order
#include <bits/stdc++.h>
using namespace std;

// Function to merge array in sorted order
void sortedMerge(int a[], int b[], int res[],
int n, int m)
{
// Sorting a[] and b[]
sort(a, a + n);
sort(b, b + m);

// Merge two sorted arrays into res[]
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) {  // Merging remaining
// elements of a[] (if any)
res[k] = a[i];
i += 1;
k += 1;
}
while (j < m) {   // Merging remaining
// elements of b[] (if any)
res[k] = b[j];
j += 1;
k += 1;
}
}

// Driver code
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]);

// Final merge list
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

```// JAVA Code for Merging two unsorted
// arrays in sorted order
import java.util.*;

class GFG {

// Function to merge array in sorted order
public static void sortedMerge(int a[], int b[],
int res[], int n,
int m)
{
// Sorting a[] and b[]
Arrays.sort(a);
Arrays.sort(b);

// Merge two sorted arrays into res[]
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) {  // Merging remaining
// elements of a[] (if any)
res[k] = a[i];
i += 1;
k += 1;
}
while (j < m) {   // Merging remaining
// elements of b[] (if any)
res[k] = b[j];
j += 1;
k += 1;
}
}

/* Driver program to test above function */
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;

// Final merge list
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]);
}
}
// This code is contributed by Arnav Kr. Mandal.
```

## Python

```# Python program to merge two unsorted lists
# in sorted order

# Function to merge array in sorted order
def sortedMerge(a, b, res, n, m):
# Sorting a[] and b[]
a.sort()
b.sort()

# Merge two sorted arrays into res[]
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):  # Merging remaining
# elements of a[] (if any)
res[k] = a[i]
i += 1
k += 1

while (j < m):  # Merging remaining
# elements of b[] (if any)
res[k] = b[j]
j += 1
k += 1

# Driver code
a = [ 10, 5, 15 ]
b = [ 20, 3, 2, 12 ]
n = len(a)
m = len(b)

# Final merge list
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],

# This code is contributed by Sachin Bisht
```

Output :

```Sorted merge list : 2 3 5 10 12 15 20
```

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.

This article is contributed by Sachin Bisht. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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