# Sort an array when two halves are sorted

Given an integer array of which both first half and second half are sorted. Task is to merge two sorted halves of array into single sorted array.

Examples:

```Input : A[] = { 2 ,3 , 8 ,-1 ,7 ,10 }
Output : -1 , 2 , 3 , 7 , 8 , 10

Input : A[] = {-4 , 6, 9 , -1 , 3 }
Output : -4 , -1 , 3 , 6 , 9
```

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

A Simple Solution is to sort the array.
Below C++ implementation of idea

```// C++ program to Merge two sorted halves of
// array Into Single Sorted Array
#include<bits/stdc++.h>
using namespace std;

void mergeTwoHalf(int A[], int n)
{
// Sort the given array using sort STL
sort(A, A+n);
}

// diver program to test above function
int main()
{
int A[] = { 2 ,3 , 8 ,-1 ,7 ,10 } ;
int n = sizeof(A)/sizeof(A[0]) ;
mergeTwoHalf(A, n);

// Print sorted Array
for (int i=0 ; i < n ; i++)
cout << A[i] << " " ;
return 0;
}
```

Output:

``` -1  2  3  7  8  10
```

Time Complexity O(nlogn) || Sort Given array using quick sort or merge sort

An efficient solution is to use an auxiliary array one half. Now whole process is same as the Merge Function of Marge sort .
Below C++ implementation of idea

```// C++ program to Merge Two Sorted Halves Of
// Array Into Single Sorted Array
#include<bits/stdc++.h>
using namespace std;

// Merge two sorted halves of Array into single
// sorted array
void mergeTwoHalf(int A[], int n)
{
int  half_i = 0; // starting index of second half

// Temp Array store sorted resultant array
int temp[n] ;

// First Find the point where array is divide
// into two half
for (int i=0; i<n-1 ; i++)
{
if (A[i] > A[i+1])
{
half_i = i+1 ;
break;
}
}

// If Given array is all-ready sorted
if (half_i == 0)
return;

// Merge two sorted arrays in single sorted array
int i = 0 , j = half_i  , k = 0;
while (i < half_i && j < n)
{
if (A[i] < A[j])
temp[k++] = A[i++];
else
temp[k++] = A[j++];
}

// Copy the remaining elements of A[i to half_! ]
while (i  < half_i)
temp[k++] = A[i++];

// Copy the remaining elements of A[ half_! to n ]
while (j < n)
temp[k++] = A[j++];

for (int i=0; i<n; i++)
A[i] = temp[i];
}

// diver program to test above function
int main()
{
int A[] = { 2 ,3 , 8 ,-1 ,7 ,10 } ;
int n = sizeof(A)/sizeof(A[0]) ;
mergeTwoHalf(A, n);

// Print sorted Array
for (int i=0 ; i < n ; i++)
cout << A[i] << " " ;
return 0;
}
```

Output:

``` -1  2  3  7  8  10
```

Time Complexity : O(n)

This article is contributed by Nishant Singh . 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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