# Merge two sorted arrays with O(1) extra space

We are given two sorted array. We need to merge these two arrays such that the initial numbers (after complete sorting) are in the first array and the remaining numbers are in the second array. Extra space allowed in O(1).

Example:

```Input: ar1[] = {10};
ar2[] = {2, 3};
Output: ar1[] = {2}
ar2[] = {3, 10}

Input: ar1[] = {1, 5, 9, 10, 15, 20};
ar2[] = {2, 3, 8, 13};
Output: ar1[] = {1, 2, 3, 5, 8, 9}
ar2[] = {10, 13, 15, 20}
```

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

This task is simple and O(m+n) if we are allowed to use extra space. But it becomes really complicated when extra space is not allowed and doesn’t look possible in less than O(m*n) worst case time.

The idea is to begin from last element of ar2[] and search it in ar1[]. If there is a greater element in ar1[], then we move last element of ar1[] to ar2[]. To keep ar1[] and ar2[] sorted, we need to place last element of ar2[] at correct place in ar1[]. We can use Insertion Sort type of insertion for this. Below is algorithm:

```1) Iterate through every element of ar2[] starting from last
element. Do following for every element ar2[i]
a) Store last element of ar1[i]: last = ar1[i]
b) Loop from last element of ar1[] while element ar1[j] is
smaller than ar2[i].
ar1[j+1] = ar1[j] // Move element one position ahead
j--
c) If any element of ar1[] was moved or (j != m-1)
ar1[j+1] = ar2[i]
ar2[i] = last
```

In above loop, elements in ar1[] and ar2[] are always kept sorted.

Below is C++ and Java implementation of above algorithm.

## C++

```// C++ program to merge two sorted arrays with O(1) extra space.
#include <bits/stdc++.h>
using namespace std;

// Merge ar1[] and ar2[] with O(1) extra space
void merge(int ar1[], int ar2[], int m, int n)
{
// Iterate through all elements of ar2[] starting from
// the last element
for (int i=n-1; i>=0; i--)
{
/* Find the smallest element greater than ar2[i]. Move all
elements one position ahead till the smallest greater
int j, last = ar1[m-1];
for (j=m-2; j >= 0 && ar1[j] > ar2[i]; j--)
ar1[j+1] = ar1[j];

// If there was a greater element
if (j != m-2 || last > ar2[i])
{
ar1[j+1] = ar2[i];
ar2[i] = last;
}
}
}

// Driver program
int main(void)
{
int ar1[] = {1, 5, 9, 10, 15, 20};
int ar2[] = {2, 3, 8, 13};
int m = sizeof(ar1)/sizeof(ar1[0]);
int n = sizeof(ar2)/sizeof(ar2[0]);
merge(ar1, ar2, m, n);

cout << "After Merging nFirst Array: ";
for (int i=0; i<m; i++)
cout << ar1[i] << " ";
cout << "nSecond Array: ";
for (int i=0; i<n; i++)
cout << ar2[i] << " ";
return 0;
}
```

## Java

```// Java program program to merge two
// sorted arrays with O(1) extra space.

import java.util.Arrays;

class Test
{
static int arr1[] = new int[]{1, 5, 9, 10, 15, 20};
static int arr2[] = new int[]{2, 3, 8, 13};

static void merge(int m, int n)
{
// Iterate through all elements of ar2[] starting from
// the last element
for (int i=n-1; i>=0; i--)
{
/* Find the smallest element greater than ar2[i]. Move all
elements one position ahead till the smallest greater
int j, last = arr1[m-1];
for (j=m-2; j >= 0 && arr1[j] > arr2[i]; j--)
arr1[j+1] = arr1[j];

// If there was a greater element
if (j != m-2 || last > arr2[i])
{
arr1[j+1] = arr2[i];
arr2[i] = last;
}
}
}

// Driver method to test the above function
public static void main(String[] args)
{
merge(arr1.length,arr2.length);
System.out.print("After Merging nFirst Array: ");
System.out.println(Arrays.toString(arr1));
System.out.print("Second Array:  ");
System.out.println(Arrays.toString(arr2));
}
}
```

Output:
```After Merging
First Array: 1 2 3 5 8 9
Second Array: 10 13 15 20 ```

Time Complexity: The worst case time complexity of code/algorithm is O(m*n). The worst case occurs when all elements of ar1[] are greater than all elements of ar2[].

Illustration:

Efficiently merging two sorted arrays with O(1) extra space   Related Articles : Merge two sorted arrays Merge k sorted arrays | Set 1 Thanks to Shubham Chauhan for suggesting above solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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