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 
       greater 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 the 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 
           element is not found */
        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 
               element is not found */
            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)); 
    } 
} 

Python3

# Python program to merge 
# two sorted arrays 
# with O(1) extra space. 
  
# Merge ar1[] and ar2[] 
# with O(1) extra space 
def merge(ar1, ar2, m, n): 
  
    # Iterate through all 
    # elements of ar2[] starting from 
    # the last element 
    for i in range(n-1, -1, -1): 
      
        # Find the smallest element 
        # greater than ar2[i]. Move all 
        # elements one position ahead 
        # till the smallest greater 
        # element is not found  
        last = ar1[m-1] 
        j=m-2
        while(j >= 0 and ar1[j] > ar2[i]): 
            ar1[j+1] = ar1[j] 
            j-=1
   
        # If there was a greater element 
        if (j != m-2 or last > ar2[i]): 
          
            ar1[j+1] = ar2[i] 
            ar2[i] = last 
   
# Driver program 
  
ar1 = [1, 5, 9, 10, 15, 20] 
ar2 = [2, 3, 8, 13] 
m = len(ar1) 
n = len(ar2) 
  
merge(ar1, ar2, m, n) 
   
print("After Merging \nFirst Array:", end="") 
for i in range(m): 
    print(ar1[i] , " ", end="") 
  
print("\nSecond Array: ", end="") 
for i in range(n): 
    print(ar2[i] , " ", end="") 
  
# This code is contributed 
# by Anant Agarwal. 

C#

// C# program program to merge two  
// sorted arrays with O(1) extra space.  
using System; 
  
// Java program program to merge two  
// sorted arrays with O(1) extra space.  
  
  
public 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  
            element is not found */
            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()  
    {  
        merge(arr1.Length,arr2.Length);  
        Console.Write("After Merging \nFirst Array: ");  
        for(int i =0; i< arr1.Length;i++){ 
            Console.Write(arr1[i]+" "); 
        } 
        Console.Write("\nSecond Array: ");  
        for(int i =0; i< arr2.Length;i++){ 
            Console.Write(arr2[i]+" "); 
        } 
    }  
}  
  
/*This code is contributed by 29AjayKumar*/

PHP

<?php  
// PHP program to merge two sorted arrays with O(1) extra space. 
   
// Merge ar1[] and ar2[] with O(1) extra space 
function merge(&$ar1, &$ar2, $m, $n) 
{ 
    // Iterate through all elements of ar2[] starting from 
    // the last element 
    for ($i = $n-1; $i >= 0; $i--) 
    { 
        /* Find the smallest element greater than ar2[i]. Move all 
           elements one position ahead till the smallest greater 
           element is not found */
        $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 
  
$ar1 = array(1, 5, 9, 10, 15, 20); 
$ar2 = array(2, 3, 8, 13); 
$m = sizeof($ar1)/sizeof($ar1[0]); 
$n = sizeof($ar2)/sizeof($ar2[0]); 
merge($ar1, $ar2, $m, $n); 
  
echo "After Merging \nFirst Array: "; 
for ($i=0; $i<$m; $i++) 
    echo $ar1[$i] . " "; 
echo "\nSecond Array: "; 
for ($i=0; $i<$n; $i++) 
    echo $ar2[$i] ." "; 
return 0; 
?> 

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:

<!— Initial Arrays:
ar1[] = {1, 5, 9, 10, 15, 20};
ar2[] = {2, 3, 8, 13};

After First Iteration:
ar1[] = {1, 5, 9, 10, 13, 15};
ar2[] = {2, 3, 8, 20};
// 20 is moved from ar1[] to ar2[]
// 13 from ar2[] is inserted in ar1[]

After Second Iteration:
ar1[] = {1, 5, 8, 9, 10, 13};
ar2[] = {2, 3, 15, 20};
// 15 is moved from ar1[] to ar2[]
// 8 from ar2[] is inserted in ar1[]

After Third Iteration:
ar1[] = {1, 3, 5, 8, 9, 10};
ar2[] = {2, 13, 15, 20};
// 13 is moved from ar1[] to ar2[]
// 3 from ar2[] is inserted in ar1[]

After Fourth Iteration:
ar1[] = {1, 2, 3, 5, 8, 9};
ar2[] = {10, 13, 15, 20};
// 10 is moved from ar1[] to ar2[]
// 2 from ar2[] is inserted in ar1[]
—!>

Efficiently merging two sorted arrays with O(1) extra space

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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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My Personal Notes

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

C++

Java

Python3

C#

PHP

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