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

Given two sorted arrays, we need to merge them in O((n+m)*log(n+m)) time with O(1) extra space into a sorted array, when n is the size of the first array, and m is the size of the second array.

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} 

We have discussed a quadratic time solution in the below post. 

In this post, a better solution is discussed.

The idea: We start comparing elements that are far from each other rather than adjacent. 
For every pass, we calculate the gap and compare the elements towards the right of the gap. Every pass, the gap reduces to the ceiling value of dividing by 2.



Examples: 

First example: 
a1[] = {3 27 38 43}, 
a2[] = {9 10 82}
Start with 
gap =  ceiling of n/2 = 4 
[This gap is for whole merged array]
3 27 38 43   9 10 82 
3 27 38 43   9 10 82
3 10 38 43   9 27 82

gap = 2:
3 10 38 43   9 27 82
3 10 38 43   9 27 82
3 10 38 43   9 27 82 
3 10 9 43   38 27 82
3 10 9 27   38 43 82

gap = 1:
3 10 9 27   38 43 82
3 10 9 27   38 43 82
3 9 10 27   38 43 82
3 9 10 27   38 43 82
3 9 10 27   38 43 82
3 9 10 27   38 43 82
Output : 3 9 10 27 38 43 82

Second Example: 
a1[] = {10 27 38 43 82}, 
a2[] = {3 9}
Start with gap = ceiling of n/2 (4):
10 27 38 43 82   3 9 
10 27 38 43 82   3 9
10 3 38 43 82   27 9
10 3 9 43 82   27 38

gap = 2:
10 3 9 43 82   27 38
9 3 10 43 82   27 38
9 3 10 43 82   27 38
9 3 10 43 82   27 38
9 3 10 27 82   43 38
9 3 10 27 38   43 82

gap = 1
9 3 10 27 38   43 82
3 9 10 27 38   43 82
3 9 10 27 38   43 82
3 9 10 27 38   43 82
3 9 10 27 38   43 82
3 9 10 27 38   43 82


Output : 3 9 10 27 38   43 82 

 Below is the implementation of the above idea:

C++

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// Merging two sorted arrays with O(1)
// extra space
#include <bits/stdc++.h>
using namespace std;
 
// Function to find next gap.
int nextGap(int gap)
{
    if (gap <= 1)
        return 0;
    return (gap / 2) + (gap % 2);
}
 
void merge(int* arr1, int* arr2, int n, int m)
{
    int i, j, gap = n + m;
    for (gap = nextGap(gap); gap > 0; gap = nextGap(gap)) {
        // comparing elements in the first array.
        for (i = 0; i + gap < n; i++)
            if (arr1[i] > arr1[i + gap])
                swap(arr1[i], arr1[i + gap]);
 
        // comparing elements in both arrays.
        for (j = gap > n ? gap - n : 0; i < n && j < m;
             i++, j++)
            if (arr1[i] > arr2[j])
                swap(arr1[i], arr2[j]);
 
        if (j < m) {
            // comparing elements in the second array.
            for (j = 0; j + gap < m; j++)
                if (arr2[j] > arr2[j + gap])
                    swap(arr2[j], arr2[j + gap]);
        }
    }
}
 
// Driver code
int main()
{
    int a1[] = { 10, 27, 38, 43, 82 };
    int a2[] = { 3, 9 };
    int n = sizeof(a1) / sizeof(int);
    int m = sizeof(a2) / sizeof(int);
 
    // Function Call
    merge(a1, a2, n, m);
 
    printf("First Array: ");
    for (int i = 0; i < n; i++)
        printf("%d ", a1[i]);
 
    printf("\nSecond Array: ");
    for (int i = 0; i < m; i++)
        printf("%d ", a2[i]);
    printf("\n");
    return 0;
}

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Java

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// Java program for Merging two sorted arrays
// with O(1) extra space
 
public class MergeTwoSortedArrays {
 
    // Function to find next gap.
    private static int nextGap(int gap)
    {
        if (gap <= 1)
            return 0;
        return (gap / 2) + (gap % 2);
    }
 
    private static void merge(int[] arr1, int[] arr2, int n,
                              int m)
    {
        int i, j, gap = n + m;
        for (gap = nextGap(gap); gap > 0;
             gap = nextGap(gap)) {
            // comparing elements in the first
            // array.
            for (i = 0; i + gap < n; i++)
                if (arr1[i] > arr1[i + gap]) {
                    int temp = arr1[i];
                    arr1[i] = arr1[i + gap];
                    arr1[i + gap] = temp;
                }
 
            // comparing elements in both arrays.
            for (j = gap > n ? gap - n : 0; i < n && j < m;
                 i++, j++)
                if (arr1[i] > arr2[j]) {
                    int temp = arr1[i];
                    arr1[i] = arr2[j];
                    arr2[j] = temp;
                }
 
            if (j < m) {
                // comparing elements in the
                // second array.
                for (j = 0; j + gap < m; j++)
                    if (arr2[j] > arr2[j + gap]) {
                        int temp = arr2[j];
                        arr2[j] = arr2[j + gap];
                        arr2[j + gap] = temp;
                    }
            }
        }
    }
 
    public static void main(String[] args)
    {
        int[] a1 = { 10, 27, 38, 43, 82 };
        int[] a2 = { 3, 9 };
 
        // Function Call
        merge(a1, a2, a1.length, a2.length);
 
        System.out.print("First Array: ");
        for (int i = 0; i < a1.length; i++) {
            System.out.print(a1[i] + " ");
        }
 
        System.out.println();
 
        System.out.print("Second Array: ");
        for (int i = 0; i < a2.length; i++) {
            System.out.print(a2[i] + " ");
        }
    }
}
 
// This code is contributed by Vinisha Shah

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Python 3

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# Merging two sorted arrays with O(1)
# extra space
 
# Function to find next gap.
 
 
def nextGap(gap):
 
    if (gap <= 1):
        return 0
    return (gap // 2) + (gap % 2)
 
 
def merge(arr1, arr2, n, m):
 
    gap = n + m
    gap = nextGap(gap)
    while gap > 0:
 
        # comparing elements in
        # the first array.
        i = 0
        while i + gap < n:
            if (arr1[i] > arr1[i + gap]):
                arr1[i], arr1[i + gap] = arr1[i + gap], arr1[i]
 
            i += 1
 
        # comparing elements in both arrays.
        j = gap - n if gap > n else 0
        while i < n and j < m:
            if (arr1[i] > arr2[j]):
                arr1[i], arr2[j] = arr2[j], arr1[i]
 
            i += 1
            j += 1
 
        if (j < m):
 
            # comparing elements in the
            # second array.
            j = 0
            while j + gap < m:
                if (arr2[j] > arr2[j + gap]):
                    arr2[j], arr2[j + gap] = arr2[j + gap], arr2[j]
 
                j += 1
 
        gap = nextGap(gap)
 
 
# Driver code
if __name__ == "__main__":
 
    a1 = [10, 27, 38, 43, 82]
    a2 = [3, 9]
    n = len(a1)
    m = len(a2)
 
    # Function Call
    merge(a1, a2, n, m)
 
    print("First Array: ", end="")
    for i in range(n):
        print(a1[i], end=" ")
    print()
 
    print("Second Array: ", end="")
    for i in range(m):
        print(a2[i], end=" ")
    print()
 
# This code is contributed
# by ChitraNayal

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C#

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// C# program for Merging two sorted arrays
// with O(1) extra space
using System;
 
class GFG {
 
    // Function to find next gap.
    static int nextGap(int gap)
    {
        if (gap <= 1)
            return 0;
        return (gap / 2) + (gap % 2);
    }
 
    private static void merge(int[] arr1, int[] arr2, int n,
                              int m)
    {
        int i, j, gap = n + m;
        for (gap = nextGap(gap); gap > 0;
             gap = nextGap(gap)) {
            // comparing elements in the first
            // array.
            for (i = 0; i + gap < n; i++)
                if (arr1[i] > arr1[i + gap]) {
                    int temp = arr1[i];
                    arr1[i] = arr1[i + gap];
                    arr1[i + gap] = temp;
                }
 
            // comparing elements in both arrays.
            for (j = gap > n ? gap - n : 0; i < n && j < m;
                 i++, j++)
                if (arr1[i] > arr2[j]) {
                    int temp = arr1[i];
                    arr1[i] = arr2[j];
                    arr2[j] = temp;
                }
 
            if (j < m) {
                // comparing elements in the
                // second array.
                for (j = 0; j + gap < m; j++)
                    if (arr2[j] > arr2[j + gap]) {
                        int temp = arr2[j];
                        arr2[j] = arr2[j + gap];
                        arr2[j + gap] = temp;
                    }
            }
        }
    }
 
    // Driver code
    public static void Main()
    {
        int[] a1 = { 10, 27, 38, 43, 82 };
        int[] a2 = { 3, 9 };
 
        // Function Call
        merge(a1, a2, a1.Length, a2.Length);
 
        Console.Write("First Array: ");
        for (int i = 0; i < a1.Length; i++) {
            Console.Write(a1[i] + " ");
        }
 
        Console.WriteLine();
 
        Console.Write("Second Array: ");
        for (int i = 0; i < a2.Length; i++) {
            Console.Write(a2[i] + " ");
        }
    }
}
 
// This code is contributed by Sam007.

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PHP

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<?php
// Merging two sorted arrays
// with O(1) extra space
 
// Function to find next gap.
function nextGap($gap)
{
    if ($gap <= 1)
        return 0;
    return ($gap / 2) +
           ($gap % 2);
}
 
function merge($arr1, $arr2,
               $n, $m)
{
    $i;
    $j;
    $gap = $n + $m;
    for ($gap = nextGap($gap);
         $gap > 0; $gap = nextGap($gap))
    {
        // comparing elements
        // in the first array.
        for ($i = 0; $i + $gap < $n; $i++)
            if ($arr1[$i] > $arr1[$i + $gap])
            {
                $tmp = $arr1[$i];
                $arr1[$i] = $arr1[$i + $gap];
                $arr1[$i + $gap] = $tmp;
            }
 
        // comparing elements
        // in both arrays.
        for ($j = $gap > $n ? $gap - $n : 0 ;
             $i < $n && $j < $m; $i++, $j++)
            if ($arr1[$i] > $arr2[$j])
            {
                $tmp = $arr1[$i];
                $arr1[$i] = $arr2[$j];
                $arr2[$j] = $tmp;
            }
 
        if ($j < $m)
        {
            // comparing elements in
            // the second array.
            for ($j = 0; $j + $gap < $m; $j++)
                if ($arr2[$j] > $arr2[$j + $gap])
                {
                    $tmp = $arr2[$j];
                    $arr2[$j] = $arr2[$j + $gap];
                    $arr2[$j + $gap] = $tmp;
                }
        }
    }
     
    echo "First Array: ";
    for ($i = 0; $i < $n; $i++)
        echo $arr1[$i]." ";
 
    echo "\nSecond Array: ";
    for ($i = 0; $i < $m; $i++)
        echo $arr2[$i] . " ";
    echo"\n";
 
}
 
// Driver code
$a1 = array(10, 27, 38, 43 ,82);
$a2 = array(3,9);
$n = sizeof($a1);
$m = sizeof($a2);
 
// Function Call
merge($a1, $a2, $n, $m);
 
// This code is contributed
// by mits.
?>

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Output

First Array: 3 9 10 27 38 
Second Array: 43 82 

Another method in O(m+n) time complexity:

Here we use the below technique:

Suppose we have a number A and we want to convert it to a number B and there is also a 
constrain that we can recover number A any time without using other variable.To achieve 
this we chose a number N which is greater than both numbers and add B*N int A.
so A --> A+B*N

To get number B out of (A+B*N) we devide (A+B*N) by N (A+B*N)/N = B.

To get number A out of (A+B*N) we take modulo with N (A+B*N)%N = A.

-> In short by taking modulo we get old number back and taking divide we new number.

We first find the maximum element of both array and increment it one-two avoid collision of 0 and maximum element during modulo operation. The idea is to we traverse both arrays from starting simultaneously. Let’s say an element in a is a[i] and in b is b[j] and k is the position at where the next minimum number will come. Now update value a[k] if k<n else b[k-n] by adding min(a[i],b[j])*maximum_element. After updating all the element divide all the elements by maximum_element so we get updated array back. 

 Below is the implementation of the above idea:

C++

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#include <bits/stdc++.h>
using namespace std;
 
void mergeArray(int a[], int b[], int n, int m)
{
    int mx = 0;
     
    // Find maximum element of both array
    for (int i = 0; i < n; i++) {
        mx = max(mx, a[i]);
    }
    for (int i = 0; i < m; i++) {
        mx = max(mx, b[i]);
    }
     
    // increment one two avoid collision of 0 and maximum
    // element of array in modulo operation
    mx++;
    int i = 0, j = 0, k = 0;
    while (i < n && j < m && k < (n + m)) {
         
        // recover back original element to compare
        int e1 = a[i] % mx;
        int e2 = b[j] % mx;
        if (e1 <= e2) {
             
            // update element by adding multiplication
            // with new number
            if (k < n)
                a[k] += (e1 * mx);
            else
                b[k - n] += (e1 * mx);
            i++;
            k++;
        }
        else {
             
            // update element by adding multiplication
            // with new number
            if (k < n)
                a[k] += (e2 * mx);
            else
                b[k - n] += (e2 * mx);
            j++;
            k++;
        }
    }
     
    // process those elements which are left in array a
    while (i < n) {
        int el = a[i] % mx;
        if (k < n)
            a[k] += (el * mx);
        else
            b[k - n] += (el * mx);
        i++;
        k++;
    }
     
    // process those elements which are left in array a
    while (j < m) {
        int el = b[j] % mx;
        if (k < n)
            b[k] += (el * mx);
        else
            b[k - n] += (el * mx);
        j++;
        k++;
    }
     
    // finally update elements by dividing
    // with maximum element
    for (int i = 0; i < n; i++)
        a[i] = a[i] / mx;
 
    // finally update elements by dividing
    // with maximum element
    for (int i = 0; i < m; i++)
        b[i] = b[i] / mx;
 
    return;
}
 
// Driver Code
int main()
{
    int a[] = { 3, 5, 6, 8, 12 };
    int b[] = { 1, 4, 9, 13 };
    int n = sizeof(a) / sizeof(int); // Length of a
    int m = sizeof(b) / sizeof(int); // length of b
 
    // Function Call
    mergeArray(a, b, n, m);
 
    cout << "First array : ";
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
    cout << endl;
 
    cout << "Second array : ";
    for (int i = 0; i < m; i++)
        cout << b[i] << " ";
    cout << endl;
 
    return 0;
}

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Output

First array : 1 3 4 5 6 
Second array : 8 9 12 13 

Time Complexity: O(m+n)

This article is contributed by Shlomi Elhaiani. 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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