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 below post.
Merge two sorted arrays with O(1) extra space

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 27 9 10   38 43 82
        3 27 9 10   38 43 82
        gap = 1:
        3 27 9 10   38 43 82
        3 27 9 10   38 43 82
        3 9 27 10   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
// 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);

    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;
}

Output:

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

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.

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