Question

Suppose that you have two sorted numerical arrays A[1 . . . m] and B[1 . . . n]. You want to compute the k’th smallest number in the merged array of all m + n elements. Please design a divide- and-conquer algorithm that can do so in O(log(m + n)). You can assume for simplicity that k is even.

Answer 1

`Hey,

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By using a divide and conquer approach, similar to the one used in binary search, we can attempt to find the k’th element in a more efficient way.

Explanation:
We compare the middle elements of arrays arr1 and arr2,
let us call these indices mid1 and mid2 respectively.

Let us assume arr1[mid1]  k, then clearly the elements after 
mid2 cannot be the required element. We then set the last
element of arr2 to be arr2[mid2].

In this way, we define a new subproblem with half the size
of one of the arrays.

Below is the implementation of function in cpp

int kth(int *arr1, int *arr2, int *end1, int *end2, int k)

{

    if (arr1 == end1)

        return arr2[k];

    if (arr2 == end2)

        return arr1[k];

    int mid1 = (end1 - arr1) / 2;

    int mid2 = (end2 - arr2) / 2;

    if (mid1 + mid2 < k)

    {

        if (arr1[mid1] > arr2[mid2])

            return kth(arr1, arr2 + mid2 + 1, end1, end2,

                k - mid2 - 1);

        else

            return kth(arr1 + mid1 + 1, arr2, end1, end2,

                k - mid1 - 1);

    }

    else

    {

        if (arr1[mid1] > arr2[mid2])

            return kth(arr1, arr2, arr1 + mid1, end2, k);

        else

            return kth(arr1, arr2, end1, arr2 + mid2, k);

    }

}

Note that in the above code, k is 0 indexed, which means if we want a k that’s 1 indexed, we have to subtract 1 when passing it to the function.
Time Complexity: O(log (n + m))

Kindly revert for any queries

Thanks.