K-th element of two sorted Arrays.cpp

/*
K-th element of two sorted Arrays
=================================

Given two sorted arrays arr1 and arr2 of size M and N respectively and an element K. The task is to find the element that would be at the k’th position of the final sorted array.

Example 1:
Input:
arr1[] = {2, 3, 6, 7, 9}
arr2[] = {1, 4, 8, 10}
k = 5
Output:
6
Explanation:
The final sorted array would be -
1, 2, 3, 4, 6, 7, 8, 9, 10
The 5th element of this array is 6.

Example 2:
Input:
arr1[] = {100, 112, 256, 349, 770}
arr2[] = {72, 86, 113, 119, 265, 445, 892}
k = 7
Output:
256
Explanation:
Final sorted array is - 72, 86, 100, 112,
113, 119, 256, 265, 349, 445, 770, 892
7th element of this array is 256.

Your Task:  
You don't need to read input or print anything. Your task is to complete the function kthElement() which takes the arrays arr1[], arr2[], its size N and M respectively and an integer K as inputs and returns the element at the Kth position.

Expected Time Complexity: O(Log(N) + Log(M))
Expected Auxiliary Space: O(Log (N))

Constraints:
1 <= N, M <= 106
1 <= arr1i, arr2i <= 106
1 <= K <= N+M
*/

int kthElement(int arr1[], int arr2[], int n, int m, int k)
{
  int i = 0, j = 0, c = 0;
  int ans = 0;

  while (i < n && j < m && c < k)
  {
    if (arr1[i] < arr2[j])
    {
      ans = arr1[i];
      i++;
      c++;
    }
    else
    {
      ans = arr2[j];
      j++;
      c++;
    }
  }

  while (i < n && c < k)
  {
    ans = arr1[i];
    i++;
    c++;
  }

  while (j < m && c < k)
  {
    ans = arr2[j];
    j++;
    c++;
  }

  return ans;
}