sol.md

Algorithm:

1. Create a new array merged of size m.length + n.length to store the merged sorted array.
2. Initialize three pointers i, j, and k. i and j will point to the first elements of the arrays m and n, respectively. k will point to the first element of the array merged.
3. While i is less than the length of m and j is less than the length of n:
   • If the element at i in m is less than or equal to the element at j in n:
     • Copy the element at i in m to the element at k in merged.
     • Increment i and k.
   • Otherwise:
     • Copy the element at j in n to the element at k in merged.
     • Increment j and k.
4. While i is less than the length of m:
   • Copy the element at i in m to the element at k in merged.
   • Increment i and k.
5. While j is less than the length of n:
   • Copy the element at j in n to the element at k in merged.
   • Increment j and k.
6. Calculate the middle index of the array merged.
7. If the length of the array merged is even:
   • Return the average of the two elements at the middle index and the middle index minus one.
8. Otherwise:
   • Return the element at the middle index.

Time complexity: O(m + n), where m and n are the lengths of the arrays m and n, respectively. This is because we only need to iterate over the two arrays once.

Space complexity: O(m + n), where m and n are the lengths of the arrays m and n, respectively. This is because we need to store the merged array merged.

for example, consider the following two arrays:

m = [1, 3, 5, 7] n = [2, 4, 6, 8]

The merged array will be:

merged = [1, 2, 3, 4, 5, 6, 7, 8]

The middle index of the merged array is 3. Therefore, the function will return the average of the two elements at index 3 and index 2, which is (3 + 4) / 2 = 3.5.