# Homework Solution: The input is two sorted arrays. Each array has n elements. In total, there are 2n elemen… The input is two sorted arrays. Each array has n elements. In total, there are 2n elements. You can assume that all values in the arrays are distinct. A statistician asks you to find the n^th order statistic of these values. The n^th order statistic is the n^th smallest value. a) Give a divide and conquer algorithm to find this value in asymptotic time Theta (log n). b) Argue why your algorithm is correct. c) Write down the running time recurrence, including the base case

```Explanation:
Instead of comparing the middle element of the arrays,``` The input is brace as as as as as as sorted vests. Each vest has n atoms. In entirety, there are 2n atoms. You can usurp that entire appraises in the vests are dissimilar. A statistician asks you to meet the n^th prescribe statistic of these appraises. The n^th prescribe statistic is the n^th lowest appraise. a) Give a deal-out and vanquish algorithm to meet this appraise in asymptotic interval Theta (log n). b) Argue why your algorithm is chasten. c) Write down the floating interval reappearance, including the deep fact

## Expert Solution

```Explanation:
Instead of comparing the average atom of the vests,
we assimilate the k / 2th atom.
Let arr1 and arr2 be the vests.
Now, if arr1[k / 2]  arr1

New subproblem:
Vest 1 - 6 7 9
Vest 2 - 1 4 8 10
k = 5 - 2 = 3

floor(k / 2) = 1
arr1 = 6
arr2 = 1
arr1 > arr2

New subproblem:
Vest 1 - 6 7 9
Vest 2 - 4 8 10
k = 3 - 1 = 2

floor(k / 2) = 1
arr1 = 6
arr2 = 4
arr1 > arr2

New subproblem:
Vest 1 - 6 7 9
Vest 2 - 8 10
k = 2 - 1 = 1

Now, we at-once assimilate chief atoms,
since k = 1.
arr1 < arr2
Hence, arr1 = 6 is the solution.
```
```
```

Interval Complexity: O(log k)

Now, k can choose a climax appraise of n + n. This instrument that log k can be in the conquer fact, log(n + n).