IN C++ USING THE CODE PROVIDED. PLEASE ALSO INCLUDE AN EXPLANATION/PSEUDO CODE

------------------------------------

Given two 0-indexed integer arrays nums1 and nums2 of equal length n and a positive integer k. You must choose a subsequence of indices from nums1 of length k.

For chosen indices i0, i1, , ik - 1, your score is defined as:

The sum of the selected elements from nums1 multiplied with the minimum of the selected elements from nums2.

It can defined simply as: (nums1[i0] + nums1[i1] ++ nums1[ik - 1]) * min(nums2[i0] , nums2[i1], ,nums2[ik - 1]).

Return the maximum possible score.

A subsequence of indices of an array is a set that can be derived from the set {0, 1, , n-1} by deleting some or no elements.

Example 1

Input: nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3

Output: 12

Explanation:

The four possible subsequence scores are:

We choose the indices 0, 1, and 2 with score = (1+3+3) * min(2,1,3) = 7.

We choose the indices 0, 1, and 3 with score = (1+3+2) * min(2,1,4) = 6.

We choose the indices 0, 2, and 3 with score = (1+3+2) * min(2,3,4) = 12.

We choose the indices 1, 2, and 3 with score = (3+3+2) * min(1,3,4) = 8.

Therefore, we return the max score, which is 12.

Example 2

Input: nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1

Output: 30

Explanation:

Choosing index 2 is optimal: nums1[2] * nums2[2] = 3 * 10 = 30 is the maximum possible score.

Constraints

n == nums1.length == nums2.length

1 <= n <= 10^5

0 <= nums1[i], nums2[j] <= 10^5

1 <= k <= n

Sample input

Enter the size of both arrays

Enter elements of the first array

Enter elements of the second array

Enter length k

4 1 3 3 2 2 1 3 4 3 

Sample output

12

------------------------------------

#include #include #include #include

using namespace std;

long long maxScore(vector& nums1, vector& nums2, int k) { //your code here }

int main(){ int sizeOfArrays; int element; vectorarray1; vectorarray2; cin>>sizeOfArrays; int i=0; int k; while(i>element; array1.push_back(element); i++; } int j= 0; while(j>element; array2.push_back(element); j++; } cin>>k; cout<

return 0; }