Max Prefix

Easy

0/40

Average time to solve is 7m

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Problem statement

You are given an array 'A' of length 'N'. You can perform the operation defined below any number of times (Possibly 0).

Choose 'l', 'r' (1 <= 'l' <= 'r' <='N') and reverse the subarray from 'l' to 'r'.

After that, Return the maximum possible sum of the prefix of this array of length 'K'.

For Example:-

Let 'N' = 5, 'K' = 3, 'A' = [4, 2, 1, 2, 2].
We can reverse the subarray from index 3 to 4 (1-based indexing).
Array becomes [4, 2, 2, 1, 2].
Our answer is 8.

Detailed explanation ( Input/output format, Notes, Images )

Input Format:-

First-line contains an integer 'T', which denotes the number of test cases.
For every test case:-
First-line contains two integers 'N' and 'K'.
Second-line contains 'N' space-separated integers, elements of array 'A'.

Output Format:-

For each test case, Return the maximum possible sum of the prefix of this array of length 'K'.

Note:-

You don’t need to print anything. Just implement the given function.

Constraints:-

1 <= 'T' <= 10
1 <= 'K' <= 'N' <= 10^5
1 <= 'A[i]' <= 10^3

The Sum of 'N' overall test cases does not exceed 10^5.
Time Limit: 1 sec

Sample Input 1:-

Sample Output 1:-

4
5

Explanation of sample input 1:-

First test case:-
We do not need to reverse any subarray.
Our answer is 4.

Second test case:-
We can reverse the subarray from index 1 to 3 (1-based indexing).
Array becomes [3, 2, 2].
Our answer is 5.

Sample Input 2:-

Sample Output 2:-

9
15

Hint 1

How can we maximize the sum of some prefixes of this array?

Approaches (1)

Approach 1

Sorting Approach

Approach:-

If we place greater elements at the start, then we can maximize the sum of prefixes.
We can perform operations in such a way:
- Index 1 to index with the largest element.
- Index 2 to index with the 2nd largest element.
- And so on.
With this, we can sort our array in non-increasing order.
So we just have to find the sum of the first 'K' largest elements.

Algorithm:-

Create 'ans'.
Sort 'A' in non-increasing order.
Iterate using 'i' from 1 to 'K':
- 'ans' += 'A[i]'.
Return 'ans' which is the required answer.

Time Complexity

O(N*log(N)), where 'N' is the length of the array 'A'.

We are sorting 'A' in non-increasing order and the complexity associated with this is O(N*log(N)). Hence, the overall time complexity is of the order O(N*log(N)).

Space Complexity

O(1).

We are using constant extra space, So the Space Complexity is O(1).

(100% EXP penalty)

Max Prefix

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