
Painters in a city are feeling bored, so they have decided to paint all boards which are visible to them. Now, they want to finish this task as soon as possible but they are unable to figure out the minimum time required to complete it. Can you help them out?
You are given ‘N’ boards that are to be painted by ‘A’ painters. Each painter takes ‘B’ units of time to paint one unit of the board. So, if the length of a board is 'X' units, then it will take a painter X * B units of time, to paint the whole board.
Your task is to return the minimum time required to paint all the boards subject to the following conditions-
1. Any painter will only paint contiguous sections of the board.
For eg, A configuration where painter 1 paints boards 1 and 3 and not 2 is invalid.
2. A board cannot be painted partially by one painter, and partially by another.
The first line of input contains an integer ‘T’ denoting the number of test cases to run. Then the test case follows.
The first line of each test case contains three single space-separated integers ‘N’, ‘A’ and ‘B’ denoting the number of boards to be painted, number of painters available and time taken to paint 1 unit of the board, respectively.
The second line of each test case contains N single space-separated integers, denoting the size of each board.
Output Format :
For each test case, print the minimum time required to paint all boards.
Output for each test case will be printed in a new line.
Note:
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 <= T <= 100
1 <= A <= 10^5
1 <= B <= 10^9
1 <= N <= 10^3
1 <= Board[i] <=10^5
Time Limit: 1sec
1
2 2 5
1 10
50
For the first test case, there are two possibilities -
If painter 1 paints both the boards, the total time required will be 55.
If painter 1 paints boards 1 and painter 2 paints board 2, the total time will be max(5, 50) = 50.
So the minimum time required will be 50.
1
4 10 1
1 8 11 3
11
Can we iterate over time to get our answer?
We can iterate through the time ‘t’ and check if all the boards can be painted within time ‘t’ using at most A painters. The smallest time t at which we will be able to complete our work will be our answer.
The algorithm will be-
O(N*sumBoards), where N denotes the total number of boards and sumBoards denotes the sum of the length of all the boards.
The time complexity due to the linear search will be O(sumBoards). For each check, we are iterating over the whole array. Thus, the total time complexity will be O(N*sumBoards).
O(1), constant space is used.