FastPrep

Description

Solutions

Find Machine Count

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There are n regions where servers are hosted. The number of machines in the i-th region is given by machineCount[i], where 0 <= i < n. Managing all these regions can be challenging, so the team decided to move some machines to exactly 3 regions. The number of machines in the i-th of these 3 target regions is given by finalMachineCount[i], where 0 <= i < 3.

There are two operations that can be performed:

Add or Remove Machines: Add or remove a machine from any existing region. The number of machines in that server should be non-zero before and after the operation. This operation costs 1 unit.

Move All Machines: Move all machines from one region to another existing region at a cost of shiftingCost units. The now-empty region is destroyed in this operation.

Objective: Find the minimum cost to shift the machines so that any 3 regions have the counts required in finalMachineCount.

Note: It is possible that there are additional machines left at the end apart from the ones placed in the final 3 regions.

Function Description

Complete the function getMinimumCost in the editor.

getMinimumCost has the following parameters:

int machineCount[]: The initial number of machines in the regions.
int finalMachineCount[]: The final number of machines required in the 3 regions.
int shiftingCost: The cost to move all machines from one region to another.

Returns

int: The minimum cost to move machines into 3 regions.

Example 1:

Input:  machineCount = [2, 3, 5, 7], finalMachineCount = [5, 10, 5], shiftingCost = 2
Output: 5 
Explanation:
      
      1. Increase the number of machines in the 1st and 2nd regions by 1 and 2 machines respectively. The new machineCount = [3, 5, 5, 7]. Total cost for these operations is 1 + 2 = 3.
      
      
      2. Move all the machines from the 4th region to the 1st region, which takes shiftingCost = 2. The new machineCount = [10, 5, 5]. The cost of this operation is 2.
      
      
      Total Cost = 3 + 2 = 5

Example 2:

Input:  machineCount = [2, 4, 5, 3], finalMachineCount = [4, 4, 4], shiftingCost = 5
Output: 2 
Explanation:
      
      Increase the number of machines in the 4th region by 1 and decrease the number of machines in the 3rd region by 1. The new machineCount = [2, 4, 4, 4]. Total cost for these operations is 1 + 1 = 2. Use the 2nd, 3rd, and 4th regions as the required servers, leaving behind the 1st region.

Example 3:

Input:  machineCount = [5, 10, 15, 20, 25], finalMachineCount = [20, 20, 20], shiftingCost = 3
Output: 8 
Explanation:
      
      On increasing the number of machines of the 5th region by 5, the new machineCount = [5, 10, 15, 20, 20]. Total cost required for these operations = 5.
      
      
      Now, moving all the machines from the 3rd region to the 1st region, which takes shiftingCost(3) dollars, the new machineCount = [20, 10, 20, 20]. Total cost required for these operations = 3.
      
      
      Use the 1st, 3rd, and 4th regions as required regions, leaving behind the 2nd region. The total cost = 5 + 3 = 8.

Constraints:

3 <= n <= 10^8
1 <= machineCount[i], finalMachineCount[i] <= 10^6
1 <= shiftingCost <= 10^6