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Problem Brief

Schedule Batch Difference

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Given n servers with their capacity, we need to schedule k batches.

Input:

  • n (number of servers)
  • Capacity array of length n
  • k (batches)
  • numServers array of length k where numServers[i] = number of servers to be included in i-th batch

    • Output:

    For each batch calculate difference between maximum and minimum server capacity, sum it over for all k batches and return it.

    1Example 1

    Input
    n = 4, capacity = [1, 2, 3, 4], k = 1, numServers = [4]
    Output
    3
    Explanation
    Heads up 👉 This and the following 2 expanations are not from the official 🦔

    There is only one batch, and it includes all servers. The difference between the maximum and minimum server capacity is 4 - 1 = 3.

    2Example 2

    Input
    n = 4, capacity = [1, 2, 3, 4], k = 4, numServers = [1, 1, 1, 1]
    Output
    0
    Explanation

    Each batch includes only one server, so the difference between the maximum and minimum server capacity for each batch is 0. The total sum is 0.

    3Example 3

    Input
    n = 4, capacity = [1, 2, 3, 4], k = 2, numServers = [2, 2]
    Output
    4
    Explanation

    The first batch can include servers with capacities 1 and 4, and the second batch can include servers with capacities 2 and 3. The difference for the first batch is 4 - 1 = 3, and for the second batch is 3 - 2 = 1. The total sum is 3 + 1 = 4.

    Constraints

    Limits and guarantees your solution can rely on.

  • Summation of numServers[i] == n
  • n <= 1e6
  • k <= n
    • public int scheduleBatchDifference(int n, int[] capacity, int k, int[] numServers) {
  // write your code here
}
    Input

    n

    4

    capacity

    [1, 2, 3, 4]

    k

    1

    numServers

    [4]

    Output

    3

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