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Classic problem #1482MediumArrayBinary Search

Minimum Number of Days to Make m Bouquets

Real-world version: Solar String Commissioning
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The statement here is our own: the same inputs, outputs and technique.

A solar field is built as one long row of panels. readyDay[i] is the day panel i is energised; the panel carries nothing before that day and stays energised from that day onwards.

The field is commissioned in strings. A string is k panels that sit next to each other in the row and are all energised, and a panel may be wired into at most one string.

Return the earliest day on which m strings can be picked out of the row at once. If the row can never supply m strings, return -1. Every readyDay[i] is at least 1, so -1 cannot be confused with a real day.

Examples

Example 1

Input
readyDay = [4, 1, 5, 1, 4, 4], m = 2, k = 2
Output
4

On day 4 the only panel still dark is panel 2, so panels 0 and 1 make one string and panels 3 and 4 make the other, leaving panel 5 spare.

Example 2

Input
readyDay = [1, 8, 2, 2, 6], m = 1, k = 2
Output
2

Panels 2 and 3 are both energised on day 2 and sit next to each other, which is the one string the field asks for.

Example 3

Input
readyDay = [2, 3], m = 2, k = 2
Output
-1

Two strings of two panels would take four panels in all and the row holds only two, so no day supplies them.

Constraints

  • readyDay.length == n
  • 1 <= n <= 10^5
  • 1 <= readyDay[i] <= 10^9
  • 1 <= m <= 10^6
  • 1 <= k <= n

The signature

The editor starts with this. It matches the parent problem's shape, so a solution that works here works on a standard judge unchanged.

Python
def earliest_commission_day(readyDay: list[int], m: int, k: int) -> int:
Java
public int earliestCommissionDay(int[] readyDay, int m, int k)

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