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1562. Find Latest Group of Size M

Given an array arr that represents a permutation of numbers from 1 to n.

You have a binary string of size n that initially has all its bits set to zero. At each step i (assuming both the binary string and arr are 1-indexed) from 1 to n, the bit at position arr[i] is set to 1.

You are also given an integer m. Find the latest step at which there exists a group of ones of length m. A group of ones is a contiguous substring of 1's such that it cannot be extended in either direction.

Return the latest step at which there exists a group of ones of length exactlym. If no such group exists, return-1.

Example 1:

Input: arr = [3,5,1,2,4], m = 1 Output: 4 Explanation: Step 1: "00100", groups: ["1"] Step 2: "00101", groups: ["1", "1"] Step 3: "10101", groups: ["1", "1", "1"] Step 4: "11101", groups: ["111", "1"] Step 5: "11111", groups: ["11111"] The latest step at which there exists a group of size 1 is step 4.

Example 2:

Input: arr = [3,1,5,4,2], m = 2 Output: -1 Explanation: Step 1: "00100", groups: ["1"] Step 2: "10100", groups: ["1", "1"] Step 3: "10101", groups: ["1", "1", "1"] Step 4: "10111", groups: ["1", "111"] Step 5: "11111", groups: ["11111"] No group of size 2 exists during any step.

Constraints:

  • n == arr.length
  • 1 <= m <= n <= 105
  • 1 <= arr[i] <= n
  • All integers in arr are distinct.

Solutions (Python)

1. Solution

fromsortedcontainersimportSortedListclassSolution: deffindLatestStep(self, arr: List[int], m: int) ->int: iflen(arr) ==m: returnmn=len(arr) groups=SortedList([(1, n+1)]) forstepinrange(n-1, -1, -1): i=groups.bisect_left((arr[step], n+2)) -1left= (groups[i][0], arr[step]) right= (arr[step] +1, groups[i][1]) groups.pop(i) ifleft[1] -left[0] ==morright[1] -right[0] ==m: returnstepifleft[0] <left[1]: groups.add(left) ifright[0] <right[1]: groups.add(right) return-1
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