Package g1501_1600.s1562_find_latest_group_of_size_m

Class Solution

  • public class Solution
extends Object
    1562 - Find Latest Group of Size M.

    Medium

    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 exactly m. 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.

    • Constructor Detail

      • Solution

        public Solution()

    • Method Detail

      • findLatestStep

        public int findLatestStep​(int[] arr,
                          int m)