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

• <code>1 <= m <= n <= 10⁵</code>

• 1 <= arr[i] <= n

• All integers in arr are distinct.