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 given an integer m and you need to find the latest step at which there exists a group of ones of length m. A group of ones is a contiguous substring of 1s 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.

Example 3:

Input: arr = [1], m = 1
Output: 1

Example 4:

Input: arr = [2,1], m = 2
Output: 2

Constraints:

• n == arr.length

• 1 <= n <= 10^5

• 1 <= arr[i] <= n

• All integers in arr are distinct.

• 1 <= m <= arr.length

class Solution {
    public int find(int[] parent, int node) {
        if (node != parent[node])
            parent[node] = find(parent, parent[node]);
        return parent[node];
    }

    public int findLatestStep(int[] arr, int m) {
        int n = arr.length;
        // Parent array for DSU
        int[] parent = new int[n];
        // Map of parent -> Size of group of that parent
        Map<Integer, Integer> map = new HashMap<>();
        char[] str = new char[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            str[i] = '0';
        }
        int ans = -1, countOfMGroups = 0;
        for (int i = 0; i < n; i++) {
            // Current parent
            int p = arr[i] - 1;
            // Initialising group size of this parent
            map.put(p, 1);
            str[p] = '1';
            // Joining process
            if (arr[i] - 2 >= 0 && str[arr[i] - 2] == '1') {
                int p1 = find(parent, arr[i] - 2);
                // This group size will be increased
                if (map.get(p1) == m)
                    countOfMGroups--;
                map.put(p1, map.get(p1) + 1);
                map.remove(p);
                parent[p] = p1;
                p = p1;
            }
            if (arr[i] < n && str[arr[i]] == '1') {
                int p2 = find(parent, arr[i]);
                // This group will be merged with others to form a bigger group
                if (map.get(p2) == m)
                    countOfMGroups--;
                map.put(p, map.get(p) + map.get(p2));
                map.remove(p2);
                parent[p2] = p;
            }
            if (map.get(p) == m)
                countOfMGroups++;
            if (countOfMGroups > 0)
                ans = i + 1;
        }
        return ans;
    }
}