# Solving Leetcode Interviews in Seconds with AI: Find Latest Group of Size M


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1562" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.  

	# Explanation
	Here's the breakdown of the approach, complexities, and the Python code:

*   **High-Level Approach:**

    *   Iterate through the `arr` array, simulating the setting of bits to 1 at each step.
    *   Use a data structure (like an array) to track the length of contiguous groups of ones. Crucially, use a *union-find* like approach to maintain connected components of 1s, and keep track of the sizes of these components.
    *   At each step, check if a group of length `m` exists. Update the latest step if such a group is found.

*   **Complexity:**
    *   Runtime Complexity: O(n)
    *   Storage Complexity: O(n)

*   **Python Code:**

	
	# Code
	```python
	def findLatestStep(arr, m):
    n = len(arr)
    if m == n:
        return n

    length = [0] * (n + 2)
    result = -1
    count = 0

    for i, pos in enumerate(arr):
        left = length[pos - 1]
        right = length[pos + 1]
        length[pos - left] = length[pos + right] = left + right + 1

        if left == m or right == m:
            count += 1

        if left + right + 1 == m:
            count -= 1

        if count == 0 and left + right + 1 > m:
            result = i + 1
        elif length[pos-left] == m:
            result = i + 1

    return result
	```