human
I found an interesting problem on https://leetcode.com/problems/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

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is Python code to find the latest step at which there exists a group of ones of length m.

#https://leetcode.com/problems/find-latest-group-of-size-m/discuss/836441/Very-Easy

class Solution:
    def findLatestStep(self, arr: List[int], m: int) -> int:
        n = len(arr)
        if n==m:
            return n
        
        memo = collections.defaultdict(lambda : 0)
        res = -1
        for idx, i in enumerate(arr):
            left = memo[i-1]
            right = memo[i+1]
            if left==m or right==m:
                res = idx
            memo[i-left] = left+right+1
            memo[i+right] = left+right+1
        # if m in memo.values():
        #     res = n
        return res