题目如下：
Given an array
arrthat represents a permutation of numbers from1ton. You have a binary string of sizenthat initially has all its bits set to zero.At each step
i(assuming both the binary string andarrare 1-indexed) from1ton, the bit at positionarr[i]is set to1. You are given an integermand you need to find the latest step at which there exists a group of ones of lengthm. 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: 1Example 4:
Input: arr = [2,1], m = 2 Output: 2Constraints:
n == arr.length1 <= n <= 10^51 <= arr[i] <= n- All integers in
arrare distinct.1 <= m <= arr.length
解题思路：我的思路是将把每个group的起始索引和结束索引记录到一个数组sub_range里面，sub_range的初始值为[[1,max(arr)]]，然后倒序处理arr，根据arr[i]的值对sub_range进行拆分，每做一次拆分，计算更新后的sub_range是否有某个元素的range的长度正好为m。
代码如下：
class Solution(object): def findLatestStep(self, arr, m): """ :type arr: List[int] :type m: int :rtype: int """ sub_range = [[1,max(arr)]] def binarySearch(a, l, r, x): if r >= l: mid = int(l + (r - l) / 2) if a[mid][0] <= x and a[mid][1] >= x: return mid elif a[mid][0] > x: return binarySearch(a, l, mid - 1, x) elif a[mid][1] < x: return binarySearch(a, mid + 1, r, x) else: return -1 if len(arr) == m:return m for i in range(len(arr)-1,-1,-1): mid = binarySearch(sub_range,0,len(sub_range)-1,arr[i]) if mid == -1:continue if sub_range[mid][1] == sub_range[mid][0]: del sub_range[mid] continue end = sub_range[mid][1] sub_range[mid][1] = arr[i] - 1 if end - (arr[i]+1) >= m: sub_range.insert(mid+1,[arr[i]+1,end]) if sub_range[mid][1] - sub_range[mid][0] + 1 == m or end - (arr[i]+1) + 1 == m: return i if sub_range[mid][1] - sub_range[mid][0] < m: del sub_range[mid] continue return -1