Solve Time-Window and Binary Swap Problems

This online assessment contained two coding problems:

1. Maximum requests in a time window You are given a non-decreasing integer array requestTimes, where requestTimes[i] is the second at which the i-th request arrives, and an integer windowSize. Return the maximum number of requests that occur within any inclusive time interval of length windowSize seconds. In other words, find the largest number of elements that can fit inside some interval [t, t + windowSize].

2. Simultaneous swaps in a binary string You are given a binary string s consisting only of '0' and '1'. Every second, all adjacent occurrences of '01' are swapped simultaneously to become '10'. Repeat this process until the string no longer contains '01'. Return the number of seconds required.

Example: s = "001011" transforms as follows: 001011 -> 010101 -> 101010 -> 110100 -> 111000 So the answer is 4 seconds.

Part 1: Maximum Requests in a Time Window

You are given a non-decreasing integer array requestTimes, where requestTimes[i] is the second at which the i-th request arrives, and a non-negative integer windowSize.

Return the maximum number of requests that occur within any inclusive time interval of length windowSize seconds. In other words, find the largest number of values that can fit inside some interval [t, t + windowSize].

Because the interval is inclusive, requests at both endpoints count.

Constraints

• 0 <= len(requestTimes) <= 200000
• 0 <= requestTimes[i] <= 10^9
• requestTimes is sorted in non-decreasing order
• 0 <= windowSize <= 10^9

Examples

Example 1

Input: ([1, 1, 2, 3, 7, 8], 2)
Output: 4

Explanation: The interval [1, 3] contains 1, 1, 2, 3, so the answer is 4.

Example 2

Input: ([], 5)
Output: 0

Explanation: There are no requests, so the maximum count is 0.

Hints

1. Since the array is already sorted, try maintaining a valid range with two pointers.
2. For a fixed right endpoint, move the left pointer forward until requestTimes[right] - requestTimes[left] <= windowSize.

Part 2: Seconds to Eliminate All '01' Pairs

You are given a binary string s containing only '0' and '1'.

Every second, all adjacent occurrences of '01' are swapped simultaneously, turning each such pair into '10'. This process repeats until the string no longer contains '01'.

Return the number of seconds required.

Example: 001011 -> 010101 -> 101010 -> 110100 -> 111000 So the answer is 4.

Constraints

• 0 <= len(s) <= 200000
• s[i] is either '0' or '1'

Hints

1. A direct simulation works, but can be too slow for long strings.
2. Track how many zeros have appeared before each '1'. A '1' may need to wait not only for zeros, but also for earlier '1' characters moving through the same zeros.