Minimum Workers to Cover All Hours

Given an array arr[] of size n, where each element arr[i] denotes the range of working hours a person at position i can cover.

• A person at index i can work and cover the time interval [i - arr[i], i + arr[i]].
• If arr[i] = -1, the person is unavailable and cannot cover any time.

Find the minimum number of people required to cover the entire interval [0, n – 1]. If it is not possible, return -1.

Examples:

Input: arr[] = [1, 2, 1, 0]
Output: 1
Explanation: The person at index 1 can cover the interval [-1, 3]. After adjusting to valid bounds, this becomes [0, 3], which fully covers the entire working day 0 to n -1. Therefore, only 1 person is required to cover the whole day.

Input: arr[] = [0, 1, 0, -1]
Output: -1
Explanation: Person at index 0 covers [0, 0].
Person at index 1 covers [0, 2].
Person at index 2 covers [2, 2].
Since the last hour cannot be covered by any person, it is impossible to cover the full working day.

[Approach] Using Greedy and Sorting - O(n log(n)) Time and O(n) Space

The idea is to always pick the worker who extends coverage the farthest when the current coverage ends. The problem is to cover the full timeline [0 … n-1] using workers’ ranges. Each worker at index i can cover [i - arr[i], i + arr[i]]. We need the minimum number of workers so that all hours are covered without gaps. This reduces to a minimum interval covering problem.

How it works?

• We need to cover the full timeline [0 … n-1] using the fewest intervals.
• Each worker gives an interval [i - arr[i], i + arr[i]].
• At each step, we track the current coverage (maxi).
• The next chosen interval must start at or before maxi + 1.
• If no such interval exists, coverage is impossible.
• Among valid intervals, pick the one with the farthest right endpoint to maximize future coverage.

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