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You are given the count of total devices ( n ) which are initially turned off, and an array of operations where each operation represents a range ([start, end]) (both inclusive). Each operation will test all devices in the specified range. You need to determine how many unique devices got tested after performing all given operations.

Clarifying Questions

  1. Range of Inputs:
    • What are the constraints for ( n ) (number of devices)?
    • What is the range for the individual operations?
    • Are these operations guaranteed to be within bounds, i.e., ( 1 \leq start \leq end \leq n )?
  2. Output:
    • Do we return the count of unique tested devices?
  3. Data Types:
    • How large can the input list of operations be?

Assuming typical constraints like ( n \leq 10^9 ) and ( len(operations) \leq 10^5 ), we focus on ensuring an efficient solution.

Strategy

To determine how many unique devices were tested, we can use a set to track all the device indices that were tested.

  1. Initialization:
    • Create an empty set to store indices of devices that were tested.
  2. Processing Operations:
    • For each operation with range ([start, end]), add all numbers in the range to the set.
  3. Result:
    • The size of the set gives the count of unique devices that were tested.

This strategy ensures uniqueness inherently because sets do not allow duplicate values.

Code

def countTestedDevices(n, operations):
    tested_devices = set()
    for start, end in operations:
        for device in range(start, end + 1):
            tested_devices.add(device)
    return len(tested_devices)

# Example Usage
n = 10
operations = [(1, 5), (2, 6), (8, 10)]
print(countTestedDevices(n, operations))  # Output should be 8

Time Complexity

This naive approach ensures correctness, but it’s important to note the potential inefficiency in case of large ( n ).

Optimized Strategy

Considering the potential inefficiency, we can optimize using a more advanced approach with a difference array or line sweep technique to avoid explicitly iterating over large ranges. An efficient approach involves:

  1. Marking Start and End Points:
    • Uses a marking strategy where additions and subtractions track the life span of the tested devices.

Here’s the optimized code for efficiency using a marking strategy:

def countTestedDevices(n, operations):
    if not operations:
        return 0

    tested_markers = [0] * (n + 2)

    for start, end in operations:
        tested_markers[start] += 1
        if end + 1 <= n:
            tested_markers[end + 1] -= 1

    tested_count = 0
    current_active_tests = 0

    for i in range(1, n + 1):
        current_active_tests += tested_markers[i]
        if current_active_tests > 0:
            tested_count += 1

    return tested_count

# Example Usage
n = 10
operations = [(1, 5), (2, 6), (8, 10)]
print(countTestedDevices(n, operations))  # Output should be 8

Optimized Time Complexity

This is much more efficient for large values of ( n ).

This improved approach balances both clarity and performance, making it suitable for a wide range of input sizes.

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