How do you solve Count Tested Devices After Test Operations on LeetCode?

Count Tested Devices After Test Operations (LeetCode #2960) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Devices can only be tested if their battery percentage is greater than the number of devices tested so far.

What is the time complexity of Count Tested Devices After Test Operations?

The optimal solution for Count Tested Devices After Test Operations runs in O(n) time and uses O(1) space. The time complexity is O(n) since we only loop through the array once, making it much more efficient than the brute force approach.

What is the brute force solution for Count Tested Devices After Test Operations?

The brute force approach for Count Tested Devices After Test Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates the testing process for each device sequentially. For each device that has a positive battery percentage, we count it as tested and reduce the battery percentage of all subsequent devices by 1, ensuring they don't go below zero. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Tested Devices After Test Operations?

Count Tested Devices After Test Operations uses the following concepts: Array, Simulation, Counting. The recommended patterns to study are: Simulation, Array Manipulation.

What are the interview tips for Count Tested Devices After Test Operations?

Always think about edge cases, like all devices having 0 battery. Explain your thought process clearly while coding to demonstrate understanding. Practice simulating the operations on paper before coding to visualize the changes.

What are common mistakes when solving Count Tested Devices After Test Operations?

Forgetting to ensure battery percentages do not go below zero. Not considering how previously tested devices affect the current device's ability to be tested.

#2960

Count Tested Devices After Test Operations

Easy

ArraySimulationCountingSimulationArray Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the observation that we can keep track of the number of previously tested devices. A device can be tested if its battery percentage is greater than the number of devices tested so far, allowing us to avoid unnecessary iterations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a count of tested devices to 0.
2Step 2: Loop through each device in the batteryPercentages array.
3Step 3: If the current device's battery percentage is greater than the count of tested devices, increment the count.
4Step 4: The battery percentage of the current device will affect the subsequent devices, so we adjust the count accordingly.

solution.py12 lines

1# Full working Python code
2
3def countTestedDevices(batteryPercentages):
4    tested_count = 0
5    n = len(batteryPercentages)
6    for i in range(n):
7        if batteryPercentages[i] > tested_count:
8            tested_count += 1
9    return tested_count
10
11# Example usage
12print(countTestedDevices([1, 1, 2, 1, 3]))

ℹ

Complexity note: The time complexity is O(n) since we only loop through the array once, making it much more efficient than the brute force approach.

1Devices can only be tested if their battery percentage is greater than the number of devices tested so far.
2The brute force method can be inefficient due to repeated updates on the battery percentages.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.