2960. Count Tested Devices After Test Operations

Problem Link

Description

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You are given a 0-indexed integer array batteryPercentages having length n, denoting the battery percentages of n 0-indexed devices.

Your task is to test each device i in order from 0 to n - 1, by performing the following test operations:

• If batteryPercentages[i] is greater than 0:

  <ul>
  	<li><strong>Increment</strong> the count of tested devices.</li>
  	<li><strong>Decrease</strong> the battery percentage of all devices with indices <code>j</code> in the range <code>[i + 1, n - 1]</code> by <code>1</code>, ensuring their battery percentage <strong>never goes below</strong> <code>0</code>, i.e, <code>batteryPercentages[j] = max(0, batteryPercentages[j] - 1)</code>.</li>
  	<li>Move to the next device.</li>
  </ul>
  </li>
  <li>Otherwise, move to the next device without performing any test.</li>
  

Return an integer denoting the number of devices that will be tested after performing the test operations in order.

 

Example 1:

Input: batteryPercentages = [1,1,2,1,3]
Output: 3
Explanation: Performing the test operations in order starting from device 0:
At device 0, batteryPercentages[0] > 0, so there is now 1 tested device, and batteryPercentages becomes [1,0,1,0,2].
At device 1, batteryPercentages[1] == 0, so we move to the next device without testing.
At device 2, batteryPercentages[2] > 0, so there are now 2 tested devices, and batteryPercentages becomes [1,0,1,0,1].
At device 3, batteryPercentages[3] == 0, so we move to the next device without testing.
At device 4, batteryPercentages[4] > 0, so there are now 3 tested devices, and batteryPercentages stays the same.
So, the answer is 3.

Example 2:

Input: batteryPercentages = [0,1,2]
Output: 2
Explanation: Performing the test operations in order starting from device 0:
At device 0, batteryPercentages[0] == 0, so we move to the next device without testing.
At device 1, batteryPercentages[1] > 0, so there is now 1 tested device, and batteryPercentages becomes [0,1,1].
At device 2, batteryPercentages[2] > 0, so there are now 2 tested devices, and batteryPercentages stays the same.
So, the answer is 2.

 

Constraints:

• 1 <= n == batteryPercentages.length <= 100
• 0 <= batteryPercentages[i] <= 100

Solution

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Python3

class Solution:
    def countTestedDevices(self, A: List[int]) -> int:
        N = len(A)
        res = 0
        
        for i in range(N):
            if A[i] == 0: continue
            res += 1
            
            for j in range(i + 1, N):
                A[j] = max(0, A[j] - 1)
        
        return res