LeetCode-in-Java

2960. Count Tested Devices After Test Operations

Easy

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

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

public class Solution {
    public int countTestedDevices(int[] batteryPercentages) {
        int count = 0;
        int diff = 0;
        for (int n : batteryPercentages) {
            if (n - diff > 0) {
                count++;
                diff++;
            }
        }
        return count;
    }
}

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