# 2960. Count Tested Devices After Test Operations

## Description

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.

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.

Constraints:

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

## Solutions

Solution 1: Simulation

Assume that the current number of devices we have tested is $ans$. When testing a new device $i$, its remaining battery is $\max(0, batteryPercentages[i] - ans)$. If the remaining battery is greater than $0$, it means this device can be tested, and we need to increase $ans$ by $1$.

Finally, return $ans$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

• class Solution {
public int countTestedDevices(int[] batteryPercentages) {
int ans = 0;
for (int x : batteryPercentages) {
x -= ans;
if (x > 0) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
int countTestedDevices(vector<int>& batteryPercentages) {
int ans = 0;
for (int x : batteryPercentages) {
x -= ans;
if (x > 0) {
++ans;
}
}
return ans;
}
};

• class Solution:
def countTestedDevices(self, batteryPercentages: List[int]) -> int:
ans = 0
for x in batteryPercentages:
x -= ans
ans += x > 0
return ans

• func countTestedDevices(batteryPercentages []int) (ans int) {
for _, x := range batteryPercentages {
x -= ans
if x > 0 {
ans++
}
}
return
}

• function countTestedDevices(batteryPercentages: number[]): number {
let ans = 0;
for (let x of batteryPercentages) {
x -= ans;
if (x > 0) {
++ans;
}
}
return ans;
}

• impl Solution {
pub fn count_tested_devices(battery_percentages: Vec<i32>) -> i32 {
let mut ans = 0;
for x in battery_percentages {
ans += if x > ans { 1 } else { 0 };
}
ans
}
}