count-tested-devices-after-test-operations

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Algorithms

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

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

<ul>
	<li>If <code>batteryPercentages[i]</code> is greater than <code>0</code>:

	<ul>
		<li>Increment the count of tested devices.</li>
		<li>Decrease 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 never goes below <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>
</ul>

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

&nbsp;
Example 1:

<pre>
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] &gt; 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] &gt; 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] &gt; 0, so there are now 3 tested devices, and batteryPercentages stays the same.
So, the answer is 3.
</pre>

Example 2:

<pre>
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] &gt; 0, so there is now 1 tested device, and batteryPercentages becomes [0,1,1].
At device 2, batteryPercentages[2] &gt; 0, so there are now 2 tested devices, and batteryPercentages stays the same.
So, the answer is 2.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= n == batteryPercentages.length &lt;= 100 </code></li>
	<li><code>0 &lt;= batteryPercentages[i] &lt;= 100</code></li>
</ul>

countTestedDevices

Count Tested Devices After Test Operations

Array

Counting

Simulation

给你一个长度为 <code>n</code> 、下标从 0 开始的整数数组 <code>batteryPercentages</code> ，表示 <code>n</code> 个设备的电池百分比。

你的任务是按照顺序测试每个设备 <code>i</code>，执行以下测试操作：

<ul>
	<li>如果 <code>batteryPercentages[i]</code> 大于 <code>0</code>：

	<ul>
		<li>增加 已测试设备的计数。</li>
		<li>将下标在 <code>[i + 1, n - 1]</code> 的所有设备的电池百分比减少 <code>1</code>，确保它们的电池百分比 不会低于 <code>0</code> ，即 <code>batteryPercentages[j] = max(0, batteryPercentages[j] - 1)</code>。</li>
		<li>移动到下一个设备。</li>
	</ul>
	</li>
	<li>否则，移动到下一个设备而不执行任何测试。</li>
</ul>

返回一个整数，表示按顺序执行测试操作后 已测试设备 的数量。

&nbsp;

示例 1：

<pre>
输入：batteryPercentages = [1,1,2,1,3]
输出：3
解释：按顺序从设备 0 开始执行测试操作：
在设备 0 上，batteryPercentages[0] &gt; 0 ，现在有 1 个已测试设备，batteryPercentages 变为 [1,0,1,0,2] 。
在设备 1 上，batteryPercentages[1] == 0 ，移动到下一个设备而不进行测试。
在设备 2 上，batteryPercentages[2] &gt; 0 ，现在有 2 个已测试设备，batteryPercentages 变为 [1,0,1,0,1] 。
在设备 3 上，batteryPercentages[3] == 0 ，移动到下一个设备而不进行测试。
在设备 4 上，batteryPercentages[4] &gt; 0 ，现在有 3 个已测试设备，batteryPercentages 保持不变。
因此，答案是 3 。
</pre>

示例 2：

<pre>
输入：batteryPercentages = [0,1,2]
输出：2
解释：按顺序从设备 0 开始执行测试操作：
在设备 0 上，batteryPercentages[0] == 0 ，移动到下一个设备而不进行测试。
在设备 1 上，batteryPercentages[1] &gt; 0 ，现在有 1 个已测试设备，batteryPercentages 变为 [0,1,1] 。
在设备 2 上，batteryPercentages[2] &gt; 0 ，现在有 2 个已测试设备，batteryPercentages 保持不变。
因此，答案是 2 。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= n == batteryPercentages.length &lt;= 100 </code></li>
	<li><code>0 &lt;= batteryPercentages[i] &lt;= 100</code></li>
</ul>