leetcode-problemset/leetcode-cn/problem (English)/使数组元素相等的最小操作次数(English) [minimum-operations-to-equalize-subarrays].html

<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>

<p>In one operation, you can <strong>increase or decrease </strong>any element of <code>nums</code> by <strong>exactly</strong> <code>k</code>.</p>

<p>You are also given a 2D integer array <code>queries</code>, where each <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>

<p>For each query, find the <strong>minimum</strong> number of operations required to make <strong>all</strong> elements in the <strong><span data-keyword="subarray-nonempty">subarray</span></strong> <code>nums[l<sub>i</sub>..r<sub>i</sub>]</code> <strong>equal</strong>. If it is impossible, the answer for that query is <code>-1</code>.</p>

<p>Return an array <code>ans</code>, where <code>ans[i]</code> is the answer for the <code>i<sup>th</sup></code> query.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,4,7], k = 3, queries = [[0,1],[0,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,2]</span></p>

<p><strong>Explanation:</strong></p>

<p>One optimal set of operations:</p>

<table style="border: 1px solid black;">
	<tbody>
		<tr>
			<th style="border: 1px solid black;"><code>i</code></th>
			<th style="border: 1px solid black;"><code>[l<sub>i</sub>, r<sub>i</sub>]</code></th>
			<th style="border: 1px solid black;"><code>nums[l<sub>i</sub>..r<sub>i</sub>]</code></th>
			<th style="border: 1px solid black;">Possibility</th>
			<th style="border: 1px solid black;">Operations</th>
			<th style="border: 1px solid black;">Final<br />
			<code>nums[l<sub>i</sub>..r<sub>i</sub>]</code></th>
			<th style="border: 1px solid black;"><code>ans[i]</code></th>
		</tr>
	</tbody>
	<tbody>
		<tr>
			<td style="border: 1px solid black;">0</td>
			<td style="border: 1px solid black;">[0, 1]</td>
			<td style="border: 1px solid black;">[1, 4]</td>
			<td style="border: 1px solid black;">Yes</td>
			<td style="border: 1px solid black;"><code>nums[0] + k = 1 + 3 = 4 = nums[1]</code></td>
			<td style="border: 1px solid black;">[4, 4]</td>
			<td style="border: 1px solid black;">1</td>
		</tr>
		<tr>
			<td style="border: 1px solid black;">1</td>
			<td style="border: 1px solid black;">[0, 2]</td>
			<td style="border: 1px solid black;">[1, 4, 7]</td>
			<td style="border: 1px solid black;">Yes</td>
			<td style="border: 1px solid black;"><code>nums[0] + k = 1 + 3 = 4 = nums[1]<br />
			nums[2] - k = 7 - 3 = 4 = nums[1]</code></td>
			<td style="border: 1px solid black;">[4, 4, 4]</td>
			<td style="border: 1px solid black;">2</td>
		</tr>
	</tbody>
</table>

<p>Thus, <code>ans = [1, 2]</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2,4], k = 2, queries = [[0,2],[0,0],[1,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[-1,0,1]</span></p>

<p><strong>Explanation:</strong></p>

<p>One optimal set of operations:</p>

<table style="border: 1px solid black;">
	<tbody>
		<tr>
			<th style="border: 1px solid black;"><code>i</code></th>
			<th style="border: 1px solid black;"><code>[l<sub>i</sub>, r<sub>i</sub>]</code></th>
			<th style="border: 1px solid black;"><code>nums[l<sub>i</sub>..r<sub>i</sub>]</code></th>
			<th style="border: 1px solid black;">Possibility</th>
			<th style="border: 1px solid black;">Operations</th>
			<th style="border: 1px solid black;">Final<br />
			<code>nums[l<sub>i</sub>..r<sub>i</sub>]</code></th>
			<th style="border: 1px solid black;"><code>ans[i]</code></th>
		</tr>
		<tr>
			<td style="border: 1px solid black;">0</td>
			<td style="border: 1px solid black;">[0, 2]</td>
			<td style="border: 1px solid black;">[1, 2, 4]</td>
			<td style="border: 1px solid black;">No</td>
			<td style="border: 1px solid black;">-</td>
			<td style="border: 1px solid black;">[1, 2, 4]</td>
			<td style="border: 1px solid black;">-1</td>
		</tr>
		<tr>
			<td style="border: 1px solid black;">1</td>
			<td style="border: 1px solid black;">[0, 0]</td>
			<td style="border: 1px solid black;">[1]</td>
			<td style="border: 1px solid black;">Yes</td>
			<td style="border: 1px solid black;">Already equal</td>
			<td style="border: 1px solid black;">[1]</td>
			<td style="border: 1px solid black;">0</td>
		</tr>
		<tr>
			<td style="border: 1px solid black;">2</td>
			<td style="border: 1px solid black;">[1, 2]</td>
			<td style="border: 1px solid black;">[2, 4]</td>
			<td style="border: 1px solid black;">Yes</td>
			<td style="border: 1px solid black;"><code>nums[1] + k = 2 + 2 = 4 = nums[2]</code></td>
			<td style="border: 1px solid black;">[4, 4]</td>
			<td style="border: 1px solid black;">1</td>
		</tr>
	</tbody>
</table>

<p>Thus, <code>ans = [-1, 0, 1]</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n == nums.length &lt;= 4 &times; 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code>​​​​​​​</li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= queries.length &lt;= 4 &times; 10<sup>4</sup></code></li>
	<li><code><sup>​​​​​​​</sup>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code></li>
	<li><code>0 &lt;= l<sub>i</sub> &lt;= r<sub>i</sub> &lt;= n - 1</code></li>
</ul>