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leetcode-cn/problem (English)/存在重复元素 III(English) [contains-duplicate-iii].html
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<p>Given an integer array <code>nums</code> and two integers <code>k</code> and <code>t</code>, return <code>true</code> if there are <strong>two distinct indices</strong> <code>i</code> and <code>j</code> in the array such that <code>abs(nums[i] - nums[j]) &lt;= t</code> and <code>abs(i - j) &lt;= k</code>.</p>
<p>You are given an integer array <code>nums</code> and two integers <code>indexDiff</code> and <code>valueDiff</code>.</p>
<p>Find a pair of indices <code>(i, j)</code> such that:</p>
<ul>
<li><code>i != j</code>,</li>
<li><code>abs(i - j) &lt;= indexDiff</code>.</li>
<li><code>abs(nums[i] - nums[j]) &lt;= valueDiff</code>, and</li>
</ul>
<p>Return <code>true</code><em> if such pair exists or </em><code>false</code><em> otherwise</em>.</p>
<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,3,1], k = 3, t = 0
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
<strong>Output:</strong> true
</pre><p><strong>Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1,0,1,1], k = 1, t = 2
<strong>Output:</strong> true
</pre><p><strong>Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [1,5,9,1,5,9], k = 2, t = 3
<strong>Output:</strong> false
<strong>Explanation:</strong> We can choose (i, j) = (0, 3).
We satisfy the three conditions:
i != j --&gt; 0 != 3
abs(i - j) &lt;= indexDiff --&gt; abs(0 - 3) &lt;= 3
abs(nums[i] - nums[j]) &lt;= valueDiff --&gt; abs(1 - 1) &lt;= 0
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
<strong>Output:</strong> false
<strong>Explanation:</strong> After trying all the possible pairs (i, j), we cannot satisfy the three conditions, so we return false.
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
<li><code>0 &lt;= k &lt;= 10<sup>4</sup></code></li>
<li><code>0 &lt;= t &lt;= 2<sup>31</sup> - 1</code></li>
<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
<li><code>1 &lt;= indexDiff &lt;= nums.length</code></li>
<li><code>0 &lt;= valueDiff &lt;= 10<sup>9</sup></code></li>
</ul>