contains-duplicate-iii

java

python

python3

mysql

mssql

oraclesql

csharp

javascript

typescript

bash

swift

kotlin

dart

golang

ruby

scala

rust

racket

erlang

elixir

pythondata

postgresql

cangjie

Algorithms

You are given an integer array <code>nums</code> and two integers <code>indexDiff</code> and <code>valueDiff</code>.

Find a pair of indices <code>(i, j)</code> such that:

<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>

Return <code>true</code> if such pair exists or <code>false</code> otherwise.

&nbsp;
Example 1:

<pre>
Input: nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
Output: true
Explanation: 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>

Example 2:

<pre>
Input: nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
Output: false
Explanation: After trying all the possible pairs (i, j), we cannot satisfy the three conditions, so we return false.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>2 &lt;= nums.length &lt;= 105</code></li>
	<li><code>-109 &lt;= nums[i] &lt;= 109</code></li>
	<li><code>1 &lt;= indexDiff &lt;= nums.length</code></li>
	<li><code>0 &lt;= valueDiff &lt;= 109</code></li>
</ul>

containsNearbyAlmostDuplicate

Contains Duplicate III

Contains Duplicate

contains-duplicate

存在重复元素

Contains Duplicate II

contains-duplicate-ii

存在重复元素 II

Array

Bucket Sort

Ordered Set

Sorting

Sliding Window

给你一个整数数组 <code>nums</code> 和两个整数 <code>indexDiff</code> 和 <code>valueDiff</code> 。

找出满足下述条件的下标对 <code>(i, j)</code>：

<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></li>
</ul>

如果存在，返回 <code>true</code> ；否则，返回 <code>false</code> 。

&nbsp;

示例 1：

<pre>
输入：nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
输出：true
解释：可以找出 (i, j) = (0, 3) 。
满足下述 3 个条件：
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>

示例 2：

<pre>
输入：nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
输出：false
解释：尝试所有可能的下标对 (i, j) ，均无法满足这 3 个条件，因此返回 false 。
</pre>

&nbsp;

提示：

<ul>
	<li><code>2 &lt;= nums.length &lt;= 105</code></li>
	<li><code>-109 &lt;= nums[i] &lt;= 109</code></li>
	<li><code>1 &lt;= indexDiff &lt;= nums.length</code></li>
	<li><code>0 &lt;= valueDiff &lt;= 109</code></li>
</ul>