leetcode-problemset/leetcode-cn/originData/contains-duplicate-iii.json

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            "categoryTitle": "Algorithms",
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            "title": "Contains Duplicate III",
            "titleSlug": "contains-duplicate-iii",
            "content": "<p>You are given an integer array <code>nums</code> and two integers <code>indexDiff</code> and <code>valueDiff</code>.</p>\n\n<p>Find a pair of indices <code>(i, j)</code> such that:</p>\n\n<ul>\n\t<li><code>i != j</code>,</li>\n\t<li><code>abs(i - j) &lt;= indexDiff</code>.</li>\n\t<li><code>abs(nums[i] - nums[j]) &lt;= valueDiff</code>, and</li>\n</ul>\n\n<p>Return <code>true</code><em> if such pair exists or </em><code>false</code><em> otherwise</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,1], indexDiff = 3, valueDiff = 0\n<strong>Output:</strong> true\n<strong>Explanation:</strong> We can choose (i, j) = (0, 3).\nWe satisfy the three conditions:\ni != j --&gt; 0 != 3\nabs(i - j) &lt;= indexDiff --&gt; abs(0 - 3) &lt;= 3\nabs(nums[i] - nums[j]) &lt;= valueDiff --&gt; abs(1 - 1) &lt;= 0\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3\n<strong>Output:</strong> false\n<strong>Explanation:</strong> After trying all the possible pairs (i, j), we cannot satisfy the three conditions, so we return false.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= indexDiff &lt;= nums.length</code></li>\n\t<li><code>0 &lt;= valueDiff &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "存在重复元素 III",
            "translatedContent": "<p>给你一个整数数组 <code>nums</code> 和两个整数 <code>indexDiff</code> 和 <code>valueDiff</code> 。</p>\n\n<p>找出满足下述条件的下标对 <code>(i, j)</code>：</p>\n\n<ul>\n\t<li><code>i != j</code>,</li>\n\t<li><code>abs(i - j) &lt;= indexDiff</code></li>\n\t<li><code>abs(nums[i] - nums[j]) &lt;= valueDiff</code></li>\n</ul>\n\n<p>如果存在，返回 <code>true</code><em> ；</em>否则，返回<em> </em><code>false</code><em> </em>。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,3,1], indexDiff = 3, valueDiff = 0\n<strong>输出：</strong>true\n<strong>解释：</strong>可以找出 (i, j) = (0, 3) 。\n满足下述 3 个条件：\ni != j --&gt; 0 != 3\nabs(i - j) &lt;= indexDiff --&gt; abs(0 - 3) &lt;= 3\nabs(nums[i] - nums[j]) &lt;= valueDiff --&gt; abs(1 - 1) &lt;= 0\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3\n<strong>输出：</strong>false\n<strong>解释：</strong>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 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= indexDiff &lt;= nums.length</code></li>\n\t<li><code>0 &lt;= valueDiff &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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