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leetcode-cn/originData/minimum-window-substring.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个字符串 <code>s</code> 、一个字符串 <code>t</code> 。返回 <code>s</code> 中涵盖 <code>t</code> 所有字符的最小子串。如果 <code>s</code> 中不存在涵盖 <code>t</code> 所有字符的子串，则返回空字符串 <code>\"\"</code> 。</p>\n\n<p> </p>\n\n<p><strong>注意：</strong></p>\n\n<ul>\n\t<li>对于 <code>t</code> 中重复字符，我们寻找的子字符串中该字符数量必须不少于 <code>t</code> 中该字符数量。</li>\n\t<li>如果 <code>s</code> 中存在这样的子串，我们保证它是唯一的答案。</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"ADOBECODEBANC\", t = \"ABC\"\n<strong>输出：</strong>\"BANC\"\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"a\", t = \"a\"\n<strong>输出：</strong>\"a\"\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong> s = \"a\", t = \"aa\"\n<strong>输出:</strong> \"\"\n<strong>解释:</strong> t 中两个字符 'a' 均应包含在 s 的子串中，\n因此没有符合条件的子字符串，返回空字符串。</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= s.length, t.length <= 10<sup>5</sup></code></li>\n\t<li><code>s</code> 和 <code>t</code> 由英文字母组成</li>\n</ul>\n\n<p> </p>\n<strong>进阶：</strong>你能设计一个在 <code>o(n)</code> 时间内解决此问题的算法吗？",
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             "hints": [
                 "Use two pointers to create a window of letters in <b>S</b>, which would have all the characters from <b>T</b>.",
                 "Since you have to find the minimum window in <b>S</b> which has all the characters from <b>T</b>, you need to expand and contract the window using the two pointers and keep checking the window for all the characters. This approach is also called Sliding Window Approach.\r\n\r\n<br><br>\r\n<pre>\r\nL ------------------------ R , Suppose this is the window that contains all characters of <b>T</b> \r\n                          \r\n&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp L----------------- R , this is the contracted window. We found a smaller window that still contains all the characters in <b>T</b>\r\n\r\nWhen the window is no longer valid, start expanding again using the right pointer. </pre>"