leetcode-problemset/leetcode-cn/originData/maximum-deletions-on-a-string.json

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        "question": {
            "questionId": "2510",
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            "categoryTitle": "Algorithms",
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            "title": "Maximum Deletions on a String",
            "titleSlug": "maximum-deletions-on-a-string",
            "content": "<p>You are given a string <code>s</code> consisting of only lowercase English letters. In one operation, you can:</p>\n\n<ul>\n\t<li>Delete <strong>the entire string</strong> <code>s</code>, or</li>\n\t<li>Delete the <strong>first</strong> <code>i</code> letters of <code>s</code> if the first <code>i</code> letters of <code>s</code> are <strong>equal</strong> to the following <code>i</code> letters in <code>s</code>, for any <code>i</code> in the range <code>1 &lt;= i &lt;= s.length / 2</code>.</li>\n</ul>\n\n<p>For example, if <code>s = &quot;ababc&quot;</code>, then in one operation, you could delete the first two letters of <code>s</code> to get <code>&quot;abc&quot;</code>, since the first two letters of <code>s</code> and the following two letters of <code>s</code> are both equal to <code>&quot;ab&quot;</code>.</p>\n\n<p>Return <em>the <strong>maximum</strong> number of operations needed to delete all of </em><code>s</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;abcabcdabc&quot;\n<strong>Output:</strong> 2\n<strong>Explanation:</strong>\n- Delete the first 3 letters (&quot;abc&quot;) since the next 3 letters are equal. Now, s = &quot;abcdabc&quot;.\n- Delete all the letters.\nWe used 2 operations so return 2. It can be proven that 2 is the maximum number of operations needed.\nNote that in the second operation we cannot delete &quot;abc&quot; again because the next occurrence of &quot;abc&quot; does not happen in the next 3 letters.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;aaabaab&quot;\n<strong>Output:</strong> 4\n<strong>Explanation:</strong>\n- Delete the first letter (&quot;a&quot;) since the next letter is equal. Now, s = &quot;aabaab&quot;.\n- Delete the first 3 letters (&quot;aab&quot;) since the next 3 letters are equal. Now, s = &quot;aab&quot;.\n- Delete the first letter (&quot;a&quot;) since the next letter is equal. Now, s = &quot;ab&quot;.\n- Delete all the letters.\nWe used 4 operations so return 4. It can be proven that 4 is the maximum number of operations needed.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;aaaaa&quot;\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> In each operation, we can delete the first letter of s.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 4000</code></li>\n\t<li><code>s</code> consists only of lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "对字母串可执行的最大删除数",
            "translatedContent": "<p>给你一个仅由小写英文字母组成的字符串 <code>s</code> 。在一步操作中，你可以：</p>\n\n<ul>\n\t<li>删除 <strong>整个字符串</strong> <code>s</code> ，或者</li>\n\t<li>对于满足&nbsp;<code>1 &lt;= i &lt;= s.length / 2</code> 的任意 <code>i</code> ，如果 <code>s</code> 中的 <strong>前</strong> <code>i</code> 个字母和接下来的 <code>i</code> 个字母 <strong>相等</strong> ，删除 <strong>前</strong> <code>i</code> 个字母。</li>\n</ul>\n\n<p>例如，如果 <code>s = \"ababc\"</code> ，那么在一步操作中，你可以删除 <code>s</code> 的前两个字母得到 <code>\"abc\"</code> ，因为 <code>s</code> 的前两个字母和接下来的两个字母都等于 <code>\"ab\"</code> 。</p>\n\n<p>返回删除 <code>s</code> 所需的最大操作数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"abcabcdabc\"\n<strong>输出：</strong>2\n<strong>解释：</strong>\n- 删除前 3 个字母（\"abc\"），因为它们和接下来 3 个字母相等。现在，s = \"abcdabc\"。\n- 删除全部字母。\n一共用了 2 步操作，所以返回 2 。可以证明 2 是所需的最大操作数。\n注意，在第二步操作中无法再次删除 \"abc\" ，因为 \"abc\" 的下一次出现并不是位于接下来的 3 个字母。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"aaabaab\"\n<strong>输出：</strong>4\n<strong>解释：</strong>\n- 删除第一个字母（\"a\"），因为它和接下来的字母相等。现在，s = \"aabaab\"。\n- 删除前 3 个字母（\"aab\"），因为它们和接下来 3 个字母相等。现在，s = \"aab\"。 \n- 删除第一个字母（\"a\"），因为它和接下来的字母相等。现在，s = \"ab\"。\n- 删除全部字母。\n一共用了 4 步操作，所以返回 4 。可以证明 4 是所需的最大操作数。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"aaaaa\"\n<strong>输出：</strong>5\n<strong>解释：</strong>在每一步操作中，都可以仅删除 s 的第一个字母。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 4000</code></li>\n\t<li><code>s</code> 仅由小写英文字母组成</li>\n</ul>\n",
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            "hints": [
                "We can use dynamic programming to find the answer. Create a 0-indexed dp array where dp[i] represents the maximum number of moves needed to remove the first i + 1 letters from s.",
                "What should we do if there is an i where it is impossible to remove the first i + 1 letters?",
                "Use a sentinel value such as -1 to show that it is impossible.",
                "How can we quickly determine if two substrings of s are equal? We can use hashing."
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