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leetcode-problemset/leetcode/originData/find-latest-group-of-size-m.json

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{
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"title": "Find Latest Group of Size M",
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"content": "<p>Given an array <code>arr</code> that represents a permutation of numbers from <code>1</code> to <code>n</code>.</p>\n\n<p>You have a binary string of size <code>n</code> that initially has all its bits set to zero. At each step <code>i</code> (assuming both the binary string and <code>arr</code> are 1-indexed) from <code>1</code> to <code>n</code>, the bit at position <code>arr[i]</code> is set to <code>1</code>.</p>\n\n<p>You are also given an integer <code>m</code>. Find the latest step at which there exists a group of ones of length <code>m</code>. A group of ones is a contiguous substring of <code>1</code>&#39;s such that it cannot be extended in either direction.</p>\n\n<p>Return <em>the latest step at which there exists a group of ones of length <strong>exactly</strong></em> <code>m</code>. <em>If no such group exists, return</em> <code>-1</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [3,5,1,2,4], m = 1\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> \nStep 1: &quot;00<u>1</u>00&quot;, groups: [&quot;1&quot;]\nStep 2: &quot;0010<u>1</u>&quot;, groups: [&quot;1&quot;, &quot;1&quot;]\nStep 3: &quot;<u>1</u>0101&quot;, groups: [&quot;1&quot;, &quot;1&quot;, &quot;1&quot;]\nStep 4: &quot;1<u>1</u>101&quot;, groups: [&quot;111&quot;, &quot;1&quot;]\nStep 5: &quot;111<u>1</u>1&quot;, groups: [&quot;11111&quot;]\nThe latest step at which there exists a group of size 1 is step 4.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [3,1,5,4,2], m = 2\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> \nStep 1: &quot;00<u>1</u>00&quot;, groups: [&quot;1&quot;]\nStep 2: &quot;<u>1</u>0100&quot;, groups: [&quot;1&quot;, &quot;1&quot;]\nStep 3: &quot;1010<u>1</u>&quot;, groups: [&quot;1&quot;, &quot;1&quot;, &quot;1&quot;]\nStep 4: &quot;101<u>1</u>1&quot;, groups: [&quot;1&quot;, &quot;111&quot;]\nStep 5: &quot;1<u>1</u>111&quot;, groups: [&quot;11111&quot;]\nNo group of size 2 exists during any step.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == arr.length</code></li>\n\t<li><code>1 &lt;= m &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= arr[i] &lt;= n</code></li>\n\t<li>All integers in <code>arr</code> are <strong>distinct</strong>.</li>\n</ul>\n",
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"code": "class Solution {\npublic:\n int findLatestStep(vector<int>& arr, int m) {\n \n }\n};",
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"code": "class Solution(object):\n def findLatestStep(self, arr, m):\n \"\"\"\n :type arr: List[int]\n :type m: int\n :rtype: int\n \"\"\"\n ",
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"code": "int findLatestStep(int* arr, int arrSize, int m) {\n \n}",
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"code": "/**\n * @param {number[]} arr\n * @param {number} m\n * @return {number}\n */\nvar findLatestStep = function(arr, m) {\n \n};",
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"lang": "Kotlin",
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"lang": "Dart",
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"code": "# @param {Integer[]} arr\n# @param {Integer} m\n# @return {Integer}\ndef find_latest_step(arr, m)\n \nend",
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"lang": "Scala",
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"code": "impl Solution {\n pub fn find_latest_step(arr: Vec<i32>, m: i32) -> i32 {\n \n }\n}",
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"code": "defmodule Solution do\n @spec find_latest_step(arr :: [integer], m :: integer) :: integer\n def find_latest_step(arr, m) do\n \n end\nend",
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"hints": [
"Since the problem asks for the latest step, can you start the searching from the end of arr?",
"Use a map to store the current “1” groups.",
"At each step (going backwards) you need to split one group and update the map."
],
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