mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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176 lines
17 KiB
JSON
176 lines
17 KiB
JSON
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"title": "House Robber IV",
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"content": "<p>There are several consecutive houses along a street, each of which has some money inside. There is also a robber, who wants to steal money from the homes, but he <strong>refuses to steal from adjacent homes</strong>.</p>\n\n<p>The <strong>capability</strong> of the robber is the maximum amount of money he steals from one house of all the houses he robbed.</p>\n\n<p>You are given an integer array <code>nums</code> representing how much money is stashed in each house. More formally, the <code>i<sup>th</sup></code> house from the left has <code>nums[i]</code> dollars.</p>\n\n<p>You are also given an integer <code>k</code>, representing the <strong>minimum</strong> number of houses the robber will steal from. It is always possible to steal at least <code>k</code> houses.</p>\n\n<p>Return <em>the <strong>minimum</strong> capability of the robber out of all the possible ways to steal at least </em><code>k</code><em> houses</em>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,3,5,9], k = 2\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> \nThere are three ways to rob at least 2 houses:\n- Rob the houses at indices 0 and 2. Capability is max(nums[0], nums[2]) = 5.\n- Rob the houses at indices 0 and 3. Capability is max(nums[0], nums[3]) = 9.\n- Rob the houses at indices 1 and 3. Capability is max(nums[1], nums[3]) = 9.\nTherefore, we return min(5, 9, 9) = 5.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,7,9,3,1], k = 2\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> There are 7 ways to rob the houses. The way which leads to minimum capability is to rob the house at index 0 and 4. Return max(nums[0], nums[4]) = 2.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>\n\t<li><code>1 <= k <= (nums.length + 1)/2</code></li>\n</ul>\n",
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"difficulty": "Medium",
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"exampleTestcases": "[2,3,5,9]\n2\n[2,7,9,3,1]\n2",
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"code": "/**\n * @param {number[]} nums\n * @param {number} k\n * @return {number}\n */\nvar minCapability = function(nums, k) {\n \n};",
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"code": "# @param {Integer[]} nums\n# @param {Integer} k\n# @return {Integer}\ndef min_capability(nums, k)\n \nend",
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"code": "object Solution {\n def minCapability(nums: Array[Int], k: Int): Int = {\n \n }\n}",
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"code": "class Solution {\n fun minCapability(nums: IntArray, k: Int): Int {\n \n }\n}",
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"code": "impl Solution {\n pub fn min_capability(nums: Vec<i32>, k: i32) -> i32 {\n \n }\n}",
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"code": "class Solution {\n\n /**\n * @param Integer[] $nums\n * @param Integer $k\n * @return Integer\n */\n function minCapability($nums, $k) {\n \n }\n}",
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"Can we use binary search to find the minimum value of a non-contiguous subsequence of a given size k?",
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"Initialize the search range with the minimum and maximum elements of the input array.",
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"Use a check function to determine if it is possible to select k non-consecutive elements that are less than or equal to the current \"guess\" value.",
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"Adjust the search range based on the outcome of the check function, until the range converges and the minimum value is found."
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