mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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176 lines
20 KiB
JSON
176 lines
20 KiB
JSON
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"title": "Earliest Second to Mark Indices I",
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"content": "<p>You are given two <strong>1-indexed</strong> integer arrays, <code>nums</code> and, <code>changeIndices</code>, having lengths <code>n</code> and <code>m</code>, respectively.</p>\n\n<p>Initially, all indices in <code>nums</code> are unmarked. Your task is to mark <strong>all</strong> indices in <code>nums</code>.</p>\n\n<p>In each second, <code>s</code>, in order from <code>1</code> to <code>m</code> (<strong>inclusive</strong>), you can perform <strong>one</strong> of the following operations:</p>\n\n<ul>\n\t<li>Choose an index <code>i</code> in the range <code>[1, n]</code> and <strong>decrement</strong> <code>nums[i]</code> by <code>1</code>.</li>\n\t<li>If <code>nums[changeIndices[s]]</code> is <strong>equal</strong> to <code>0</code>, <strong>mark</strong> the index <code>changeIndices[s]</code>.</li>\n\t<li>Do nothing.</li>\n</ul>\n\n<p>Return <em>an integer denoting the <strong>earliest second</strong> in the range </em><code>[1, m]</code><em> when <strong>all</strong> indices in </em><code>nums</code><em> can be marked by choosing operations optimally, or </em><code>-1</code><em> if it is impossible.</em></p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1]\n<strong>Output:</strong> 8\n<strong>Explanation:</strong> In this example, we have 8 seconds. The following operations can be performed to mark all indices:\nSecond 1: Choose index 1 and decrement nums[1] by one. nums becomes [1,2,0].\nSecond 2: Choose index 1 and decrement nums[1] by one. nums becomes [0,2,0].\nSecond 3: Choose index 2 and decrement nums[2] by one. nums becomes [0,1,0].\nSecond 4: Choose index 2 and decrement nums[2] by one. nums becomes [0,0,0].\nSecond 5: Mark the index changeIndices[5], which is marking index 3, since nums[3] is equal to 0.\nSecond 6: Mark the index changeIndices[6], which is marking index 2, since nums[2] is equal to 0.\nSecond 7: Do nothing.\nSecond 8: Mark the index changeIndices[8], which is marking index 1, since nums[1] is equal to 0.\nNow all indices have been marked.\nIt can be shown that it is not possible to mark all indices earlier than the 8th second.\nHence, the answer is 8.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,3], changeIndices = [1,1,1,2,1,1,1]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> In this example, we have 7 seconds. The following operations can be performed to mark all indices:\nSecond 1: Choose index 2 and decrement nums[2] by one. nums becomes [1,2].\nSecond 2: Choose index 2 and decrement nums[2] by one. nums becomes [1,1].\nSecond 3: Choose index 2 and decrement nums[2] by one. nums becomes [1,0].\nSecond 4: Mark the index changeIndices[4], which is marking index 2, since nums[2] is equal to 0.\nSecond 5: Choose index 1 and decrement nums[1] by one. nums becomes [0,0].\nSecond 6: Mark the index changeIndices[6], which is marking index 1, since nums[1] is equal to 0.\nNow all indices have been marked.\nIt can be shown that it is not possible to mark all indices earlier than the 6th second.\nHence, the answer is 6.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,1], changeIndices = [2,2,2]\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> In this example, it is impossible to mark all indices because index 1 isn't in changeIndices.\nHence, the answer is -1.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= n == nums.length <= 2000</code></li>\n\t<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>\n\t<li><code>1 <= m == changeIndices.length <= 2000</code></li>\n\t<li><code>1 <= changeIndices[i] <= n</code></li>\n</ul>\n",
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"exampleTestcases": "[2,2,0]\n[2,2,2,2,3,2,2,1]\n[1,3]\n[1,1,1,2,1,1,1]\n[0,1]\n[2,2,2]",
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"code": "impl Solution {\n pub fn earliest_second_to_mark_indices(nums: Vec<i32>, change_indices: Vec<i32>) -> i32 {\n \n }\n}",
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"Consider using binary search.",
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"Suppose the <code>answer <= x</code>; we can mark each index as late as possible. Namely, mark each index at the last occurrence in the array <code>changeIndices[1..x]</code>.",
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"When marking an index, which is the last occurrence at the second <code>i</code>, we check whether we have a sufficient number of decrement operations to mark all the previous indices whose last occurrences have already been marked, and the current index, i.e., <code>i - sum_of_marked_indices_values - cnt_of_marked_indices >= nums[changeIndices[i]]</code>.",
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"The answer is the earliest second when all indices can be marked after running the binary search or <code>-1</code> if there is no such second."
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