You are given two 1-indexed integer arrays, nums and, changeIndices, having lengths n and m, respectively.
Initially, all indices in nums are unmarked. Your task is to mark all indices in nums.
In each second, s, in order from 1 to m (inclusive), you can perform one of the following operations:
i in the range [1, n] and decrement nums[i] by 1.nums[changeIndices[s]] is equal to 0, mark the index changeIndices[s].Return an integer denoting the earliest second in the range [1, m] when all indices in nums can be marked by choosing operations optimally, or -1 if it is impossible.
\n
Example 1:
\n\n\nInput: nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1]\nOutput: 8\nExplanation: 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\n\n
Example 2:
\n\n\nInput: nums = [1,3], changeIndices = [1,1,1,2,1,1,1]\nOutput: 6\nExplanation: 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\n\n
Example 3:
\n\n\nInput: nums = [0,1], changeIndices = [2,2,2]\nOutput: -1\nExplanation: In this example, it is impossible to mark all indices because index 1 isn't in changeIndices.\nHence, the answer is -1.\n\n\n
\n
Constraints:
\n\n1 <= n == nums.length <= 20000 <= nums[i] <= 1091 <= m == changeIndices.length <= 20001 <= changeIndices[i] <= n给你两个下标从 1 开始的整数数组 nums 和 changeIndices ，数组的长度分别为 n 和 m 。
一开始，nums 中所有下标都是未标记的，你的任务是标记 nums 中 所有 下标。
从第 1 秒到第 m 秒（包括 第 m 秒），对于每一秒 s ，你可以执行以下操作 之一 ：
[1, n] 中的一个下标 i ，并且将 nums[i] 减少 1 。nums[changeIndices[s]] 等于 0 ，标记 下标 changeIndices[s] 。请你返回范围 [1, m] 中的一个整数，表示最优操作下，标记 nums 中 所有 下标的 最早秒数 ，如果无法标记所有下标，返回 -1 。
\n\n
示例 1：
\n\n\n输入：nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1]\n输出：8\n解释：这个例子中，我们总共有 8 秒。按照以下操作标记所有下标：\n第 1 秒：选择下标 1 ，将 nums[1] 减少 1 。nums 变为 [1,2,0] 。\n第 2 秒：选择下标 1 ，将 nums[1] 减少 1 。nums 变为 [0,2,0] 。\n第 3 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [0,1,0] 。\n第 4 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [0,0,0] 。\n第 5 秒，标记 changeIndices[5] ，也就是标记下标 3 ，因为 nums[3] 等于 0 。\n第 6 秒，标记 changeIndices[6] ，也就是标记下标 2 ，因为 nums[2] 等于 0 。\n第 7 秒，什么也不做。\n第 8 秒，标记 changeIndices[8] ，也就是标记下标 1 ，因为 nums[1] 等于 0 。\n现在所有下标已被标记。\n最早可以在第 8 秒标记所有下标。\n所以答案是 8 。\n\n\n
示例 2：
\n\n\n输入：nums = [1,3], changeIndices = [1,1,1,2,1,1,1]\n输出：6\n解释：这个例子中，我们总共有 7 秒。按照以下操作标记所有下标：\n第 1 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [1,2] 。\n第 2 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [1,1] 。\n第 3 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [1,0] 。\n第 4 秒：标记 changeIndices[4] ，也就是标记下标 2 ，因为 nums[2] 等于 0 。\n第 5 秒：选择下标 1 ，将 nums[1] 减少 1 。nums 变为 [0,0] 。\n第 6 秒：标记 changeIndices[6] ，也就是标记下标 1 ，因为 nums[1] 等于 0 。\n现在所有下标已被标记。\n最早可以在第 6 秒标记所有下标。\n所以答案是 6 。\n\n\n
示例 3：
\n\n\nInput: nums = [0,1], changeIndices = [2,2,2]\nOutput: -1\nExplanation: 这个例子中，无法标记所有下标，因为下标 1 不在 changeIndices 中。\n所以答案是 -1 。\n\n\n
\n\n
提示：
\n\n1 <= n == nums.length <= 20000 <= nums[i] <= 1091 <= m == changeIndices.length <= 20001 <= changeIndices[i] <= nanswer <= x; we can mark each index as late as possible. Namely, mark each index at the last occurrence in the array changeIndices[1..x].",
"When marking an index, which is the last occurrence at the second i, 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., i - sum_of_marked_indices_values - cnt_of_marked_indices >= nums[changeIndices[i]].",
"The answer is the earliest second when all indices can be marked after running the binary search or -1 if there is no such second."
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