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]] to any non-negative value.i in the range [1, n], where nums[i] is equal to 0, and mark index i.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 = [3,2,3], changeIndices = [1,3,2,2,2,2,3]\nOutput: 6\nExplanation: In this example, we have 7 seconds. The following operations can be performed to mark all indices:\nSecond 1: Set nums[changeIndices[1]] to 0. nums becomes [0,2,3].\nSecond 2: Set nums[changeIndices[2]] to 0. nums becomes [0,2,0].\nSecond 3: Set nums[changeIndices[3]] to 0. nums becomes [0,0,0].\nSecond 4: Mark index 1, since nums[1] is equal to 0.\nSecond 5: Mark index 2, since nums[2] is equal to 0.\nSecond 6: Mark index 3, since nums[3] 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 2:
\n\n\nInput: nums = [0,0,1,2], changeIndices = [1,2,1,2,1,2,1,2]\nOutput: 7\nExplanation: In this example, we have 8 seconds. The following operations can be performed to mark all indices:\nSecond 1: Mark index 1, since nums[1] is equal to 0.\nSecond 2: Mark index 2, since nums[2] is equal to 0.\nSecond 3: Decrement index 4 by one. nums becomes [0,0,1,1].\nSecond 4: Decrement index 4 by one. nums becomes [0,0,1,0].\nSecond 5: Decrement index 3 by one. nums becomes [0,0,0,0].\nSecond 6: Mark index 3, since nums[3] is equal to 0.\nSecond 7: Mark index 4, since nums[4] 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 7th second.\nHence, the answer is 7.\n\n\n
Example 3:
\n\n\nInput: nums = [1,2,3], changeIndices = [1,2,3]\nOutput: -1\nExplanation: In this example, it can be shown that it is impossible to mark all indices, as we don't have enough seconds. \nHence, the answer is -1.\n\n\n
\n
Constraints:
\n\n1 <= n == nums.length <= 50000 <= nums[i] <= 1091 <= m == changeIndices.length <= 50001 <= 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]] 设置成任意的 非负 整数。[1, n] 中的一个下标 i ， 满足 nums[i] 等于 0, 并 标记 下标 i 。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 = [3,2,3], changeIndices = [1,3,2,2,2,2,3]\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：将 nums[changeIndices[1]] 变为 0 。nums 变为 [0,2,3] 。\n第 2 秒：将 nums[changeIndices[2]] 变为 0 。nums 变为 [0,2,0] 。\n第 3 秒：将 nums[changeIndices[3]] 变为 0 。nums 变为 [0,0,0] 。\n第 4 秒：标记下标 1 ，因为 nums[1] 等于 0 。\n第 5 秒：标记下标 2 ，因为 nums[2] 等于 0 。\n第 6 秒：标记下标 3 ，因为 nums[3] 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 2：
\n\n\n输入：nums = [0,0,1,2], changeIndices = [1,2,1,2,1,2,1,2]\n输出：7\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：标记下标 1 ，因为 nums[1] 等于 0 。\n第 2 秒：标记下标 2 ，因为 nums[2] 等于 0 。\n第 3 秒：将 nums[4] 减少 1 。nums 变为 [0,0,1,1] 。\n第 4 秒：将 nums[4] 减少 1 。nums 变为 [0,0,1,0] 。\n第 5 秒：将 nums[3] 减少 1 。nums 变为 [0,0,0,0] 。\n第 6 秒：标记下标 3 ，因为 nums[3] 等于 0 。\n第 7 秒：标记下标 4 ，因为 nums[4] 等于 0 。\n现在所有下标已被标记。\n最早可以在第 7 秒标记所有下标。\n所以答案是 7 。\n\n\n
示例 3：
\n\n\n输入：nums = [1,2,3], changeIndices = [1,2,3]\n输出：-1\n解释：这个例子中，无法标记所有下标，因为我们没有足够的秒数。\n所以答案是 -1 。\n\n\n
\n\n
提示：
\n\n1 <= n == nums.length <= 50000 <= nums[i] <= 1091 <= m == changeIndices.length <= 50001 <= changeIndices[i] <= nn seconds, and at most sum(nums[i]) + n seconds.",
"We can binary search the earliest second where all indices can be marked.",
"If there is an operation where we change nums[changeIndices[i]] to a non-negative value, it is best for it to satisfy the following constraints:nums[changeIndices[i]] should not be equal to 0.nums[changeIndices[i]] should be changed to 0.changeIndices[i] occurs in changeIndices.j, where changeIndices[i] will be marked. j is in the range [i + 1, m].time_needed = sum(nums[i]) + n. To check if we can mark all indices at some second x, we need to make time_needed <= x, using non-negative change operations as described previously.",
"Using a non-negative change operation on some nums[changeIndices[i]] that satisfies the constraints described previously reduces time_needed by nums[changeIndices[i]] - 1. So, we need to maximize the sum of (nums[changeIndices[i]] - 1) while ensuring that the non-negative change operations still satisfy the constraints.",
"Maximizing the sum of (nums[changeIndices[i]] - 1) can be done greedily using a min-priority queue and going in reverse starting from second x to second 1, maximizing the sum of the values in the priority queue and ensuring that for every non-negative change operation on nums[changeIndices[i]] chosen, there is another second j in the range [i + 1, x] where changeIndices[i] can be marked.",
"The answer is the first value of x in the range [1, m] where it is possible to make time_needed <= x, or -1 if there is no such second."
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