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] <= 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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