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] <= 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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