Class Solution

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    public final class Solution
    
                        

    3049 - Earliest Second to Mark Indices II.

    Hard

    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:

    • Choose an index i in the range [1, n] and decrement nums[i] by 1.

    • Set nums[changeIndices[s]] to any non-negative value.

    • Choose an index i in the range [1, n], where nums[i] is equal to 0, and mark index i.

    • Do nothing.

    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.

    Example 1:

    Input: nums = 3,2,3, changeIndices = 1,3,2,2,2,2,3

    Output: 6

    Explanation: In this example, we have 7 seconds. The following operations can be performed to mark all indices:

    Second 1: Set nums[changeIndices1] to 0. nums becomes 0,2,3.

    Second 2: Set nums[changeIndices2] to 0. nums becomes 0,2,0.

    Second 3: Set nums[changeIndices3] to 0. nums becomes 0,0,0.

    Second 4: Mark index 1, since nums1 is equal to 0.

    Second 5: Mark index 2, since nums2 is equal to 0.

    Second 6: Mark index 3, since nums3 is equal to 0.

    Now all indices have been marked.

    It can be shown that it is not possible to mark all indices earlier than the 6th second.

    Hence, the answer is 6.

    Example 2:

    Input: nums = 0,0,1,2, changeIndices = 1,2,1,2,1,2,1,2

    Output: 7

    Explanation: In this example, we have 8 seconds. The following operations can be performed to mark all indices:

    Second 1: Mark index 1, since nums1 is equal to 0.

    Second 2: Mark index 2, since nums2 is equal to 0.

    Second 3: Decrement index 4 by one. nums becomes 0,0,1,1.

    Second 4: Decrement index 4 by one. nums becomes 0,0,1,0.

    Second 5: Decrement index 3 by one. nums becomes 0,0,0,0.

    Second 6: Mark index 3, since nums3 is equal to 0.

    Second 7: Mark index 4, since nums4 is equal to 0.

    Now all indices have been marked.

    It can be shown that it is not possible to mark all indices earlier than the 7th second.

    Hence, the answer is 7.

    Example 3:

    Input: nums = 1,2,3, changeIndices = 1,2,3

    Output: -1

    Explanation: In this example, it can be shown that it is impossible to mark all indices, as we don't have enough seconds. Hence, the answer is -1.

    Constraints:

    • 1 <= n == nums.length <= 5000

    • <code>0 <= numsi<= 10<sup>9</sup></code>

    • 1 &lt;= m == changeIndices.length &lt;= 5000

    • 1 &lt;= changeIndices[i] &lt;= n

    • Nested Class Summary

      Nested Classes 
      Modifier and Type Class Description
    • Field Summary

      Fields 
      Modifier and Type Field Description
    • Constructor Summary

      Constructors 
      Constructor Description
      Solution()
    • Enum Constant Summary

      Enum Constants 
      Enum Constant Description
    • Method Summary

      Modifier and Type Method Description
      final Integer earliestSecondToMarkIndices(IntArray nums, IntArray changeIndices)
      • Methods inherited from class java.lang.Object

        clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait