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 lengthsn
andm
, respectively.Initially, all indices in
nums
are unmarked. Your task is to mark all indices innums
.In each second,
s
, in order from1
tom
( inclusive ), you can perform one of the following operations:Choose an index
i
in the range[1, n]
and decrementnums[i]
by1
.Set
nums[changeIndices[s]]
to any non-negative value.Choose an index
i
in the range[1, n]
, wherenums[i]
is equal to0
, and mark indexi
.Do nothing.
Return an integer denoting the earliest second in the range
[1, m]
when all indices innums
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 <= m == changeIndices.length <= 5000
1 <= changeIndices[i] <= n
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final Integer
earliestSecondToMarkIndices(IntArray nums, IntArray changeIndices)
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Method Detail
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earliestSecondToMarkIndices
final Integer earliestSecondToMarkIndices(IntArray nums, IntArray changeIndices)
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