Earliest Second to Mark Indices II

Earliest Second to Mark Indices II - Problem

Earliest Second to Mark Indices II is a challenging optimization problem that simulates a time-based marking system.

You are given two arrays: nums (1-indexed, length n) containing positive integers, and changeIndices (1-indexed, length m) containing indices that reference positions in nums.

Goal: Mark all indices in nums in the minimum number of seconds.

Rules for each second s (from 1 to m):
1. Decrement: Choose any index i and reduce nums[i] by 1
2. Reset: Set nums[changeIndices[s]] to any non-negative value
3. Mark: Mark index i if nums[i] == 0
4. Do nothing

An index can only be marked when its value reaches 0. The challenge is to optimally use reset operations (which can instantly zero out values) and decrement operations to mark everything as quickly as possible.

Return: The earliest second when all indices can be marked, or -1 if impossible.

Input & Output

example_1.py — Basic Case

$ Input: nums = [2, 2, 0], changeIndices = [2, 2, 2, 2, 3, 2, 2, 2]

› Output: 8

💡 Note: We can mark all indices by second 8. Use reset operations on indices 2 and 3 at their last available times, then decrement and mark index 1 (which has no reset opportunity).

example_2.py — Reset Required

$ Input: nums = [1, 3], changeIndices = [1, 1, 1, 2, 1, 1, 1]

› Output: 6

💡 Note: Reset index 2 at second 4, mark it at second 5. For index 1, decrement once and mark, which takes 2 more operations. Total minimum time is 6 seconds.

example_3.py — Impossible Case

$ Input: nums = [0, 1, 1, 2], changeIndices = [1, 1, 2]

› Output: -1

💡 Note: Index 4 (with value 2) never appears in changeIndices, so we need 3 operations (2 decrements + 1 mark). But we only have 3 seconds total, and we also need to handle other indices. Impossible.

Constraints

1 ≤ n == nums.length ≤ 5000
0 ≤ nums[i] ≤ 10⁹
1 ≤ m == changeIndices.length ≤ 5000
1 ≤ changeIndices[i] ≤ n
Time limit: 2000ms for optimal solution

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8 ⊞ Microsoft 6

The optimal approach uses binary search on the answer combined with greedy simulation. Key insights: (1) Binary search the minimum time needed from 1 to m, (2) For each candidate time, use greedy strategy to delay reset operations until their latest possible moment, (3) Calculate if all required operations can fit within the time limit. Time complexity is O(m log m) making it efficient for the given constraints.

Common Approaches

✓ Math

⏱️ Time: N/A Space: N/A

Binary Search + Greedy Simulation

⏱️ Time: O(m log m) Space: O(n + m)

Use binary search to find the minimum number of seconds needed. For each candidate answer, use a greedy simulation to determine if it's possible to mark all indices within that time limit. The key insight is to delay reset operations until the latest possible moment.

Brute Force Simulation

⏱️ Time: O(4^m * n) Space: O(n)

For each possible ending second from 1 to m, simulate all possible sequences of actions to see if we can mark all indices by that time. This involves trying all combinations of operations at each second.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

⚡ Linearithmic

Space Complexity

⚡ Linearithmic Space

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