#3048

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Earliest Second to Mark Indices I

The problem asks to determine the earliest second to mark all indices in an array using a set of operations.

Problem Statement

You are given two 1-indexed integer arrays, nums and changeIndices, with lengths n and m, respectively. Initially, all indices in nums are unmarked. Your task is to mark all indices in nums as fast as possible using a set of operations.

In each second s, from 1 to m (inclusive), you can perform one of the following operations: decrement one element from nums or mark an index from changeIndices if the corresponding element in nums is zero. Your goal is to find the earliest second when all indices are marked.

Examples

Example 1

Input: nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1]

Output: 8

In this example, we have 8 seconds. The following operations can be performed to mark all indices: Second 1: Choose index 1 and decrement nums[1] by one. nums becomes [1,2,0]. Second 2: Choose index 1 and decrement nums[1] by one. nums becomes [0,2,0]. Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [0,1,0]. Second 4: Choose index 2 and decrement nums[2] by one. nums becomes [0,0,0]. Second 5: Mark the index changeIndices[5], which is marking index 3, since nums[3] is equal to 0. Second 6: Mark the index changeIndices[6], which is marking index 2, since nums[2] is equal to 0. Second 7: Do nothing. Second 8: Mark the index changeIndices[8], which is marking index 1, since nums[1] 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 8th second. Hence, the answer is 8.

Example 2

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

Output: 6

In this example, we have 7 seconds. The following operations can be performed to mark all indices: Second 1: Choose index 2 and decrement nums[2] by one. nums becomes [1,2]. Second 2: Choose index 2 and decrement nums[2] by one. nums becomes [1,1]. Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [1,0]. Second 4: Mark the index changeIndices[4], which is marking index 2, since nums[2] is equal to 0. Second 5: Choose index 1 and decrement nums[1] by one. nums becomes [0,0]. Second 6: Mark the index changeIndices[6], which is marking index 1, since nums[1] 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 3

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

Output: -1

In this example, it is impossible to mark all indices because index 1 isn't in changeIndices. Hence, the answer is -1.

Constraints

1 <= n == nums.length <= 2000
0 <= nums[i] <= 109
1 <= m == changeIndices.length <= 2000
1 <= changeIndices[i] <= n

Solution Approach

Simulate the Marking Process

We need to simulate marking all indices, checking how many seconds it takes to fully mark nums. Each second, decrement nums or mark an index from changeIndices if its corresponding value in nums is zero. If we mark all indices within a certain time, it indicates that a potential solution exists.

Apply Binary Search

We can use binary search over the valid answer space (seconds) to efficiently find the minimum time required to mark all indices. The lower bound is 1 (the first second), and the upper bound is m (the total number of seconds). Using binary search helps narrow down the earliest time faster than a brute force solution.

Check Feasibility at Each Step

At each step of binary search, check if it's possible to mark all indices by simulating the marking process for the current midpoint. If it's feasible, narrow the search to earlier times; otherwise, extend the search to later times.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the binary search, which has a time complexity of O(log m), combined with the simulation of marking all indices, which takes O(n) time. Hence, the total time complexity is O(n log m). The space complexity is O(n) due to the storage of the nums array and the changeIndices array.

What Interviewers Usually Probe

The candidate is expected to demonstrate understanding of binary search as a tool for narrowing down the solution space.
The candidate should simulate the process to check for valid solutions at each time step, ensuring they can handle edge cases.
The ability to analyze time complexity and use efficient algorithms (like binary search) is a key factor in evaluating the candidate.

Common Pitfalls or Variants

Common pitfalls

Failing to account for the edge case where it's impossible to mark all indices, which should return -1.
Not using binary search effectively and resorting to a brute force approach, which would not scale well with larger inputs.
Misunderstanding the marking mechanism, leading to incorrect simulations and premature conclusions.

Follow-up variants

Consider variations where multiple indices can be marked in the same second.
Handling larger arrays or larger values in nums could require adjustments in both time complexity and space complexity.
Introducing constraints on the order of operations could lead to different approaches to binary search or simulation.

FAQ

How does binary search help in solving this problem?

Binary search helps by narrowing down the range of possible answers (the earliest second) and checking if it's feasible to mark all indices at that time.

What happens if it's impossible to mark all indices?

If it's impossible to mark all indices, the answer should be -1, which can be detected by simulating the process and checking if all indices are marked within the given time.

What is the time complexity of the solution?

The time complexity is O(n log m), where n is the length of nums and m is the length of changeIndices. Binary search narrows the search space, and the simulation takes O(n) time.

How do we handle edge cases like missing indices in changeIndices?

Edge cases where an index is missing in changeIndices should be handled by checking if all elements in nums can be decremented to zero, and if not, returning -1.

Can this problem be solved without binary search?

It is possible to solve the problem without binary search, but binary search significantly reduces the time complexity compared to a brute force approach, making it more efficient.

terminal

Solution

Solution 1: Binary Search

We notice that if we can mark all indices within $t$ seconds, then we can also mark all indices within $t' \geq t$ seconds. Therefore, we can use binary search to find the earliest seconds.

class Solution:
    def earliestSecondToMarkIndices(
        self, nums: List[int], changeIndices: List[int]
    ) -> int:
        def check(t: int) -> bool:
            decrement = 0
            marked = 0
            last = {i: s for s, i in enumerate(changeIndices[:t])}
            for s, i in enumerate(changeIndices[:t]):
                if last[i] == s:
                    if decrement < nums[i - 1]:
                        return False
                    decrement -= nums[i - 1]
                    marked += 1
                else:
                    decrement += 1
            return marked == len(nums)

        m = len(changeIndices)
        l = bisect_left(range(1, m + 2), True, key=check) + 1
        return -1 if l > m else l

Continue Practicing

#3049 earliest second to mark indices ii #3113 find the number of subarrays where boundary elements are maximum #3116 kth smallest amount with single denomination combination