How do you solve Earliest Second to Mark Indices I on LeetCode?

Earliest Second to Mark Indices I (LeetCode #3048) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m log m) time complexity and O(n) space complexity. Key insight: Understanding the last occurrence of indices helps optimize marking.

What is the brute force solution for Earliest Second to Mark Indices I?

The brute force approach for Earliest Second to Mark Indices I is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we simulate each second and decrement the values in the nums array based on the changeIndices. We check if we can mark indices as we go, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n + m log m).

What algorithm or data structure is used to solve Earliest Second to Mark Indices I?

Earliest Second to Mark Indices I uses the following concepts: Array, Binary Search. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Earliest Second to Mark Indices I?

Always clarify the problem constraints and input format. Think aloud while solving; it shows your thought process. Practice similar problems to recognize patterns and improve efficiency.

What are common mistakes when solving Earliest Second to Mark Indices I?

Failing to account for the 1-indexed nature of the input arrays. Not considering edge cases where marking might be impossible.

#3048

Earliest Second to Mark Indices I

Medium

ArrayBinary SearchHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m log m)
Space	O(1)	O(n)

💡

Intuition

Time O(n + m log m)Space O(n)

This approach uses binary search to efficiently determine the earliest second when all indices can be marked. We can mark indices as late as possible, leveraging the last occurrence of each index in changeIndices.

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Algorithm

3 steps

1Step 1: Create a last occurrence array to store the last index where each index in nums can be marked.
2Step 2: For each index in nums, find the last occurrence in changeIndices using binary search.
3Step 3: Check if the last occurrence is valid and return the maximum of these values.

solution.py13 lines

1def earliest_time_to_mark(nums, changeIndices):
2    from bisect import bisect_right
3    n = len(nums)
4    last_occurrence = [-1] * n
5    for second in range(len(changeIndices)):
6        idx = changeIndices[second] - 1
7        last_occurrence[idx] = second
8    max_time = 0
9    for i in range(n):
10        if nums[i] > last_occurrence[i] + 1:
11            return -1
12        max_time = max(max_time, last_occurrence[i] + 1 - nums[i])
13    return max_time

ℹ

Complexity note: The time complexity is O(n + m log m) due to the need to fill the last occurrence array and check conditions, while the space complexity is O(n) for storing the last occurrences.

1Understanding the last occurrence of indices helps optimize marking.
2Binary search can be a powerful tool for efficiently finding positions in sorted data.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.