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

Earliest Second to Mark Indices II (LeetCode #3049) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(n)) time complexity and O(n) space complexity. Key insight: We need at least n seconds to mark all indices, as each index must be marked once.

What is the time complexity of Earliest Second to Mark Indices II?

The optimal solution for Earliest Second to Mark Indices II runs in O(n log(n)) time and uses O(n) space. The time complexity is O(n log(n)) due to the binary search over the seconds (log(n)) and checking the marking process (O(n)).

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

The brute force approach for Earliest Second to Mark Indices II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating every second and checking if we can mark all indices by the end of the given seconds. We will try every possible operation at each second and see if we can mark all indices by the end. The optimal Optimal Solution approach improves this to O(n log(n)).

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

Earliest Second to Mark Indices II uses the following concepts: Array, Binary Search, Greedy, Heap (Priority Queue). The recommended patterns to study are: Binary Search, Greedy Algorithms.

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

Always think about edge cases, such as when nums already contains zeros. Explain your thought process clearly while coding; it helps the interviewer understand your reasoning. Practice binary search problems to gain confidence in optimizing search space.

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

Not considering the case where it's impossible to mark all indices. Failing to optimize the search space and using brute force unnecessarily.

#3049

Earliest Second to Mark Indices II

Hard

ArrayBinary SearchGreedyHeap (Priority Queue)Binary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log(n))Space O(n)

The optimal solution leverages binary search to find the earliest second where all indices can be marked. We calculate if it's possible to mark all indices by a certain second using a greedy approach.

⚙️

Algorithm

3 steps

1Step 1: Define a function to check if all indices can be marked by a given second.
2Step 2: Use binary search on the range [n, sum(nums) + n] to find the earliest second.
3Step 3: For each mid-point in the binary search, simulate the marking process and check if all indices can be marked.

solution.py27 lines

1# Full working Python code
2
3def can_mark_all(nums, changeIndices, second):
4    n = len(nums)
5    marked = [False] * n
6    available = 0
7    for s in range(second):
8        idx = changeIndices[s] - 1
9        nums[idx] = 0
10        available += 1
11        for i in range(n):
12            if nums[i] == 0 and not marked[i]:
13                marked[i] = True
14                available -= 1
15    return all(marked) and available >= 0
16
17
18def earliest_second(nums, changeIndices):
19    left, right = len(nums), sum(nums) + len(nums)
20    while left < right:
21        mid = (left + right) // 2
22        if can_mark_all(nums[:], changeIndices, mid):
23            right = mid
24        else:
25            left = mid + 1
26    return left if can_mark_all(nums[:], changeIndices, left) else -1
27

ℹ

Complexity note: The time complexity is O(n log(n)) due to the binary search over the seconds (log(n)) and checking the marking process (O(n)).

1We need at least n seconds to mark all indices, as each index must be marked once.
2The maximum possible seconds needed is sum(nums) + n, as we need to decrement values before marking.

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