What are the key insights for solving Earliest Second to Mark Indices II?

Insight 1: We can use binary search to find the earliest second. The check function determines if all indices can be marked within a given time frame. Insight 2: The 'check' function should use a greedy approach. For a given second 's', we need to decide when to use the 'changeIndices' to set a specific index of 'nums' to zero, and when to decrement other indices in 'nums'.

Which companies ask LeetCode 3049. Earliest Second to Mark Indices II in interviews?

This problem is frequently asked at MathWorks.

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Earliest Second to Mark Indices II - LeetCode 3049 Solution

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

The optimal solution has a time complexity of O(n + m + n * log(m)) or O(m + nlogm). The binary search contributes a factor of O(log m). Inside each binary search iteration, the `check` function iterates through 's' seconds in the worst case and find the last indices. Finding the last indices takes O(s), where s <= m and s is the number of seconds considered in the simulation. The rest of the check function runs in O(n). The overall complexity is dominated by the final check in each binary search step..

Q: What is the space complexity of Earliest Second to Mark Indices II?

The space complexity is O(n). The `last` dictionary in the `final_check` function stores at most n elements, and we use O(1) extra space in the check function. The original `nums` array is not modified. Other variables are of constant size..

Q: What data structures are used to solve Earliest Second to Mark Indices II?

This problem uses Array, Binary Search, Greedy, Heap (Priority Queue) (Although not explicitly used in the given solution, a heap could potentially be useful for optimization). Understanding these concepts is essential for solving this Hard problem efficiently.

Earliest Second to Mark Indices II - Complete Solution Guide

Earliest Second to Mark Indices II is LeetCode problem 3049, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3.

Problem Statement

You are given two 1-indexed integer arrays, nums and, changeIndices , having lengths n and m , respectively. Initially, all indices in nums are unmarked. Your task is to mark all indices in nums . In each second, s , in order from 1 to m ( inclusive ), you can perform one of the following operations: Choose an index i in the range [1, n] and decrement nums[i] by 1 . Set nums[changeIndices[s]] to any non-negative value. Choose an index i in the range [1, n] , where nums[i] is equal to 0 , and mar

Detailed Explanation

The problem requires us to find the earliest second 's' within the range [1, m] such that we can mark all 'n' indices in the 'nums' array by strategically performing operations up to second 's'. We are given 'nums' (the values to be reduced to zero) and 'changeIndices' (which indicates the index in 'nums' that can be set to any non-negative value at a given second). The allowed operations are: decrementing an element in 'nums', setting an element in 'nums' to any non-negative value according to 'changeIndices', marking an index in 'nums' if its value is 0, and doing nothing. If it's impossible to mark all indices, return -1.

Solution Approach

The solution uses binary search to find the earliest second. The `check` function (more accurately, the `final_check` function) within the binary search validates whether all indices can be marked before a given time. Inside `final_check`, the last occurrence of each index within the `changeIndices` array is tracked. Then, the available operations (either marking or decrementing) are simulated from second 1 to the target second `s`. The greedy approach prioritizes decrementing the indices that have their last chance to be zeroed out. This simulation determines if all indices can indeed be marked within the time limit.

Step-by-Step Algorithm

Step 1: Define the `check` (or `final_check`) function which takes an integer `s` as input. This function determines if all indices can be marked before the given second `s`.
Step 2: In the check function, find the last occurrence of each index in `changeIndices` up to second `s`. If any index is not present, it means it cannot be marked before `s`, so return `False`.
Step 3: Simulate the operations from second 1 to `s`. Prioritize using operations for indices with their last occurrences in `changeIndices`. If not a last occurence, use those operations to decrement other indexes
Step 4: Maintain a count of available operations. For each second, we either decrement a `nums` index, set a `nums` index to 0 (at its last occurrence), do nothing or perform marking.
Step 5: If at any point, it becomes impossible to decrement all necessary `nums` indices, return `False`. If the simulation completes successfully and all indices can be marked before `s`, return `True`.
Step 6: Perform binary search between `n` (a lower bound, minimum number of operations) and `m + 1` (maximum number of operations) to find the smallest `s` for which the `check` function returns `True`.

Key Insights

Insight 1: We can use binary search to find the earliest second. The check function determines if all indices can be marked within a given time frame.
Insight 2: The 'check' function should use a greedy approach. For a given second 's', we need to decide when to use the 'changeIndices' to set a specific index of 'nums' to zero, and when to decrement other indices in 'nums'.
Insight 3: The crucial optimization is realizing the importance of last occurrences. You can't go backwards in time, so the last occurence to change an index in `changeIndices` is where the `num` at that index *must* be set to zero for that time frame to be viable.

Complexity Analysis

Time Complexity: O(n + m + n * log(m)) or O(m + nlogm). The binary search contributes a factor of O(log m). Inside each binary search iteration, the `check` function iterates through 's' seconds in the worst case and find the last indices. Finding the last indices takes O(s), where s <= m and s is the number of seconds considered in the simulation. The rest of the check function runs in O(n). The overall complexity is dominated by the final check in each binary search step.

Space Complexity: O(n). The `last` dictionary in the `final_check` function stores at most n elements, and we use O(1) extra space in the check function. The original `nums` array is not modified. Other variables are of constant size.

Topics

This problem involves: Array, Binary Search, Greedy, Heap (Priority Queue) (Although not explicitly used in the given solution, a heap could potentially be useful for optimization).

Companies

Asked at: MathWorks.

Earliest Second to Mark Indices II | LeetCode 3049

Hard

Array Binary Search Greedy Heap (Priority Queue) (Although not explicitly used in the given solution, a heap could potentially be useful for optimization)

Time: O(n + m + n * log(m)) or O(m + nlogm). The binary search contributes a factor of O(log m). Inside each binary search iteration, the `check` function iterates through 's' seconds in the worst case and find the last indices. Finding the last indices takes O(s), where s <= m and s is the number of seconds considered in the simulation. The rest of the check function runs in O(n). The overall complexity is dominated by the final check in each binary search step.Space: O(n). The `last` dictionary in the `final_check` function stores at most n elements, and we use O(1) extra space in the check function. The original `nums` array is not modified. Other variables are of constant size.

Solution Code

Available in:

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Problem Info

Description

Problem 3049: Earliest Second to Mark Indices II - Detailed explanation and solution approaches.

Topics

Array Binary Search Greedy Heap (Priority Queue) (Although not explicitly used in the given solution, a heap could potentially be useful for optimization)

Companies

MathWorks

Stats

Difficulty:Hard

Languages:2

Companies:1

Problem #:3049

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Define the `check` (or `final_check`) function which takes an integer `s` as input. This function determines if all indices can be marked before the given second `s`.
2
Step 2: In the check function, find the last occurrence of each index in `changeIndices` up to second `s`. If any index is not present, it means it cannot be marked before `s`, so return `False`.
3
Step 3: Simulate the operations from second 1 to `s`. Prioritize using operations for indices with their last occurrences in `changeIndices`. If not a last occurence, use those operations to decrement other indexes
4
Step 4: Maintain a count of available operations. For each second, we either decrement a `nums` index, set a `nums` index to 0 (at its last occurrence), do nothing or perform marking.
5
Step 5: If at any point, it becomes impossible to decrement all necessary `nums` indices, return `False`. If the simulation completes successfully and all indices can be marked before `s`, return `True`.
6
Step 6: Perform binary search between `n` (a lower bound, minimum number of operations) and `m + 1` (maximum number of operations) to find the smallest `s` for which the `check` function returns `True`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n + m + n * log(m)) or O(m + nlogm). The binary search contributes a factor of O(log m). Inside each binary search iteration, the `check` function iterates through 's' seconds in the worst case and find the last indices. Finding the last indices takes O(s), where s <= m and s is the number of seconds considered in the simulation. The rest of the check function runs in O(n). The overall complexity is dominated by the final check in each binary search step.

🚀 Logarithmic - Very efficient! Halves the problem each step.👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n). The `last` dictionary in the `final_check` function stores at most n elements, and we use O(1) extra space in the check function. The original `nums` array is not modified. Other variables are of constant size.

💪 Constant space - Uses same memory regardless of input!📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived by the binary search combined with the operations performed within the `check` function. The `check` function initializes a dictionary with a size of at most O(n) in order to store last occurrences. Space is largely determined by the size of changeIndices and nums.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Heap (Priority Queue) based Greedy

Instead of using a plain greedy and binary search, you can use a heap to prioritize marking/setting to zero indices based on some criteria (e.g., the number of operations required). Although the overall time complexity might not improve drastically, it could offer better performance in certain scenarios by making more informed decisions at each step.

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

Trade-offs: The heap approach might be more complex to implement. While it *could* be faster for some inputs due to its prioritisation, in the worst case might be slightly slower than the binary search approach because of heap operations. The simpler approach will be more predictable.

✨

Key Insights

1Insight 1: We can use binary search to find the earliest second. The check function determines if all indices can be marked within a given time frame.
2Insight 2: The 'check' function should use a greedy approach. For a given second 's', we need to decide when to use the 'changeIndices' to set a specific index of 'nums' to zero, and when to decrement other indices in 'nums'.
3Insight 3: The crucial optimization is realizing the importance of last occurrences. You can't go backwards in time, so the last occurence to change an index in `changeIndices` is where the `num` at that index *must* be set to zero for that time frame to be viable.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: When it's impossible to mark all indices because the 'changeIndices' array doesn't cover all indices in 'nums'. The solution explicitly checks for `len(last) < n` in the `final_check` function to handle this case.
Edge case 2: When the operations required for decrements and marks exceed the number of available seconds, even if all indices are covered in `changeIndices`. The solution accurately calculates available ops to test viability.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider the time constraints. A naive greedy approach might attempt to mark indices in an arbitrary order without considering the limited number of operations available at each second.

✗Common mistake 2: Not handling the 'impossible' cases correctly. It's essential to verify that the 'changeIndices' array covers all indices in 'nums' and that there are enough operations available to decrement the `nums` to zero for each index.

✗Common mistake 3: Overlooking last occurrences. Focusing on first occurrences leads to a more complicated and less efficient algorithm as the last occurrence in the given time limit is what ultimately matters.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation optimization, where you have a limited amount of time and resources to complete multiple tasks, each with its own dependencies and deadlines. The problem is to find the earliest possible deadline to complete all tasks within the given constraints.

🏦Real-world scenario 2: Scheduling and task management where you need to schedule tasks with dependencies and constraints, aiming to minimize the overall completion time.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 17, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Earliest Second to Mark Indices II"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?