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

The optimal solution has a time complexity of O(m * (n + m)).

What data structures are used to solve Earliest Second to Mark Indices I?

This problem uses Array, Binary Search. Understanding these concepts is essential for solving this Medium problem efficiently.

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

Insight 1: The problem can be solved using binary search because we are looking for the *earliest* second. If it's possible to mark all indices by second `k`, then it's also possible by any second greater than `k` (though not necessarily optimally). Insight 2: A crucial step is devising a `check` function that determines if it's possible to mark all indices within a given number of seconds `k`. This function simulates the process of decrementing and marking, trying to optimize the operations.

Which companies ask LeetCode 3048. Earliest Second to Mark Indices I in interviews?

This problem is frequently asked at MathWorks.

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Earliest Second to Mark Indices I - LeetCode 3048 Solution

Earliest Second to Mark Indices I - Complete Solution Guide

Earliest Second to Mark Indices I is LeetCode problem 3048, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 . If nums[changeIndices[s]] is equal to 0 , mark the index changeIndices[s] . Do nothing. Return an integer denoting the earliest sec

Detailed Explanation

The problem asks us to find the earliest second `s` in the `changeIndices` array such that all indices in the `nums` array can be marked. We are given `nums` (representing the initial values at each index) and `changeIndices` (representing the index that can be marked at each second if its corresponding value in `nums` is 0). In each second, we can either decrement a value in `nums`, mark an index if its corresponding value in `nums` is 0 and it appears in `changeIndices` at that second, or do nothing. The goal is to find the smallest `s` such that all indices in `nums` are marked by second `s`. If it is not possible to mark all indices, return -1.

Solution Approach

The solution uses binary search to find the earliest second. The `check` function simulates the marking process for a given number of seconds. It first finds the last occurrence of each index in `changeIndices` within the range [0, k-1]. If any index is missing, it's impossible to mark all indices, so the function returns `false`. Then, it iterates through the first `k` seconds. If the current second is the last time an index appears, it *must* be marked at that time. Before marking, we need to ensure enough decrement operations have been performed to reduce the corresponding `nums` value to 0. If not, return `false`. Otherwise, use decrement operations to reduce the value. Seconds where the current index is *not* the last occurrence can be used for decrement operations.

Step-by-Step Algorithm

Step 1: Initialize `low` to 1 and `high` to `m` (the length of `changeIndices`). Initialize `ans` to -1.
Step 2: Perform binary search between `low` and `high`.
Step 3: In each iteration of the binary search, calculate `mid` as the middle point.
Step 4: Call the `check` function with `mid`. If `check(mid)` returns `true`, update `ans` to `mid` and set `high` to `mid - 1` to search for an earlier second. Otherwise, set `low` to `mid + 1`.
Step 5: The `check` function does the following:
- Create a `last_pos` map to store the last occurrence of each index in `changeIndices` within the first `k` seconds.
- If the number of unique indices found is not equal to `n` (the length of `nums`), return `false` (not all indices can be marked).
- Initialize `free_slots` to 0.
- Iterate through the first `k` seconds. For each second `s`:
- If `s` is the last occurrence of `changeIndices[s]` in the first `k` seconds:
- Calculate `decrements_needed` as `nums[changeIndices[s] - 1]`.
- If `free_slots` is less than `decrements_needed`, return `false` (not enough decrement operations available).
- Subtract `decrements_needed` from `free_slots`.
- Otherwise, increment `free_slots` by 1 (use this second for a decrement operation).
- If the loop completes without returning `false`, return `true` (all indices can be marked).
Step 6: After the binary search completes, return `ans`.

Key Insights

Insight 1: The problem can be solved using binary search because we are looking for the *earliest* second. If it's possible to mark all indices by second `k`, then it's also possible by any second greater than `k` (though not necessarily optimally).
Insight 2: A crucial step is devising a `check` function that determines if it's possible to mark all indices within a given number of seconds `k`. This function simulates the process of decrementing and marking, trying to optimize the operations.
Insight 3: We need to keep track of the *last* time each index appears in `changeIndices` within the considered time frame (first `k` seconds). This is important because that second is when that index *must* be marked if possible.

Complexity Analysis

Time Complexity: O(m * (n + m))

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: MathWorks.

Earliest Second to Mark Indices I | LeetCode 3048

Medium

Array Binary Search

Time: O(m * (n + m))Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int earliestSecondToMarkIndices(int[] nums, int[] changeIndices) {
        int n = nums.length;
        int m = changeIndices.length;

        // k is a 1-based second. We are checking using the first k seconds.
        // This corresponds to changeIndices[0...k-1].
        
        
        int low, high;
        int ans;
        
        low = 1;
        high = m;
        ans = -1;
                
        while(low <= high) {
            int mid = low + (high - low) / 2;
            
            if (check(nums, changeIndices, mid)) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        
        return ans;
    }

    private boolean check(int[] nums, int[] changeIndices, int k) {
        int n = nums.length;
        Map<Integer, Integer> last_pos = new HashMap<>();
        for (int s = 0; s < k; s++) {
            // changeIndices values are 1-based.
            last_pos.put(changeIndices[s], s);
        }

        if (last_pos.size() != n) {
            return false;
        }

        int free_slots = 0;
        for (int s = 0; s < k; s++) {
            int idx_val = changeIndices[s];
            // If current second s is the last time this index is mentioned in changeIndices[0...k-1]
            if (last_pos.get(idx_val) == s) {
                // This is the designated second to mark index idx_val.
                // We must have performed nums[idx_val - 1] decrements by this time.
                int decrements_needed = nums[idx_val - 1];
                if (free_slots < decrements_needed) {
                    return false;
                }
                // Use the free slots for decrements.
                free_slots -= decrements_needed;
            } else {
                // This second is not a designated marking day for changeIndices[s].
                // It can be used as a free slot for a decrement operation.
                free_slots += 1;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 3048: Earliest Second to Mark Indices I - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

MathWorks

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3048

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

16 steps

1
Step 1: Initialize `low` to 1 and `high` to `m` (the length of `changeIndices`). Initialize `ans` to -1.
2
Step 2: Perform binary search between `low` and `high`.
3
Step 3: In each iteration of the binary search, calculate `mid` as the middle point.
4
Step 4: Call the `check` function with `mid`. If `check(mid)` returns `true`, update `ans` to `mid` and set `high` to `mid - 1` to search for an earlier second. Otherwise, set `low` to `mid + 1`.
5
Step 5: The `check` function does the following:
6
- Create a `last_pos` map to store the last occurrence of each index in `changeIndices` within the first `k` seconds.
7
- If the number of unique indices found is not equal to `n` (the length of `nums`), return `false` (not all indices can be marked).
8
- Initialize `free_slots` to 0.
9
- Iterate through the first `k` seconds. For each second `s`:
10
- If `s` is the last occurrence of `changeIndices[s]` in the first `k` seconds:
11
- Calculate `decrements_needed` as `nums[changeIndices[s] - 1]`.
12
- If `free_slots` is less than `decrements_needed`, return `false` (not enough decrement operations available).
13
- Subtract `decrements_needed` from `free_slots`.
14
- Otherwise, increment `free_slots` by 1 (use this second for a decrement operation).
15
- If the loop completes without returning `false`, return `true` (all indices can be marked).
16
Step 6: After the binary search completes, return `ans`.

💡

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(m * (n + m))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The binary search divides the search space in half with each iteration, leading to a logarithmic factor. The check function's time complexity is dominated by the loops that iterate through `changeIndices`. The `last_pos` map stores at most `n` unique keys, hence the space complexity.

📖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

Greedy Approach (Incorrect)

A naive greedy approach might try to mark each index as soon as possible. However, this is unlikely to be optimal because it might exhaust decrement operations too early, preventing other indices from being marked.

Time: Potentially O(m*n) or worse if not carefully implemented.Space: O(n)

Trade-offs: This is generally not a valid approach. The binary search approach is much more reliable and correct.

✨

Key Insights

1Insight 1: The problem can be solved using binary search because we are looking for the *earliest* second. If it's possible to mark all indices by second `k`, then it's also possible by any second greater than `k` (though not necessarily optimally).
2Insight 2: A crucial step is devising a `check` function that determines if it's possible to mark all indices within a given number of seconds `k`. This function simulates the process of decrementing and marking, trying to optimize the operations.
3Insight 3: We need to keep track of the *last* time each index appears in `changeIndices` within the considered time frame (first `k` seconds). This is important because that second is when that index *must* be marked if possible.

📚

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!

Edge Cases

Edge case 1: If not all indices in `nums` appear in `changeIndices`, it is impossible to mark all indices. The `check` function handles this by verifying if the size of `last_pos` is equal to `n`. If it's not, it returns `false`.
Edge case 2: When `nums[i]` is very large, and there are not enough decrement operations available before the last occurrence of `changeIndices[s]`, it's impossible to mark all indices. The `check` function handles this by tracking `free_slots` and returning `false` if it's less than the required decrements.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if all indices appear in `changeIndices`. If an index doesn't appear, it's impossible to mark it.

✗Common mistake 2: Not handling the case where the available `free_slots` are insufficient to decrement `nums[i]` to zero before marking the index.

✗Common mistake 3: Off-by-one errors when dealing with 1-based indexing in `nums` and `changeIndices`.

✅

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 with deadlines. Imagine `nums` represents the initial workload of different tasks, `changeIndices` represents the days when a task can be completed (marked), and you have limited resources (represented by `free_slots`) to reduce the workload before the task's deadline. Finding the earliest completion time is analogous to the problem.

🏦Real-world scenario 2: Software testing and bug fixing. `nums` can represent the severity of different bugs, `changeIndices` represents the days when a particular bug can be fixed, and `free_slots` represents the available development effort to reduce the bug's severity. The problem then becomes finding the earliest time all bugs can be resolved.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 16, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

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

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

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

For this medium-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?