What is the time complexity of Maximum Number of Consecutive Values You Can Make?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Maximum Number of Consecutive Values You Can Make?

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

What are the key insights for solving Maximum Number of Consecutive Values You Can Make?

Insight 1: Sorting the coins is crucial because it allows us to greedily determine the maximum reachable value. We process the coins in ascending order. Insight 2: The key idea is to maintain a 'reachable' variable, which represents the maximum consecutive value we can make so far (starting from 0). If the next coin's value is greater than 'reachable', we can no longer extend the consecutive sequence.

Which companies ask LeetCode 1798. Maximum Number of Consecutive Values You Can Make in interviews?

This problem is frequently asked at Infosys.

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Maximum Number of Consecutive Values You Can Make - LeetCode 1798 Solution

Maximum Number of Consecutive Values You Can Make - Complete Solution Guide

Maximum Number of Consecutive Values You Can Make is LeetCode problem 1798, 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 an integer array coins of length n which represents the n coins that you own. The value of the i th coin is coins[i] . You can make some value x if you can choose some of your n coins such that their values sum up to x . Return the maximum number of consecutive integer values that you can make with your coins starting from and including 0 . Note that you may have multiple coins of the same value. Example 1: Input: coins = [1,3] Output: 2 Explanation: You can make the following valu

Detailed Explanation

The problem asks us to find the maximum number of consecutive integer values that can be made using a given set of coins, starting from 0. We are given an array of coin values, and we can choose any combination of these coins to form a sum. The goal is to find the largest consecutive sequence of numbers (starting from 0) that we can create using these sums. For example, if coins = [1, 3], we can make 0 (no coins), 1 (using the 1 coin), but we cannot make 2. So the answer is 2. If coins = [1, 1, 1, 4], we can make 0, 1, 2, 3, 4, 5, 6, 7. Thus the answer is 8.

Solution Approach

The solution uses a greedy approach. First, the coins array is sorted in ascending order. This ensures that we consider smaller denominations first, which is crucial for the greedy strategy. We maintain a variable 'reachable' to represent the maximum consecutive value we can make starting from 0. We iterate through the sorted coins. For each coin, if its value is greater than 'reachable', it means we cannot extend the consecutive sequence any further, and we can stop. Otherwise, we add the coin's value to 'reachable' to extend the sequence.

Step-by-Step Algorithm

Step 1: Sort the `coins` array in ascending order.
Step 2: Initialize a variable `reachable` to 1. This represents the fact that we can always make 0 (by taking no coins), so the consecutive range starts at 0, making the maximum consecutive value at least 1 (the value 0).
Step 3: Iterate through the sorted `coins` array.
Step 4: For each `coin`, check if `coin > reachable`. If it is, break out of the loop because we can no longer extend the consecutive range.
Step 5: If `coin <= reachable`, add `coin` to `reachable`. This extends the consecutive range because now we can make all values from 0 to `reachable + coin - 1`.
Step 6: After iterating through all the coins (or breaking the loop), return the final value of `reachable`.

Key Insights

Insight 1: Sorting the coins is crucial because it allows us to greedily determine the maximum reachable value. We process the coins in ascending order.
Insight 2: The key idea is to maintain a 'reachable' variable, which represents the maximum consecutive value we can make so far (starting from 0). If the next coin's value is greater than 'reachable', we can no longer extend the consecutive sequence.
Insight 3: If a coin's value is less than or equal to the current 'reachable' value, we can add the coin to all the values from 0 to 'reachable' to extend the range of consecutive values we can make.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: Infosys.

Maximum Number of Consecutive Values You Can Make | LeetCode 1798

Medium

Array Greedy Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int getMaximumConsecutive(int[] coins) {
        Arrays.sort(coins);
        int reachable = 1;
        for (int coin : coins) {
            if (coin > reachable) {
                break;
            }
            reachable += coin;
        }
        return reachable;
    }
}

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Problem Info

Description

Problem 1798: Maximum Number of Consecutive Values You Can Make - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

Infosys

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1798

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: Sort the `coins` array in ascending order.
2
Step 2: Initialize a variable `reachable` to 1. This represents the fact that we can always make 0 (by taking no coins), so the consecutive range starts at 0, making the maximum consecutive value at least 1 (the value 0).
3
Step 3: Iterate through the sorted `coins` array.
4
Step 4: For each `coin`, check if `coin > reachable`. If it is, break out of the loop because we can no longer extend the consecutive range.
5
Step 5: If `coin <= reachable`, add `coin` to `reachable`. This extends the consecutive range because now we can make all values from 0 to `reachable + coin - 1`.
6
Step 6: After iterating through all the coins (or breaking the loop), return the final value of `reachable`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the sorting algorithm. Common sorting algorithms like merge sort or quicksort have a time complexity of O(n log n). The space complexity is O(1) because we only use a few extra variables to store the current maximum reachable value and the coin value during iteration. No additional data structures that scale with input 'n' are utilized.

📖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

Dynamic Programming

We can use dynamic programming to determine all the possible sums that can be formed. Create a boolean array `dp` where `dp[i]` is true if the sum `i` can be formed. Initialize `dp[0]` to true. Then, for each coin, iterate through the `dp` array and update the reachable sums. However, this might take too much time.

Time: O(n * sum), where n is the number of coins and sum is the total sum of all coins.Space: O(sum), where sum is the total sum of all coins.

Trade-offs: The dynamic programming approach can be useful if you need to know *all* possible sums that can be formed, not just the maximum consecutive range. However, it is generally less efficient than the greedy approach because the 'sum' can be very large.

✨

Key Insights

1Insight 1: Sorting the coins is crucial because it allows us to greedily determine the maximum reachable value. We process the coins in ascending order.
2Insight 2: The key idea is to maintain a 'reachable' variable, which represents the maximum consecutive value we can make so far (starting from 0). If the next coin's value is greater than 'reachable', we can no longer extend the consecutive sequence.
3Insight 3: If a coin's value is less than or equal to the current 'reachable' value, we can add the coin to all the values from 0 to 'reachable' to extend the range of consecutive values we can make.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty `coins` array. The code implicitly handles this. If the input `coins` array is empty, the loop will not execute, and the initial value of `reachable`, which is 1, will be returned, as we can always make zero using no coins.
Edge case 2: All coins are 1. In this case, the `reachable` value will keep increasing by 1 for each coin, and the final `reachable` value will be the number of coins plus 1.
Edge case 3: Large coin values early in the array. Sorting ensures that these large coin values are considered later, allowing smaller coins to contribute to a larger consecutive range before encountering the large coin values.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the `coins` array. Without sorting, the greedy approach will not work correctly, and you will not be able to find the maximum consecutive range.

✗Common mistake 2: Assuming that the coins are always unique. The problem statement mentions that there may be multiple coins of the same value, so the solution must correctly handle duplicate coin values.

✗Common mistake 3: Incorrectly initializing the `reachable` variable. It should be initialized to 1 because we can always make the value 0 by taking no coins.

✅

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: Optimizing resource allocation. Imagine you have a set of resources with different sizes. This algorithm can help you determine the maximum consecutive sizes you can create by combining these resources.

🏦Real-world scenario 2: Inventory management. Suppose you have a set of items with different weights. This algorithm can help to determine the maximum consecutive weights that can be achieved by combining these items.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 26, 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 verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

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 "Maximum Number of Consecutive Values You Can Make"?

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 greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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