What is the time complexity of Maximum Number of Integers to Choose From a Range I?

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

What is the space complexity of Maximum Number of Integers to Choose From a Range I?

The space complexity is O(len(banned)).

What data structures are used to solve Maximum Number of Integers to Choose From a Range I?

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

What are the key insights for solving Maximum Number of Integers to Choose From a Range I?

Insight 1: Using a set (or hash table) to store the banned numbers allows for O(1) lookup to quickly check if a number is banned. Insight 2: Iterating sequentially from 1 to 'n' is a greedy approach that prioritizes smaller numbers. Since we want to maximize the *count* of numbers, this approach is optimal.

Which companies ask LeetCode 2554. Maximum Number of Integers to Choose From a Range I in interviews?

This problem is frequently asked at PayPal.

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Maximum Number of Integers to Choose From a Range I - LeetCode 2554 Solution

Maximum Number of Integers to Choose From a Range I - Complete Solution Guide

Maximum Number of Integers to Choose From a Range I is LeetCode problem 2554, 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 banned and two integers n and maxSum . You are choosing some number of integers following the below rules: The chosen integers have to be in the range [1, n] . Each integer can be chosen at most once . The chosen integers should not be in the array banned . The sum of the chosen integers should not exceed maxSum . Return the maximum number of integers you can choose following the mentioned rules . Example 1: Input: banned = [1,6,5], n = 5, maxSum = 6 Output: 2 Expl

Detailed Explanation

The problem asks us to find the maximum number of integers we can choose from the range [1, n], given a list of banned integers and a maximum sum constraint. We can only choose integers within the range [1, n], and each integer can be chosen at most once. The chosen integers must not be present in the 'banned' list, and their sum must not exceed 'maxSum'. The goal is to maximize the number of integers chosen under these constraints.

Solution Approach

The solution uses a greedy approach. It iterates through the numbers from 1 to 'n'. For each number, it checks if it is in the 'banned' list. If it's not, it checks if adding the number to the current sum exceeds 'maxSum'. If it doesn't, the number is added to the sum and the count is incremented. The loop terminates when adding the next number would exceed 'maxSum'. A set is used to store the banned numbers to allow for efficient lookup.

Step-by-Step Algorithm

Step 1: Create a set (or hash table) from the 'banned' array for O(1) lookup of banned numbers.
Step 2: Initialize a 'count' variable to 0 (to store the number of chosen integers) and a 'current_sum' variable to 0.
Step 3: Iterate through numbers from 1 to 'n' (inclusive).
Step 4: For each number 'i', check if 'i' is present in the 'banned' set.
Step 5: If 'i' is not in the 'banned' set, check if 'current_sum + i <= maxSum'.
Step 6: If 'current_sum + i <= maxSum', add 'i' to 'current_sum' and increment 'count'.
Step 7: If 'current_sum + i > maxSum', break the loop, as adding any further numbers will exceed the limit.
Step 8: Return the 'count'.

Key Insights

Insight 1: Using a set (or hash table) to store the banned numbers allows for O(1) lookup to quickly check if a number is banned.
Insight 2: Iterating sequentially from 1 to 'n' is a greedy approach that prioritizes smaller numbers. Since we want to maximize the *count* of numbers, this approach is optimal.
Insight 3: Terminating the loop as soon as the current sum exceeds 'maxSum' is crucial for efficiency.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(len(banned))

Topics

This problem involves: Array, Hash Table, Binary Search, Greedy, Sorting.

Companies

Asked at: PayPal.

Maximum Number of Integers to Choose From a Range I | LeetCode 2554

Medium

Array Hash Table Binary Search Greedy Sorting

Time: O(n)Space: O(len(banned))

Solution Code

Available in:

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int maxCount(int[] banned, int n, int maxSum) {
        Set<Integer> banned_set = new HashSet<>();
        for (int num : banned) {
            banned_set.add(num);
        }
        
        int count = 0;
        int current_sum = 0;
        
        for (int i = 1; i <= n; i++) {
            if (!banned_set.contains(i)) {
                if (current_sum + i <= maxSum) {
                    current_sum += i;
                    count += 1;
                } else {
                    break;
                }
            }
        }
        
        return count;
    }
}

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Problem Info

Description

Problem 2554: Maximum Number of Integers to Choose From a Range I - Detailed explanation and solution approaches.

Topics

Array Hash Table Binary Search Greedy Sorting

Companies

PayPal

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2554

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

8 steps

1
Step 1: Create a set (or hash table) from the 'banned' array for O(1) lookup of banned numbers.
2
Step 2: Initialize a 'count' variable to 0 (to store the number of chosen integers) and a 'current_sum' variable to 0.
3
Step 3: Iterate through numbers from 1 to 'n' (inclusive).
4
Step 4: For each number 'i', check if 'i' is present in the 'banned' set.
5
Step 5: If 'i' is not in the 'banned' set, check if 'current_sum + i <= maxSum'.
6
Step 6: If 'current_sum + i <= maxSum', add 'i' to 'current_sum' and increment 'count'.
7
Step 7: If 'current_sum + i > maxSum', break the loop, as adding any further numbers will exceed the limit.
8
Step 8: Return the 'count'.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(len(banned))

📝

Why?

The time complexity is dominated by the loop iterating from 1 to 'n'. The set lookup within the loop takes O(1) time on average. The space complexity is determined by the size of the set used to store the banned numbers. In the C example, we initialize an array of size `n+1` to simulate the set, thus resulting in O(n) space.

📖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

Sorting and Binary Search

Instead of a set, the 'banned' array could be sorted, and binary search could be used to check if a number is banned. This would be less efficient, unless you need to do this checking many times. It may improve space complexity in some extreme scenarios.

Time: O(k log k + n log k), where k is the length of banned array (sorting plus binary search inside the loop)Space: O(1) if sorting is done in place; otherwise O(k)

Trade-offs: Generally slower due to the O(log k) cost of binary search in the inner loop. Use only if space complexity is critical and k is very small.

Mathematical Approach (for large n and small banned)

If 'n' is very large and the number of banned integers is small, one could calculate the sum of all integers from 1 to 'n' using the formula n*(n+1)/2. Then iterate through the 'banned' array, subtracting each banned number from this sum. Keep track of how many numbers were subtracted, and use that count as a starting point, adjusting as needed based on maxSum.

Time: O(k), where k is the size of the banned array.Space: O(1)

Trade-offs: This approach is beneficial only when 'n' is very large compared to 'k'. It would need extra logic to handle the maxSum constraint effectively and prevent exceeding it when adding numbers from 1 to 'n' sequentially.

✨

Key Insights

1Insight 1: Using a set (or hash table) to store the banned numbers allows for O(1) lookup to quickly check if a number is banned.
2Insight 2: Iterating sequentially from 1 to 'n' is a greedy approach that prioritizes smaller numbers. Since we want to maximize the *count* of numbers, this approach is optimal.
3Insight 3: Terminating the loop as soon as the current sum exceeds 'maxSum' is crucial for efficiency.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

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.

Sorting

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

Edge Cases

Edge case 1: 'banned' is empty. The solution correctly chooses as many integers as possible until 'maxSum' is reached.
Edge case 2: 'maxSum' is very small (e.g., 0 or 1). The solution correctly handles this by potentially choosing no integers.
Edge case 3: All numbers from 1 to 'n' are in 'banned'. The solution returns 0.
Edge case 4: n is smaller than banned numbers. C version will cause a problem because if banned contains number x > n, then you are attempting to write outside of your declared space of banned_set. Solution is to only add banned numbers to the set if they're less than or equal to n.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a set (or hash table) for 'banned' numbers. This results in O(n*k) time complexity instead of O(n) in the best implementations.

✗Common mistake 2: Not stopping the loop when 'current_sum' exceeds 'maxSum'. This leads to unnecessary iterations and potential integer overflow issues if 'maxSum' is close to the maximum integer value.

✗Common mistake 3: Forgetting to consider the constraint that the chosen integers must be within the range [1, n].

✅

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. Imagine you have a limited budget ('maxSum') and a list of tasks with associated costs (integers from 1 to 'n'). Some tasks are blocked ('banned'). This algorithm helps you choose the maximum number of tasks you can complete within your budget, avoiding the blocked tasks.

🏦Real-world scenario 2: Network routing. Consider a network where you need to choose the maximum number of network paths from a set of available paths (1 to 'n'). Some paths are unavailable ('banned'), and there's a limit on the total bandwidth ('maxSum') that can be used. This algorithm could help select the maximum number of paths within the bandwidth constraint.

💡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: July 5, 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 14

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

@DevExpert92Jul 7

❓

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.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 Integers to Choose From a Range 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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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