What is the time complexity of Maximum Value at a Given Index in a Bounded Array?

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

What data structures are used to solve Maximum Value at a Given Index in a Bounded Array?

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

What are the key insights for solving Maximum Value at a Given Index in a Bounded Array?

Insight 1: The array forms a 'mountain' shape centered at the given index, with values decreasing as you move away from the index. The key is to efficiently compute the sum of this mountain. Insight 2: Binary search can be used to find the maximum possible value for `nums[index]`. This is because the problem exhibits a monotonic property: if a value `v` for `nums[index]` is feasible (i.e., the sum of the array is less than or equal to `maxSum`), then any value less than `v` is also feasible.

Which companies ask LeetCode 1802. Maximum Value at a Given Index in a Bounded Array in interviews?

This problem is frequently asked at ByteDance.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Maximum Value at a Given Index in a Bounded Array - LeetCode 1802 Solution

Maximum Value at a Given Index in a Bounded Array - Complete Solution Guide

Maximum Value at a Given Index in a Bounded Array is LeetCode problem 1802, 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 three positive integers: n , index , and maxSum . You want to construct an array nums ( 0-indexed ) that satisfies the following conditions: nums.length == n nums[i] is a positive integer where 0 <= i < n . abs(nums[i] - nums[i+1]) <= 1 where 0 <= i < n-1 . The sum of all the elements of nums does not exceed maxSum . nums[index] is maximized . Return nums[index] of the constructed array . Note that abs(x) equals x if x >= 0 , and -x otherwise. Example 1: Input: n = 4, index = 2, ma

Detailed Explanation

The problem requires us to construct an array of length `n` such that each element is a positive integer, the absolute difference between adjacent elements is at most 1, the sum of all elements does not exceed `maxSum`, and the element at the given `index` is maximized. The goal is to return the maximum possible value for `nums[index]` under these constraints.

Solution Approach

The solution utilizes binary search to find the maximum possible value for `nums[index]`. For each potential value `mid` within the binary search, we compute the sum of the array assuming `nums[index]` is equal to `mid`. This sum is calculated by considering the 'mountain' shape around the `index`. If the calculated sum is less than or equal to `maxSum`, we update our answer and search for a larger value; otherwise, we search for a smaller value. The binary search converges to the maximum possible value for `nums[index]`.

Step-by-Step Algorithm

Step 1: Initialize `low` to 1 and `high` to `maxSum`. These represent the lower and upper bounds for the binary search.
Step 2: While `low` is less than or equal to `high`, calculate the middle value `mid = low + (high - low) / 2`.
Step 3: Define a helper function `get_sum(mid)` to calculate the sum of the array when `nums[index] = mid`. This function considers the left and right sides of the 'mountain'. It checks if the slope of the mountain reaches 1 before the array edges. If it does, it calculates the sum as a sum of an arithmetic series plus the sum of 1's. Otherwise, it calculates the sum as the sum of a truncated arithmetic series.
Step 4: Call `get_sum(mid)` to calculate the total sum for the given `mid`.
Step 5: If the calculated sum is less than or equal to `maxSum`, update the answer `ans = mid` and set `low = mid + 1` to search for a larger value.
Step 6: Otherwise, if the calculated sum is greater than `maxSum`, set `high = mid - 1` to search for a smaller value.
Step 7: After the binary search completes, return the final answer `ans`.

Key Insights

Insight 1: The array forms a 'mountain' shape centered at the given index, with values decreasing as you move away from the index. The key is to efficiently compute the sum of this mountain.
Insight 2: Binary search can be used to find the maximum possible value for `nums[index]`. This is because the problem exhibits a monotonic property: if a value `v` for `nums[index]` is feasible (i.e., the sum of the array is less than or equal to `maxSum`), then any value less than `v` is also feasible.
Insight 3: The calculation of the array sum needs to consider two cases: when the slope of the 'mountain' reaches 1 before the edges of the array, and when it doesn't. This is crucial for an efficient sum calculation.

Complexity Analysis

Time Complexity: O(log(maxSum))

Space Complexity: O(1)

Topics

This problem involves: Math, Binary Search, Greedy.

Companies

Asked at: ByteDance.

Maximum Value at a Given Index in a Bounded Array | LeetCode 1802

Medium

Math Binary Search Greedy

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

Solution Code

Available in:

class Solution {
    public int maxValue(int n, int index, int maxSum) {
        
        // Helper function to calculate the sum
        
        
        long low = 1, high = maxSum;
        long ans = 0;
        
        while (low <= high) {
            long mid = low + (high - low) / 2;
            
            // mid must be at least 1, as nums[i] are positive
            if (mid < 1) {
                continue;
            }

            long current_sum = get_sum(mid, n, index);
            
            if (current_sum <= maxSum) {
                ans = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        
        return (int) ans;
    }
    
    private long get_sum(long val, int n, int index) {
        int left_len = index;
        int right_len = n - 1 - index;
        
        long total_sum = val;
        long m = val - 1;
        
        // Sum for the left side
        long left_sum;
        if (left_len >= m) {
            // Full slope: val-1, val-2, ..., 1, followed by 1s
            left_sum = m * (m + 1) / 2 + (left_len - m);
        } else {
            // Truncated slope: val-1, val-2, ..., val-left_len
            left_sum = left_len * (2 * val - left_len - 1) / 2;
        }
        
        // Sum for the right side
        long right_sum;
        if (right_len >= m) {
            // Full slope
            right_sum = m * (m + 1) / 2 + (right_len - m);
        } else {
            // Truncated slope
            right_sum = right_len * (2 * val - right_len - 1) / 2;
        }
        
        total_sum += left_sum + right_sum;
        return total_sum;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1802: Maximum Value at a Given Index in a Bounded Array - Detailed explanation and solution approaches.

Topics

Math Binary Search Greedy

Companies

ByteDance

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1802

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

7 steps

1
Step 1: Initialize `low` to 1 and `high` to `maxSum`. These represent the lower and upper bounds for the binary search.
2
Step 2: While `low` is less than or equal to `high`, calculate the middle value `mid = low + (high - low) / 2`.
3
Step 3: Define a helper function `get_sum(mid)` to calculate the sum of the array when `nums[index] = mid`. This function considers the left and right sides of the 'mountain'. It checks if the slope of the mountain reaches 1 before the array edges. If it does, it calculates the sum as a sum of an arithmetic series plus the sum of 1's. Otherwise, it calculates the sum as the sum of a truncated arithmetic series.
4
Step 4: Call `get_sum(mid)` to calculate the total sum for the given `mid`.
5
Step 5: If the calculated sum is less than or equal to `maxSum`, update the answer `ans = mid` and set `low = mid + 1` to search for a larger value.
6
Step 6: Otherwise, if the calculated sum is greater than `maxSum`, set `high = mid - 1` to search for a smaller value.
7
Step 7: After the binary search completes, return the final answer `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(log(maxSum))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search range is from 1 to maxSum. The get_sum function consists of arithmetic operations which take constant time. The variables used in the function occupy constant space. Thus, the space complexity is O(1).

📖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

Mathematical Formula

A direct mathematical formula can be derived to calculate the maximum possible value. This requires careful analysis of different cases based on the relative values of `n`, `index`, and `maxSum` to avoid the binary search.

Time: O(1)Space: O(1)

Trade-offs: While it has O(1) time complexity, deriving and implementing the formula correctly can be complex and error-prone. The binary search approach is more straightforward to implement and less likely to contain subtle bugs. The O(1) approach might be faster for some extremely large inputs, but for most practical cases the binary search is sufficiently efficient.

✨

Key Insights

1Insight 1: The array forms a 'mountain' shape centered at the given index, with values decreasing as you move away from the index. The key is to efficiently compute the sum of this mountain.
2Insight 2: Binary search can be used to find the maximum possible value for `nums[index]`. This is because the problem exhibits a monotonic property: if a value `v` for `nums[index]` is feasible (i.e., the sum of the array is less than or equal to `maxSum`), then any value less than `v` is also feasible.
3Insight 3: The calculation of the array sum needs to consider two cases: when the slope of the 'mountain' reaches 1 before the edges of the array, and when it doesn't. This is crucial for an efficient sum calculation.

📚

Key Terms Explained

For Students

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 `maxSum` is very small (e.g., equal to `n`). The binary search handles this case correctly, as it explores all possible values within the range [1, maxSum].
Edge case 2: When `index` is close to 0 or `n-1`. The `get_sum` function correctly calculates the sum of the array even when the 'mountain' is truncated due to the index being near the boundaries.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the sum of the array. Specifically, failing to handle the cases where the slope of the 'mountain' reaches 1 before the edges of the array, or using integer division where floating-point division is needed.

✗Common mistake 2: Not handling potential integer overflow when calculating the sum. The intermediate calculations, like m * (m + 1), can exceed the maximum value of a 32-bit integer, necessitating the use of 64-bit integers (long long in C++, long in Java).

✗Common mistake 3: Off-by-one errors in the binary search condition, leading to incorrect bounds or infinite loops.

✅

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 distributing resources across a set of servers such that one server has the maximum amount and the distribution decreases by at most one unit for adjacent servers. You want to maximize the resource on the main server while staying within a total resource limit.

🏦Real-world scenario 2: Signal processing. Creating a signal with a peak at a specific frequency index and gradually decreasing amplitudes on either side. The constraints might be a maximum signal strength or a maximum total power of the signal.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: June 3, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

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 "Maximum Value at a Given Index in a Bounded Array"?

For this medium-level Math 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?