How do you solve Maximum Value at a Given Index in a Bounded Array on LeetCode?

Maximum Value at a Given Index in a Bounded Array (LeetCode #1802) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log(maxSum) * n) time complexity and O(1) space complexity. Key insight: The maximum value at the given index is constrained by the total sum allowed and the adjacent values.

What is the brute force solution for Maximum Value at a Given Index in a Bounded Array?

The brute force approach for Maximum Value at a Given Index in a Bounded Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible arrays that meet the conditions and checking their sums. This is inefficient but helps us understand the problem's constraints. The optimal Optimal Solution approach improves this to O(log(maxSum) * n).

What algorithm or data structure is used to solve Maximum Value at a Given Index in a Bounded Array?

Maximum Value at a Given Index in a Bounded Array uses the following concepts: Math, Binary Search, Greedy. The recommended patterns to study are: Binary Search, Greedy Algorithms.

What are the interview tips for Maximum Value at a Given Index in a Bounded Array?

Always clarify the constraints and edge cases before diving into coding. Think about how you can optimize your solution early on, especially if the problem involves large inputs. Practice explaining your thought process clearly, as communication is key during interviews.

What are common mistakes when solving Maximum Value at a Given Index in a Bounded Array?

Failing to account for edge cases where the index is at the start or end of the array. Overlooking the constraints of the problem when calculating the sum, leading to incorrect assumptions.

#1802

Maximum Value at a Given Index in a Bounded Array

Medium

MathBinary SearchGreedyBinary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log(maxSum) * n)
Space	O(1)	O(1)

💡

Intuition

Time O(log(maxSum) * n)Space O(1)

Instead of generating arrays, we can use binary search to find the maximum possible value for nums[index] that satisfies the conditions, leveraging the constraints to minimize the sum.

⚙️

Algorithm

3 steps

1Step 1: Define a binary search range from 1 to maxSum.
2Step 2: For each mid value in the binary search, calculate the minimum sum required to create a valid array with nums[index] = mid.
3Step 3: If the calculated sum is less than or equal to maxSum, it means we can potentially increase mid; otherwise, decrease it.

solution.py14 lines

1def maxValue(n, index, maxSum):
2    def canConstruct(target):
3        left = max(0, target - index)
4        right = max(0, target - (n - index - 1))
5        total = (target * (target + 1)) // 2 - (left * (left + 1)) // 2 - (right * (right + 1)) // 2
6        return total <= maxSum
7    low, high = 1, maxSum
8    while low < high:
9        mid = (low + high + 1) // 2
10        if canConstruct(mid):
11            low = mid
12        else:
13            high = mid - 1
14    return low

ℹ

Complexity note: The binary search runs in logarithmic time relative to maxSum, and for each mid value, we calculate the sum in linear time, leading to an overall efficient complexity.

1The maximum value at the given index is constrained by the total sum allowed and the adjacent values.
2Using binary search allows us to efficiently narrow down the maximum possible value without generating all possible arrays.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.