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Maximum Value at a Given Index in a Bounded Array

Difficulty: Medium

Problem Description

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.

Key Insights

  • The array must consist of positive integers and should maintain a difference of at most 1 between consecutive elements.
  • The value at the specified index should be maximized while ensuring that the total sum remains within the provided limit (maxSum).
  • A binary search approach can be utilized to efficiently determine the maximum possible value at the given index.

Space and Time Complexity

Time Complexity: O(log(maxSum)) - due to the binary search on the possible values at the index. Space Complexity: O(1) - only a few variables are used for calculations.

Solution

To solve this problem, we can use a binary search approach to find the maximum value that can be assigned to nums[index]. The key steps are:

  1. Define the search range for the possible values at nums[index] from 1 to maxSum.
  2. For each midpoint value in the binary search, calculate the total sum of the array that can be constructed with this midpoint as nums[index].
  3. Ensure that the sum does not exceed maxSum while respecting the constraints of the problem.
  4. Adjust the binary search bounds based on whether the total sum exceeds maxSum or not.

This approach ensures that we efficiently find the maximum possible value for nums[index] while adhering to all the given constraints.

Code Solutions

def maxValue(n, index, maxSum):
    def canAllocate(mid):
        left = max(0, mid - index)
        right = max(0, mid - (n - index - 1))
        total = mid + (mid * (mid - 1)) // 2 - (left * (left - 1)) // 2 - (right * (right - 1)) // 2
        return total <= maxSum

    low, high = 1, maxSum
    while low < high:
        mid = (low + high + 1) // 2
        if canAllocate(mid):
            low = mid
        else:
            high = mid - 1
    return low
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