maximum-value-at-a-given-index-in-a-bounded-array

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1802. Maximum Value at a Given Index in a Bounded Array (Medium)

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,  maxSum = 6
Output: 2
Explanation: nums = [1,2,2,1] is one array that satisfies all the conditions.
There are no arrays that satisfy all the conditions and have nums[2] == 3, so 2 is the maximum nums[2].

Example 2:

Input: n = 6, index = 1,  maxSum = 10
Output: 3

 

Constraints:

• 1 <= n <= maxSum <= 109
• 0 <= index < n

Related Topics

[Binary Search] [Greedy]

Hints

Hint 1

What if the problem was instead determining if you could generate a valid array with nums[index] == target?

Hint 2

To generate the array, set nums[index] to target, nums[index-i] to target-i, and nums[index+i] to target-i. Then, this will give the minimum possible sum, so check if the sum is less than or equal to maxSum.

Hint 3

n is too large to actually generate the array, so you can use the formula 1 + 2 + ... + n = n * (n+1) / 2 to quickly find the sum of nums[0...index] and nums[index...n-1].

Hint 4

Binary search for the target. If it is possible, then move the lower bound up. Otherwise, move the upper bound down.