Solution: Maximum Value at a Given Index in a Bounded Array
Let's solve the Maximum Value at a Given Index in a Bounded Array problem using the modified binary search pattern.
We'll cover the following
Statement
Given three positive integers, n
, index
, and maxSum
, output the nums[index]
by constructing an array of nums
with the length of n
, which satisfies the following conditions:
The length of the array
nums
is equal ton
.Each element
nums[i]
is a positive integer, wherei
n
.The absolute difference between two consecutive elements,
nums[i]
andnums[i+1]
, is at most. The sum of all elements in
nums
does not exceedmaxSum
.The element at
nums[index]
contains the maximum value.
Constraints:
n
maxSum
index
n
Solution
The solution to maximizing the value at a given index in a bounded array involves using a binary search approach combined with mathematical calculations of arithmetic sequences. To achieve this, we need to ensure that the values on both sides of nums[index]
decrease by at most
The algorithm to solve this problem is as follows:
Initialization
Initialize two variables,
left
andright
, with the valuesand maxSum
, respectively. We can assign a minimum value ofto an element, and the maximum value of maxSum
.Start a loop and Iterate until
left
is less thanright
, and calculate the sum of left and right elements as listed below:
Calculate mid
Calculate
mid
as.
Determine values on the left side
If
mid
is greater thanindex
, there will be a sequence starting frommid
to the(right to left). This sequence will form a series , which is an arithmetic series. This arithmetic series is calculated as
For example, if n = 5
, index = 4
, and maxSum = 20
, then mid
will equal
Otherwise, if
mid
is less than or equal toindex
, then there will be a sequence starting frommid
to(right to left). This sequence will form a series , which is an arithmetic series. In mathematics, this arithmetic series is calculated as . In addition to the arithmetic series, there will also be a sequence of s of length .
For example, if n = 8
, index = 7
, maxSum = 9
, then mid
will equal
Determine values on the right side
If
mid
is greater than or equal ton - index
, there will be a sequence starting frommid
to(left to right). This sequence will form a series , which is an arithmetic series. In mathematics, this arithmetic series is calculated as .
For example, if n = 10
, index = 7
, and maxSum = 11
, then mid
will equal
Otherwise, if
mid
is less thann - index
, there will be a sequence starting frommid
to1
. This sequence will form a series, which is an arithmetic series. In mathematics, this arithmetic series is calculated as . In addition to the arithmetic series, there will also be a sequence of s of length .
For example, if n = 8
, index = 2
, maxSum = 9
, then mid
will equal
Sum calculation
Calculate the total sum of the array by combining the sum of both sides and subtracting the value of
mid
. The reason for subtractingmid
is that it is calculated twice (in the sum of the left and the sum of the right).
Updating variables
If the sum of the array is less than or equal to the
maxSum
, we will use the right half of the array by updating theleft
with the value ofmid
.Otherwise, we will go with the left half of the array by updating
right
with the value ofmid - 1
.
After the loop terminates, we will return the value of left
because it is the index of the maximum value of nums
.
Let’s look at the following illustration to get a better understanding of the solution:
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