# 1802. Maximum Value at a Given Index in a Bounded Array

## 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.

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

## Solutions

Solution 1: Binary Search

According to the problem description, if we determine the value of $nums[index]$ as $x$, we can find a minimum array sum. That is, the elements on the left side of $index$ in the array decrease from $x-1$ to $1$, and if there are remaining elements, the remaining elements are all $1$; similarly, the elements at $index$ and on the right side of the array decrease from $x$ to $1$, and if there are remaining elements, the remaining elements are all $1$.

In this way, we can calculate the sum of the array. If the sum is less than or equal to $maxSum$, then the current $x$ is valid. As $x$ increases, the sum of the array will also increase, so we can use the binary search method to find the maximum $x$ that meets the conditions.

To facilitate the calculation of the sum of the elements on the left and right sides of the array, we define a function $sum(x, cnt)$, which represents the sum of an array with $cnt$ elements and a maximum value of $x$. The function $sum(x, cnt)$ can be divided into two cases:

• If $x \geq cnt$, then the sum of the array is $\frac{(x + x - cnt + 1) \times cnt}{2}$
• If $x \lt cnt$, then the sum of the array is $\frac{(x + 1) \times x}{2} + cnt - x$

Next, define the left boundary of the binary search as $left = 1$, the right boundary as $right = maxSum$, and then binary search for the value $mid$ of $nums[index]$. If $sum(mid - 1, index) + sum(mid, n - index) \leq maxSum$, then the current $mid$ is valid, we can update $left$ to $mid$, otherwise we update $right$ to $mid - 1$.

Finally, return $left$ as the answer.

The time complexity is $O(\log M)$, where $M=maxSum$. The space complexity is $O(1)$.

• class Solution {
public int maxValue(int n, int index, int maxSum) {
int left = 1, right = maxSum;
while (left < right) {
int mid = (left + right + 1) >>> 1;
if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}

private long sum(long x, int cnt) {
return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
}
}

• class Solution {
public:
int maxValue(int n, int index, int maxSum) {
auto sum = [](long x, int cnt) {
return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
};
int left = 1, right = maxSum;
while (left < right) {
int mid = (left + right + 1) >> 1;
if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}
};

• class Solution:
def maxValue(self, n: int, index: int, maxSum: int) -> int:
def sum(x, cnt):
return (
(x + x - cnt + 1) * cnt // 2 if x >= cnt else (x + 1) * x // 2 + cnt - x
)

left, right = 1, maxSum
while left < right:
mid = (left + right + 1) >> 1
if sum(mid - 1, index) + sum(mid, n - index) <= maxSum:
left = mid
else:
right = mid - 1
return left


• func maxValue(n int, index int, maxSum int) int {
sum := func(x, cnt int) int {
if x >= cnt {
return (x + x - cnt + 1) * cnt / 2
}
return (x+1)*x/2 + cnt - x
}
return sort.Search(maxSum, func(x int) bool {
x++
return sum(x-1, index)+sum(x, n-index) > maxSum
})
}