# Program to find maximum value at a given index in a bounded array in Python

Suppose we have three values, n, index and maxSum. Consider an array called nums we have to find nums[index] and nums satisfies the following conditions −

• size of nums is n

• All elements in n is positive.

• |nums[i] - nums[i+1]| <= 1 for all i, 0 <= i < n-1.

• The sum of all the elements of nums does not exceed maxSum.

• nums[index] is maximized.

So, if the input is like n = 6, index = 3, maxSum = 8, then the output will be 2 because, we can get an array like [1,2,2,2,1,1], that satisfies all conditions, and here nums[3] is maximized.

To solve this, we will follow these steps −

• left := quotient of maxSum/n, right := maxSum + 1

• ans := 0

• while left < right, do

• mid := left + quotient of (right-left)/2

• ind_l := (mid-1 + maximum of 1 and (mid-index)) * quotient of (minimum of index and (mid-1) /2 + |minimum of 0, mid-index-1|

• ind_r = (mid + maximum of 1 and (mid-(n-index-1))) * quotient of (minimum of (n-index) and mid)/2 + |minimum of 0 and (mid-(n-index-1)-1)|

• if ind_l + ind_r <= maxSum, then

• ans := mid

• left := mid+1

• otherwise,

• right := mid

• return ans

## Example

Let us see the following implementation to get better understanding −

def solve(n, index, maxSum):
left, right = maxSum//n, maxSum+1
ans = 0
while(left<right):
mid = left + (right-left)//2
ind_l = (mid-1+max(1,mid-index))*min(index,mid-1)//2 + abs(min(0,mid-index-1))
ind_r = (mid+max(1,mid-(n-index-1)))*min(n-index, mid)//2+ abs(min(0,mid-(n-index-1)-1))

if ind_l + ind_r <=maxSum:
ans = mid
left = mid+1
else:
right = mid
return ans

n = 6
index = 3
maxSum = 8
print(solve(n, index, maxSum))

## Input

6, 3, 8


## Output

2