Find Maximum Value at Given Index in 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