Maximum Value at a Given Index in a Bounded Array | Leetcode - 1802

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[@MGtvMusic]

The nums[i] is a positive integer where i is 0 to

[@sakarmichel5556]

Hi,
Thank you for taking the time to make this detail video!
You have put in a lot of work in making this video and that deserves praise.
I am having a hard time understand how the division operator is use to calculate the remaining increment.
Is it possible you can provide another example of how to calculate the remaining increment with the division operator please?
Timestamp 13:12
Never Give Up,

[@netanelkomm5636]

Thank you, you helped me a lot.
I got the unoptimized solution before watching your video and was going insane about the TLE and didn't know how to improve it :P

[@AlgorithmsMadeEasy]

Glad to hear !!

[@Prem_Parmar]

TLE Step approach:
class Solution {
public int maxValue(int n, int index, int maxSum) {
int maxLeft = index;
int maxRight = n - index - 1;
int sum = n;
int maxIndexValue = 1;
for (int i = 1; sum < maxSum ;i++) {
maxIndexValue++; sum++;
sum += (maxLeft - i > 0 ? i : maxLeft);
sum += (maxRight - i > 0 ? i : maxRight);
}
return maxIndexValue;
}
}

[@Aryan91191]

python equivalent O(n)
def maxValue(self, n: int, index: int, maxSum: int) -> int:
value = 1
maxSum -= n # initalize array by 1
left, right = 0, 0
leftMax = index # length of subarray to the left of index
rightMax = n - index - 1
while maxSum > 0 :
value += 1
leftVal = min(left, leftMax) # [1, 2, 2, 1]
rightVal = min(right, rightMax)
left += 1
right += 1
maxSum -= (1 + leftVal + rightVal)
if leftVal == leftMax and rightVal == rightMax:
break
if maxSum > 0:
value += maxSum // n
return value-1 if maxSum < 0 else value