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The binary search approach focuses on maximizing `nums[index]` while maintaining the constraints. We consider `nums[index]` as the peak of a potential 'pyramid' in the array. By conducting a binary search on possible values for `nums[index]`, we iterate through and check whether the constraints on both sides of the index and the sum are maintained.
Each mid value tested in the binary search represents the height of the peak at index. For each mid value, we compute the minimum sum required for both sides of the index.
If the calculated sum is within `maxSum`, the peak can be higher, leading us to the right half (mid + 1) of our binary search space. Otherwise, we try smaller values by searching the left half.
Time complexity: O(log(maxSum)) - Binary search divides the range of possible peak values logarithmically.
Space complexity: O(1) - Uses a constant amount of space.
1public class MaximumValue {
2 private static long sumLeft(int index, int value) {
3 if (value > index) {
4
The Java solution employs two helper functions to calculate the required sums for either side of the index around `nums[index]`. Binary search is used to maximize the value at the specified index while keeping within `maxSum`.
This approach involves incrementally building the array by focusing solely on maximizing `nums[index]` using a greedy construction. This directly extends from index to balance the array's condition and sum constraints effectively, utilizing the available sum step-by-step. We increment each value symmetrically from the index point, considering both left and right constraints and ensuring all steps preserve the `abs(nums[i] - nums[i+1]) <= 1` condition.
We keep track of increments needed and subtract from `maxSum` accordingly, thereby implicitly maximizing `nums[index]` while maintaining valid conditions.
Time complexity: O(n) - Incrementally checks and adjusts values from center to borders of the array.
Space complexity: O(1) - Constant space used without additional data structures.
1def max_value(n, index, max_sum):
2 left, right = index, index
3 current_value, sum_ = 1, n
4
5 while sum_ + right - left <= max_sum:
6 sum_ += (right - left + 1)
7 current_value += 1
8 if left > 0: left -= 1
9 if right < n - 1: right += 1
10
11 if left == 0 and right == n - 1:
12 added = (max_sum - sum_) // n
13 return current_value + added
14
15 return current_value
16
17n = 6
18index = 1
19max_sum = 10
20print(max_value(n, index, max_sum))The Python solution applies a logical approach to extend the array incrementally, ensuring conditions are met and maximizing within available sum across boundaries centered at index.