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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.
1#include <stdio.h>
2#include <stdlib.h>
3
4int sumLeft(int index, int value) {
5 if (value
The implementation defines two helper functions to calculate the sum of the left and right parts of the array relative to the index.
We use a binary search within the range of 1 to `maxSum` to find the maximum possible value at `nums[index]` that satisfies the given constraints.
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.
1#include <iostream>
2using namespace std;
3
4int maxValue(int n, int index, int maxSum) {
5 int left = index, right = index;
6 int currentValue = 1, sum = n;
7
8 while (sum + right - left <= maxSum) {
9 sum += (right - left + 1);
10 currentValue++;
11 if (left > 0) left--;
12 if (right < n - 1) right++;
13 if (left == 0 && right == n - 1) {
14 int added = (maxSum - sum) / n;
15 return currentValue + added;
16 }
17 }
18
19 return currentValue;
20}
21
22int main() {
23 int n = 6, index = 1, maxSum = 10;
24 cout << maxValue(n, index, maxSum) << endl;
25 return 0;
26}In C++, the greedy approach extends symmetrically about the index, maximizing possible additions according to the total available sum.