Sponsored
Sponsored
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 <iostream>
2using namespace std;
3
4long long sumLeft(int index, int value) {
5 if (value > index) {
6 long long count = value - index;
7 return static_cast<long long>(value + (value - index - 1)) * index / 2 + count;
8 } else {
9 return static_cast<long long>(value * (value + 1)) / 2 + (index - value + 1);
10 }
11}
12
13long long sumRight(int n, int index, int value) {
14 int rightIndex = n - index - 1;
if (value > rightIndex) {
long long count = value - rightIndex;
return static_cast<long long>(value + (value - rightIndex - 1)) * rightIndex / 2 + count;
} else {
return static_cast<long long>(value * (value + 1)) / 2 + (rightIndex - value + 1);
}
}
int maxValue(int n, int index, int maxSum) {
int left = 1, right = maxSum + 1;
while (left < right) {
int mid = left + (right - left) / 2;
long long sum = mid + sumLeft(index, mid) + sumRight(n, index, mid);
if (sum <= maxSum) {
left = mid + 1;
} else {
right = mid;
}
}
return left - 1;
}
int main() {
int n = 6, index = 1, maxSum = 10;
cout << maxValue(n, index, maxSum) << endl;
return 0;
}The C++ solution follows a similar approach to the C solution, using helper functions to compute required segment sums and performing binary search to maximize `nums[index]` within 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.
1public class Solution {
2 public static int maxValue(int n, int index, int maxSum) {
3 int left = index, right = index;
4 int currentValue = 1, sum = n;
5
6 while (sum + right - left <= maxSum) {
7 sum += (right - left + 1);
8 currentValue++;
9 if (left > 0) left--;
10 if (right < n - 1) right++;
11 if (left == 0 && right == n - 1) {
12 int added = (maxSum - sum) / n;
13 return currentValue + added;
14 }
15 }
16
17 return currentValue;
18 }
19
20 public static void main(String[] args) {
21 int n = 6, index = 1, maxSum = 10;
22 System.out.println(maxValue(n, index, maxSum));
23 }
24}The Java approach methodically widens from the pivotal index, maintaining constraints and adjusting sum across the array's bounds until maximization is achieved.