You are given three positive integers: n, index, and maxSum. You want to construct an array nums (0-indexed) that satisfies the following conditions:
nums.length == nnums[i] is a positive integer where 0 <= i < n.abs(nums[i] - nums[i+1]) <= 1 where 0 <= i < n-1.nums does not exceed maxSum.nums[index] is maximized.Return nums[index] of the constructed array.
Note that abs(x) equals x if x >= 0, and -x otherwise.
Example 1:
Input: n = 4, index = 2, maxSum = 6 Output: 2 Explanation: nums = [1,2,2,1] is one array that satisfies all the conditions. There are no arrays that satisfy all the conditions and have nums[2] == 3, so 2 is the maximum nums[2].
Example 2:
Input: n = 6, index = 1, maxSum = 10 Output: 3
Constraints:
1 <= n <= maxSum <= 1090 <= index < nThis problem asks for the maximum possible value at a specific index while keeping the array positive and ensuring the total sum does not exceed maxSum. A key observation is that if we fix a candidate value for the target index, the surrounding values must decrease by at most 1 as we move away from that index while staying at least 1.
An efficient strategy is to use Binary Search on the answer. We guess a value for the target index and check whether it is feasible under the sum constraint. To validate a guess, we compute the required contribution on the left and right sides using arithmetic progression logic, ensuring values never drop below 1. This calculation can be done in constant time.
If the computed total sum is within maxSum, the value is feasible and we try a larger candidate; otherwise we reduce the search range. This approach avoids constructing the full array and efficiently narrows the optimal value.
Time Complexity: O(log(maxSum)) for binary search checks. Space Complexity: O(1).
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Binary Search with Greedy Sum Calculation | O(log(maxSum)) | O(1) |
Cherry Coding [IIT-G]
Use these hints if you're stuck. Try solving on your own first.
What if the problem was instead determining if you could generate a valid array with nums[index] == target?
To generate the array, set nums[index] to target, nums[index-i] to target-i, and nums[index+i] to target-i. Then, this will give the minimum possible sum, so check if the sum is less than or equal to maxSum.
n is too large to actually generate the array, so you can use the formula 1 + 2 + ... + n = n * (n+1) / 2 to quickly find the sum of nums[0...index] and nums[index...n-1].
Binary search for the target. If it is possible, then move the lower bound up. Otherwise, move the upper bound down.
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
long long sumRight(int n, int index, int value) {
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.
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Explore More ProblemsJot down your thoughts, approach, and key learnings
Binary search works because the feasibility of a value is monotonic. If a certain value at the index is valid under the sum constraint, then all smaller values are also valid, allowing efficient narrowing of the answer range.
Yes, this problem or similar variations appear in technical interviews. It tests understanding of binary search on answers, greedy reasoning, and efficient arithmetic calculations under constraints.
No special data structure is required. The solution mainly relies on mathematical calculations and binary search to evaluate whether a candidate value satisfies the constraints.
The optimal approach uses binary search on the possible value at the given index. For each candidate value, we calculate the minimum required sum for the rest of the array using a greedy decreasing pattern and check if it stays within maxSum.
1using System;
2
3public class Solution {
4 public static int 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
22 public static void Main() {
23 int n = 6, index = 1, maxSum = 10;
24 Console.WriteLine(MaxValue(n, index, maxSum));
25 }
26}C# solution systematically incrementally reinforces the value at the given index, wideness around it, while observing all conditions and maximum sum.