Leetcode 1802: Maximum Value at a Given Index in a Bounded Array
You are tasked with constructing an array
nums of length n such that: all elements are positive integers, the difference between adjacent elements is at most 1, the sum of elements does not exceed maxSum, and the value at index index is maximized. Return the value at index in the array nums.Problem
Approach
Steps
Complexity
Input: The input consists of three integers: `n`, `index`, and `maxSum`. These are the length of the array, the index whose value is to be maximized, and the maximum allowable sum of all elements in the array, respectively.
Example: n = 5, index = 2, maxSum = 10
Constraints:
• 1 <= n <= maxSum <= 10^9
• 0 <= index < n
Output: Return the value at the index `index` in the array `nums` after constructing it under the given conditions.
Example: Output: 3
Constraints:
• The value at `index` must be maximized while ensuring that the sum of the elements in the array does not exceed `maxSum`.
Goal: Construct the array such that the value at `index` is maximized, while adhering to the constraints of adjacent element differences and sum limits.
Steps:
• Use binary search to determine the maximum possible value for `nums[index]`.
• For each candidate value, calculate if the sum of the array can remain within `maxSum` while keeping the difference between adjacent elements at most 1.
Goal: The solution must efficiently handle the large constraints, especially for `n` and `maxSum` as they can be as large as 10^9.
Steps:
• Time complexity should be optimized for large inputs.
• The solution should use binary search to find the optimal value for `nums[index]`.
Assumptions:
• We assume the values are large enough to require an efficient solution using binary search.
• Input: n = 5, index = 2, maxSum = 10
• Explanation: In this case, the array `[1, 2, 3, 2, 1]` satisfies all conditions, and the value at index 2 is maximized as 3, with a total sum of 10.
Approach: The approach uses binary search to determine the maximum possible value for the element at `index`, checking if the sum of the array remains within `maxSum`.
Observations:
• The problem involves finding the maximum possible value at a specific index in an array, while satisfying sum and adjacent element difference constraints.
• Binary search is the most efficient method to maximize the value at the index since it allows us to narrow down the possible values for `nums[index]`.
Steps:
• Use binary search to find the maximum value that `nums[index]` can take, starting from 1 and going up to `maxSum`.
• For each midpoint in the binary search, calculate the sum of the array assuming the value at `index` is fixed at that midpoint, and check if the sum is within `maxSum`.
Empty Inputs:
• There are no empty inputs since `n` is always at least 1.
Large Inputs:
• For large values of `n` and `maxSum`, binary search ensures the solution runs efficiently even when the input values approach their upper bounds.
Special Values:
• Consider the case where `index` is at either end of the array, or when `n` is equal to `maxSum`.
Constraints:
• Ensure the solution efficiently handles the maximum constraints.
bool can(long long pk, long long n, long long i, long long sum) {
sum -= n;
long j = n - i - 1;
long long need = pk * pk - ((pk > i ? (pk - i - 1) * (pk - i) : 0) + (pk > j? (pk - j - 1) * (pk - j): 0)) / 2;
return sum - need >= 0;
}
int maxValue(int n, int index, int mx) {
int ans = 0, l = 0, r = mx;
while(l <= r) {
long mid = l + (r - l + 1) / 2;
// cout << mid << " ";
if(can(mid, n, index, mx)) {
ans = mid;
l = mid + 1;
} else r = mid - 1;
}
return ans + 1;
}
1 : Function Definition
bool can(long long pk, long long n, long long i, long long sum) {
Define the 'can' function which takes parameters pk, n, i, and sum to check if a certain value can be achieved based on the constraints.
2 : Mathematical Operations
sum -= n;
Adjust the sum by subtracting the total number of elements.
3 : Variable Initialization
long j = n - i - 1;
Initialize the variable 'j' to represent the index of the second element in the range.
4 : Mathematical Operations
Blank line for separation.
5 : Mathematical Operations
long long need = pk * pk - ((pk > i ? (pk - i - 1) * (pk - i) : 0) + (pk > j? (pk - j - 1) * (pk - j): 0)) / 2;
Calculate the amount needed based on the value of pk and the indices i and j.
6 : Return
return sum - need >= 0;
Return whether the remaining sum is greater than or equal to the calculated need.
7 : Function Definition
int maxValue(int n, int index, int mx) {
Define the 'maxValue' function which implements a binary search to find the maximum value.
8 : Variable Initialization
int ans = 0, l = 0, r = mx;
Initialize variables for binary search: ans (answer), l (lower bound), and r (upper bound).
9 : Loop and Recursion
while(l <= r) {
Start the binary search loop.
10 : Mathematical Operations
long mid = l + (r - l + 1) / 2;
Calculate the midpoint for the binary search.
11 : Flow Control
if(can(mid, n, index, mx)) {
If the 'can' function returns true for the current mid value, adjust the binary search bounds.
12 : Variable Initialization
ans = mid;
Update the answer with the current value of mid.
13 : Mathematical Operations
l = mid + 1;
Move the lower bound up to search the higher range.
14 : Flow Control
} else r = mid - 1;
Otherwise, move the upper bound down to search the lower range.
15 : Return
return ans + 1;
Return the final answer, incremented by 1 to match the problem's requirements.
Best Case: O(log(maxSum))
Average Case: O(log(maxSum))
Worst Case: O(log(maxSum))
Description: The binary search over the possible values for `nums[index]` results in a time complexity of O(log(maxSum)).
Best Case: O(1)
Worst Case: O(1)
Description: The solution only requires a constant amount of extra space.
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