grid47 Exploring patterns and algorithms
Sep 28, 2024
6 min read
Solution to LeetCode 396: Rotate Function Problem
You are given an integer array nums of length n. For each rotation k, we define a function F(k) which is calculated as: F(k) = 0 * arrk[0] + 1 * arrk[1] + … + (n-1) * arrk[n-1]. The goal is to return the maximum value of F(k) for all possible k in the range from 0 to n-1.
Problem
Approach
Steps
Complexity
Input: The input consists of an array of integers `nums` and an integer `n` representing the length of the array.
Example: Input: nums = [1, 2, 3, 4]
Constraints:
• 1 <= n <= 10^5
• -100 <= nums[i] <= 100
Output: The output is an integer representing the maximum value of F(k) for all possible values of `k`.
Example: Output: 20
Constraints:
• The solution must return the maximum value of F(k) among all rotations of the array.
Goal: The goal is to compute F(k) for all `k` and return the maximum value.
Steps:
• Compute the sum of the array and the initial value of F(0).
• Use a rolling approach to compute subsequent F(k) values by adjusting the previous F(k-1).
• Return the maximum value of F(k).
Goal: The solution must handle arrays with up to 10^5 elements efficiently and handle values of `nums[i]` between -100 and 100.
Steps:
• The solution should handle arrays with lengths up to 10^5.
• Ensure correct handling of edge cases like arrays with only one element.
Assumptions:
• The input array `nums` is a valid integer array of length `n`.
• Input: Input: nums = [1, 2, 3, 4]
• Explanation: We calculate F(k) for all rotations: F(0), F(1), F(2), and F(3). The maximum value is 20.
• Input: Input: nums = [100]
• Explanation: With a single element, there is only one possible rotation, and the value of F(k) is 0.
Approach: The approach uses an efficient way to compute F(k) using a rolling sum, so that we do not need to recompute the entire sum for each rotation.
Observations:
• The problem involves rotating the array and computing a function based on the rotated array.
• We need an optimized solution to avoid recalculating the entire function for every rotation.
Steps:
• Start by calculating F(0) and the sum of all elements in the array.
• For each subsequent rotation, calculate F(k) using the previous value of F(k-1) and adjust it by subtracting the contribution of the element that is removed from the front and adding the contribution of the element that is added to the back.
Empty Inputs:
• The input array `nums` will not be empty as it has a minimum length of 1.
Large Inputs:
• Ensure the solution handles input arrays with lengths up to 10^5 efficiently.
Special Values:
• Handle cases where the array has only one element or all elements are the same.
Constraints:
• Handle edge cases such as small or large values in the array and rotations for arrays of size 1.
intmaxRotateFunction(vector<int>& nums) {
long sum =0, fn =0;
int n = nums.size();
for(int i =0; i < n; i++) {
sum += nums[i];
fn += i * nums[i];
}
long l =1, r;
long newfn = fn;
while(l < n) {
r = l + n -1;
long removed = (l -1) *nums[l -1];
long added = r * nums[r % n];
newfn = newfn - removed + added - sum;
fn = max(fn, newfn);
l++;
}
return fn;
}
1 : Function Definition
intmaxRotateFunction(vector<int>& nums) {
The function `maxRotateFunction` takes a vector of integers `nums` and returns the maximum possible value of the rotate function.
2 : Variable Initialization
long sum =0, fn =0;
Two variables are initialized: `sum` will store the sum of all elements in the array, and `fn` will store the initial value of the rotate function.
3 : Array Length
int n = nums.size();
The variable `n` stores the size of the `nums` array.
4 : Initial Calculation
for(int i =0; i < n; i++) {
A loop iterates over each element of the array to compute the initial `sum` and `fn`.
5 : Sum Calculation
sum += nums[i];
The sum of all elements in the array is computed.
6 : Initial Function Calculation
fn += i * nums[i];
The initial value of the rotate function is calculated by multiplying each element by its index and adding it to `fn`.
7 : Variable Initialization
long l =1, r;
Two variables, `l` and `r`, are initialized. `l` will track the left index for the next rotation, and `r` will be used for the right index.
8 : Function Update
long newfn = fn;
A new variable `newfn` is initialized to store the updated value of the rotate function during each iteration.
9 : Rotation Loop
while(l < n) {
The loop continues as long as `l` is less than the size of the array, ensuring each rotation is processed.
10 : Right Index Calculation
r = l + n -1;
The right index `r` is calculated based on `l` and the size of the array.
11 : Element Removal
long removed = (l -1) *nums[l -1];
The value of the element at `l - 1` is removed from the calculation, as it is no longer part of the rotated array.
12 : Element Addition
long added = r * nums[r % n];
The value of the element at index `r % n` is added to the calculation.
13 : Function Update
newfn = newfn - removed + added - sum;
The value of `newfn` is updated by removing the contribution of the previous element and adding the contribution of the new element.
14 : Max Function Update
fn = max(fn, newfn);
The value of `fn` is updated to be the maximum of its current value and the updated `newfn`.
15 : Index Update
l++;
The left index `l` is incremented to move to the next rotation.
16 : Final Result
return fn;
The function returns the maximum value of the rotate function after all rotations are processed.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The solution runs in linear time O(n) because we only need to compute the sum and adjust the function in each step.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) because we only use a few variables to keep track of the sum and function values.
