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Leetcode 396. Rotate Function

Problem Statement

LeetCode 396: Rotate Function

You are given an integer array nums of length n.

Assume F(k) is a function that calculates the sum of the products of each element in nums with their respective indexes after nums is rotated k times.

The rotation is done in such a manner that:

• When k = 1, the last element moves to the first position, and all other elements are shifted to the right.
• When k = n-1, the same rightward shift continues.

Define F(k) as:

F(k) = 0 * nums[k] + 1 * nums[k+1] + ... + (n-1) * nums[k-1]

The task is to find out the maximum value after calculating F(k) for all k in the range [0, n-1].

Clarifying Questions

1. Input Constraints:
   • What is the range of the input size n?
   • What is the range of the elements in the nums array?
2. Output Requirements:
   • Should the function return an integer representing the maximum value of F(k)?

Strategy

1. Initial Observations:
   • We need to calculate F(k) for all rotations k from 0 to n-1.
   • Directly computing F(k) would involve a lot of repeated calculations, so we need an optimal approach.
2. Mathematical Insight:
   • Notice that the transition from F(k) to F(k+1) can be derived using the relationship between consecutive rotations rather than recalculating from scratch each time.
   • Define F(0) as the initial state:

     F(0) = 0 * nums[0] + 1 * nums[1] + ... + (n-1) * nums[n-1]
     

   • For F(k+1), the value can be derived from F(k):

     F(k+1) = F(k) + sum(nums) - n * nums[n-k]
     

   • Where sum(nums) is the sum of all elements in nums.
3. Algorithm:
   • Calculate sum(nums) and F(0) initially.
   • Iterate from 1 to n-1 and use the relationship to compute F(k+1) from F(k).
   • Track the maximum value during these iterations.

Code

public class RotateFunction {
    public int maxRotateFunction(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;

        int sum = 0;
        int f = 0;

        // Compute sum of all numbers and the initial F(0)
        for (int i = 0; i < n; i++) {
            sum += nums[i];
            f += i * nums[i];
        }

        int max = f;

        // Compute F(k) iteratively and update the maximum found
        for (int i = 1; i < n; i++) {
            f = f + sum - n * nums[n - i];
            max = Math.max(max, f);
        }

        return max;
    }

    public static void main(String[]args) {
        RotateFunction rf = new RotateFunction();
        int[] nums = {4, 3, 2, 6};
        System.out.println(rf.maxRotateFunction(nums)); // Output: 26
    }
}

Time Complexity

• Time Complexity: O(n)
  • We iterate through the array a few times but the operations within each iteration are O(1). Hence, the overall complexity is O(n).
• Space Complexity: O(1)
  • We use a few extra variables for tracking sums and the current value of F(k), which does not depend on the input size n.

This solution efficiently computes the desired maximum value of the rotate function by leveraging the mathematical relationship between consecutive rotations, avoiding redundant calculations.

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