algoadvance

Leetcode 396. Rotate Function

Problem Statement

You are given an integer array nums of size n.

The rotate function F is defined as follows:

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

Return the maximum value of F(k).

The test cases are generated so that at least one valid k exists.

Clarifying Questions

  1. Input Constraints:
    • The array nums will have at least one element (n >= 1).
    • Elements in nums will fit within the range of a 32-bit integer.
  2. Edge Cases:
    • What if nums has only one element? (F(k) should always be 0 regardless of k)
    • Are there any assumptions about the values (positive, negative, or zero) within nums?
  3. Expected Output:
    • Return the maximum integer value of the rotate function F(k) for 0 <= k < n.

Strategy

  1. Initial Calculation:
    • Calculate F(0), which is the sum 0 * nums[0] + 1 * nums[1] + ... + (n-1) * nums[n-1].
  2. Subsequent Rotations:
    • Notice how transitioning from F(k) to F(k+1) affects the sum. We can derive that:
      • F(k + 1) = F(k) + sum(nums) - n * nums[n - k] after rotating one step.
    • Use this relation to compute F(k+1) efficiently from F(k).
  3. Optimize:
    • Use a single loop to update the value of F(k) for all possible k and keep track of the maximum value.

Code

#include <vector>
#include <numeric> // for accumulate
#include <algorithm> // for max

class Solution {
public:
    int maxRotateFunction(std::vector<int>& nums) {
        int n = nums.size();
        int F = 0;
        
        // Sum of all elements in nums
        int sum_nums = accumulate(nums.begin(), nums.end(), 0);

        // Calculate F(0)
        for (int i = 0; i < n; ++i) {
            F += i * nums[i];
        }
        
        int maxF = F;

        // Calculate F(k) for k = 1 to n-1 using the observed relationship
        for (int k = 1; k < n; ++k) {
            F = F + sum_nums - n * nums[n - k];
            maxF = std::max(maxF, F);
        }
        
        return maxF;
    }
};

Time Complexity

This results in a linear time complexity solution, which is efficient given the constraints typically observed in interview questions.

Summary

By using an efficient update relationship between successive rotations, we avoid recalculating the rotate function from scratch for each possible rotation, thus optimizing the solution to O(n) time complexity.

Cut your prep time in half and DOMINATE your interview with AlgoAdvance AI