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396. Rotate Function

Description

You are given an integer array nums of length n.

Assume arrk to be an array obtained by rotating nums by k positions clock-wise. We define the rotation function F on nums as follow:

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

Return the maximum value of F(0), F(1), ..., F(n-1).

The test cases are generated so that the answer fits in a 32-bit integer.

 

Example 1:

Input: nums = [4,3,2,6]
Output: 26
Explanation:
F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

Example 2:

Input: nums = [100]
Output: 0

 

Constraints:

  • n == nums.length
  • 1 <= n <= 105
  • -100 <= nums[i] <= 100

Solutions

Solution 1

Thinking

\(F(k)=\sum i\cdot \textit{nums}[(i+k)\bmod n]\); maximize \(F\). Recomputing each rotation is \(O(n^2)\). Consecutive \(F\) values differ by a closed form.

\(F(k+1)=F(k)+\sum nums-n\cdot nums[n-1-k]\). Compute \(F(0)\), then roll \(n-1\) times and keep the max.

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class Solution:
    def maxRotateFunction(self, nums: List[int]) -> int:
        f = sum(i * v for i, v in enumerate(nums))
        n, s = len(nums), sum(nums)
        ans = f
        for i in range(1, n):
            f = f + s - n * nums[n - i]
            ans = max(ans, f)
        return ans

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class Solution {
    public int maxRotateFunction(int[] nums) {
        int f = 0;
        int s = 0;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            f += i * nums[i];
            s += nums[i];
        }
        int ans = f;
        for (int i = 1; i < n; ++i) {
            f = f + s - n * nums[n - i];
            ans = Math.max(ans, f);
        }
        return ans;
    }
}

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class Solution {
public:
    int maxRotateFunction(vector<int>& nums) {
        int f = 0, s = 0, n = nums.size();
        for (int i = 0; i < n; ++i) {
            f += i * nums[i];
            s += nums[i];
        }
        int ans = f;
        for (int i = 1; i < n; ++i) {
            f = f + s - n * nums[n - i];
            ans = max(ans, f);
        }
        return ans;
    }
};

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func maxRotateFunction(nums []int) int {
    f, s, n := 0, 0, len(nums)
    for i, v := range nums {
        f += i * v
        s += v
    }
    ans := f
    for i := 1; i < n; i++ {
        f = f + s - n*nums[n-i]
        ans = max(ans, f)
    }
    return ans
}

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function maxRotateFunction(nums: number[]): number {
    const n = nums.length;
    const sum = nums.reduce((r, v) => r + v);
    let res = nums.reduce((r, v, i) => r + v * i, 0);
    let pre = res;
    for (let i = 1; i < n; i++) {
        pre = pre - (sum - nums[i - 1]) + nums[i - 1] * (n - 1);
        res = Math.max(res, pre);
    }
    return res;
}

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impl Solution {
    pub fn max_rotate_function(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let sum: i32 = nums.iter().sum();
        let mut pre: i32 = nums.iter().enumerate().map(|(i, &v)| (i as i32) * v).sum();
        (0..n)
            .map(|i| {
                let res = pre;
                pre = pre - (sum - nums[i]) + nums[i] * ((n - 1) as i32);
                res
            })
            .max()
            .unwrap_or(0)
    }
}

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