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.length1 <= n <= 105-100 <= nums[i] <= 100
Solutions
Solution 1
PythonJavaC++GoTypeScriptRust
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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]
if ans < f {
ans = f
}
}
return ans
}(code-box)
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;
}(code-box)
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)
}
}(code-box)
