Leetcode 396 - Rotate Function

Note:

  • By the example, it feels like there might be some relations between each state.
  • Bingo! It’s a DP question.
  • I soon found that by adding sum into each state and just subtract length * endNum from it, we can get the new dp[i].
  • dp[i] = dp[i - 1] + sum - length * sum.

Question:

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:

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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.

Code:

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/**
* @param {number[]} nums
* @return {number}
*/
var maxRotateFunction = function(nums) {
const sum = nums.reduce((acc, cur) => acc + cur);
let dp = [...Array(nums.length).fill(0)];
dp[0] = nums.reduce((acc, cur, i) => acc + cur * i, 0);
let max = dp[0];
let end = nums[nums.length - 1];
const length = nums.length;
for (let i = 1; i < nums.length; i++) {
dp[i] = dp[i - 1] + sum - length * end;
end = nums[length - i - 1];
max = Math.max(max, dp[i]);
}
return max;
};