Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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] <= 100Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math · Dynamic Programming
[4,3,2,6]
[100]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #396: Rotate Function
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;
}
}
// Accepted solution for LeetCode #396: Rotate Function
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
}
# Accepted solution for LeetCode #396: Rotate Function
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
// Accepted solution for LeetCode #396: Rotate Function
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)
}
}
// Accepted solution for LeetCode #396: Rotate Function
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;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.