Rotate Function

Rotate Function - Problem

You are given an integer array nums of length n. Assume arr_k 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 * arr_k[0] + 1 * arr_k[1] + ... + (n - 1) * arr_k[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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,3,2,6]

› Output: 26

💡 Note: F(0) = 0×4+1×3+2×2+3×6 = 25, F(1) = 0×3+1×2+2×6+3×4 = 26, F(2) = 0×2+1×6+2×4+3×3 = 23, F(3) = 0×6+1×4+2×3+3×2 = 16. Maximum is F(1) = 26.

Example 2 — Single Element

$ Input: nums = [100]

› Output: 0

💡 Note: Only one rotation possible: F(0) = 0×100 = 0

Example 3 — All Same Elements

$ Input: nums = [1,1,1,1]

› Output: 6

💡 Note: All rotations give same result: F(0) = 0×1+1×1+2×1+3×1 = 6

Constraints

1 ≤ nums.length ≤ 10⁵
-100 ≤ nums[i] ≤ 100

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is recognizing the mathematical relationship between consecutive rotations. Instead of recalculating each F(k) from scratch, we can derive F(k+1) from F(k) using the formula: F(k+1) = F(k) + sum - n×nums[n-k-1]. Best approach uses this mathematical optimization with Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Calculate All Rotations

⏱️ Time: O(n²) Space: O(1)

For each possible rotation k, physically create the rotated array and calculate the weighted sum. Compare all results to find the maximum.

Mathematical Optimization

⏱️ Time: O(n) Space: O(1)

Instead of calculating each F(k) from scratch, derive a formula to compute F(k+1) from F(k) using the mathematical relationship between rotations.

Brute Force - Calculate All Rotations — Algorithm Steps

For each rotation k from 0 to n-1
Create rotated array arr_k
Calculate F(k) = sum of i * arr_k[i]
Track maximum value

Visualization

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Step-by-Step Walkthrough

Original Array

Start with nums = [4,3,2,6]

Calculate F(0), F(1), F(2), F(3)

For each rotation, compute weighted sum

Find Maximum

Return the largest F(k) value

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int solution(int* nums, int n) {
    int maxVal = INT_MIN;
    
    for (int k = 0; k < n; k++) {
        int currentSum = 0;
        for (int i = 0; i < n; i++) {
            currentSum += i * nums[(i + k) % n];
        }
        if (currentSum > maxVal) {
            maxVal = currentSum;
        }
    }
    
    return maxVal;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing - remove brackets and split by comma
    int nums[1000];
    int n = 0;
    char *token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n rotations, we calculate the sum over n elements

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track current sum and maximum

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Calculate All Rotations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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