LeetCode-in-All

189. Rotate Array

Medium

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3

Output: [5,6,7,1,2,3,4]

Explanation:

rotate 1 steps to the right: [7,1,2,3,4,5,6]

rotate 2 steps to the right: [6,7,1,2,3,4,5]

rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: nums = [-1,-100,3,99], k = 2

Output: [3,99,-1,-100]

Explanation:

rotate 1 steps to the right: [99,-1,-100,3]

rotate 2 steps to the right: [3,99,-1,-100]

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

Follow up:

• Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

Solution

#include <stdio.h>

// Helper function to reverse a subarray in place
void reverse(int* nums, int l, int r) {
    while (l < r) {
        int temp = nums[l];
        nums[l] = nums[r];
        nums[r] = temp;
        l++;
        r--;
    }
}

// Function to rotate the array to the right by k steps
void rotate(int* nums, int numsSize, int k) {
    int n = numsSize;
    k = k % n;  // Handle cases where k is greater than n
    int t = n - k;
    
    // Reverse different parts of the array in three steps
    reverse(nums, 0, t - 1);   // Reverse the first part (0 to t - 1)
    reverse(nums, t, n - 1);   // Reverse the second part (t to n - 1)
    reverse(nums, 0, n - 1);   // Reverse the whole array (0 to n - 1)
}

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