Problem Description:
Given an integer array nums, 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 - 10 <= k <= 105
Problem Link : Rotate Array – LeetCode
Solution(s):
Approach 1: Using temporary array
- Create a separate array named tempArr[] of size k.
- First copy last k elements from original array to tempArr.
- Now move first n-k elements from original array to last of original array.
- Then copy all elements from tempArr to original array.
Java Code:
public void rotate(int[] nums, int k) {
k = k % nums.length;
//Create a temporary array of size K
int[] tempArr = new int[k];
//Copy last K elemets from original array to tempArr
for (int i = 0; i < k; i++) {
tempArr[i] = nums[nums.length-k+i];
}
//Move first n-k elements of original array to the end of it.
for (int i = nums.length-1; i >= k; i--) {
nums[i] = nums[i-k];
}
//Copy all the elements of tempArr to orignal array
for (int i = 0; i < k; i++) {
nums[i] = tempArr[i];
}
}
}Complexity:
- Time-Complexity : O(n)
- Space Complexity : O(K)
Approach 2 : In-Place
- the array is first reversed in its entirety.
- then the first “k” elements are reversed, and
- finally the remaining elements are reversed.
Java Code :
public void rotate(int[] nums, int k) {
k = k % nums.length;
reverseArr(nums, 0, nums.length-1);
reverseArr(nums, 0, k-1);
reverseArr(nums, k, nums.length-1);
}
static void reverseArr(int[] nums, int start, int end) {
while(start <= end) {
int temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
start++;
end--;
}
}Complexity:
- Time-Complexity: O(n)
- Space Complexity: O(1)