189. Rotate Array

 

Intuition

we can solve it by finding the right possition for each elements

Approach

first find mod of k with length of array, than we have to reverse a array only 3 times first from whole than 0 to k-1 and lastly from k to last index

Complexity

  • Time complexity:O(n)
  • Space complexity:O(1)

Code

Java
class Solution {
    public void rotate(int[] nums, int k) {
        k %= nums.length;
        // for(int i =0; i<k; i++){
        //     int temp = nums[nums.length-1];
        //     for(int j =nums.length-1 ; j>0;j--){
        //         nums[j] = nums[j-1];
        //     }
        //     nums[0] = temp;
        // }
        reverse(nums, 0, nums.length-1);
        reverse(nums, 0, k-1);
        reverse(nums, k, nums.length-1);
    }
    public void reverse(int[] nums, int start, int end){
        while(start < end){
            int temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start++;
            end--;
        }
    }
}

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