Today I am going to show how to solve the Leetcode Rotate Array algorithm problem.

Here is the problem:

There are several ways to solve this problem. The most basic solution is to remove the last element from the array and place it in front of the array K number of times.

var rotate = function(nums, k) {
    for (let i = 0; i < k; i++) {
        nums.unshift(nums.pop()) 
    }
    return nums;
};

But I decided to use a different approach. In this solution, I first reverse all elements of the array. Afterwards, I reverse again the first k elements and then reverse the rest of elements.

Original List : 1 2 3 4 5 6 7
After reversing all numbers : 7 6 5 4 3 2 1
After reversing first k numbers : 5 6 7 4 3 2 1
After revering last n-k numbers (Result) : 5 6 7 1 2 3 4

1) First I create my own reverse function.

var rotate = function(nums, k) {

    var reverse = function(nums, start, end) {
        while (start < end) {
            var temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start ++;
            end --;
        }
    }
};

2) Then I can start reversing elements of the array.

var rotate = function(nums, k) {

    var reverse = function(nums, start, end) {
        while (start < end) {
            var temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start ++;
            end --;
        }
    }

    k = k % nums.length
    reverse(nums, 0, nums.length - 1)
    reverse(nums, 0, k-1)
    reverse(nums, k, nums.length - 1)
};

You might be wondering why we need to do k = k % nums.length

Imagine we have an array [1,2,3,4,5]. What happens when you rotate it by 5 steps? All of the elements stay in place. Essentially, no change occurred because the size of the array is also 5.
So rotating it by 5 (the size of the array) gives you the same result as rotating it by 0. Similarly, rotating it by 6 gives you the same result as rotating it by 1. Rotating it by 7 gives you the same result as rotating it by 2.