added Rotate Array (easy)

Author: minhyeong-joe

Easy/RotateArray/README.md

@@ -0,0 +1,37 @@
+
+# Rotate Array
+[Leetcode Link](https://leetcode.com/problems/rotate-array/)
+
+## Problem:
+
+Given an array, rotate the array to the right by *k* steps, where *k* is non-negative.
+
+**Follow up:**
+
+- Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
+- Could you do it in-place with O(1) extra space?
+ 
+
+## Example:
+
+```
+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]
+```
+```
+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]
+```
+
+## Note:
+
+- `1 <= nums.length <= 2 * 10^4`
+- It's guaranteed that `nums[i]` fits in a 32 bit-signed integer.
+- `k >= 0`

Easy/RotateArray/solution.py

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+from typing import List
+
+class Solution:
+ def rotate(self, nums: List[int], k: int) -> None:
+ """
+ Do not return anything, modify nums in-place instead.
+ """
+ # rotations repeat in modulo
+ # k %= len(nums)
+ i = 0
+ count = 0
+ while count < len(nums):
+ start = i
+ copy = nums[i]
+ while True:
+ j = (i + k) % len(nums) # allows end of array "cycle" back to beginning of array
+ copy, nums[j] = nums[j], copy
+ i = j
+ count += 1
+ if i == start:
+ break
+ i += 1
+
+
+
+# test driver
+sol = Solution()
+nums = [1,2]
+k = 3
+sol.rotate(nums, k)
+print("Output:", nums)

README.md

@@ -41,6 +41,7 @@ Languages used: Java and Python
  - [Longest Continuous Increasing Subsequence](Easy/LongestIncreasingSubsequence)
  - [Happy Number](Easy/HappyNumber)
  - [Assign Cookies](Easy/AssignCookies)
+ - [Rotate Array](Easy/RotateArray)
 - Medium
  - [Minimum Add to Make Parentheses Valid](Medium/MinimumAddtoMakeParenthesesValid)
  - [Distribute Coins in Binary Tree](Medium/DistributionCoinsInBinaryTree)