solution(java): 189. Rotate Array

189. Rotate Array
- Java

Medium/189. Rotate Array/README.md

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+# [189. Rotate Array](https://leetcode.com/problems/rotate-array/)
+
+<div class="xFUwe" data-track-load="description_content"><p>Given an integer array <code>nums</code>, rotate the array to the right by <code>k</code> steps, where <code>k</code> is non-negative.</p>
+
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 3
+<strong>Output:</strong> [5,6,7,1,2,3,4]
+<strong>Explanation:</strong>
+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]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> nums = [-1,-100,3,99], k = 2
+<strong>Output:</strong> [3,99,-1,-100]
+<strong>Explanation:</strong> 
+rotate 1 steps to the right: [99,-1,-100,3]
+rotate 2 steps to the right: [3,99,-1,-100]
+</pre>
+
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
+	<li><code>0 &lt;= k &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+<p><strong>Follow up:</strong></p>
+
+<ul>
+	<li>Try to come up with as many solutions as you can. There are at least <strong>three</strong> different ways to solve this problem.</li>
+	<li>Could you do it in-place with <code>O(1)</code> extra space?</li>
+</ul>
+</div>
+
+---
+
+### Explaination of the approach:
+
+**Step 1:** n is assigned the length of the input array nums. This determines the number of elements in the array.
+
+**Step 2:** k is taken modulo n to ensure that it is within the range (0, n). This is because if k is larger than n, rotating the array by k positions is equivalent to rotating it by k % n positions.
+
+**Step 3:** If k is negative, it is adjusted by adding n to it. This is to handle cases where k is a negative number. Adding n to a negative k effectively rotates the array to the left.
+
+**Step 4:** An auxiliary array temp of the same length as nums is created. This array will be used to store the rotated elements.
+
+**Step 5:** The loop copies the elements from nums to temp with a right shift of k positions. The modulo operation is used to ensure that the indices wrap around if they exceed the length of the array.
+
+**Step 6:** Finally, another loop copies the rotated elements from the temp array back to the original nums array.
+
+### Time Complexity:
+
+The code performs two separate loops through the array, each with n iterations. Therefore, the time complexity is O(n).
+
+### Space Complexity:
+
+The code uses an additional temp array of the same size as the input array nums. So the space complexity is O(n) as well.

Medium/189. Rotate Array/Solution.java

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+class Solution {
+    public void rotate(int[] nums, int k) {
+        int n=nums.length;
+        k=k%nums.length;
+        if(k<0){
+        k=k+nums.length;
+        }
+        int[] temp = new int[n];
+         // Copy elements to the temp array with the right shift
+         for (int i = 0; i < n; i++) {
+            temp[(i + k) % n] = nums[i];
+        }
+        // Copy elements back to the original array
+        for (int i = 0; i < n; i++) {
+            nums[i] = temp[i];
+        }
+    }
+}