Added leetcode solution for 198

dsa-solutions/lc-solutions/0100-0199/0198-house-robber.md

@@ -0,0 +1,176 @@
+---
+id: rotate-array
+title: Rotate Array
+sidebar_label: 0189 Rotate Array
+tags:
+  - Array
+  - LeetCode
+  - Java
+  - Python
+  - C++
+description: "This is a solution to the Rotate Array problem on LeetCode."
+---
+
+## Problem Description
+
+Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.
+
+### Examples
+
+**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
+
+- $1 <= nums.length <= 105$
+- $-231 <= nums[i] <= 231 - 1$
+- $0 <= k <= 105$
+
+
+## Solution for Candy Distribution Problem
+
+### Intuition And Approach
+
+To rotate an array to the right by k steps, we need to move the last k elements to the front and shift the rest of the elements to the right. A straightforward way to achieve this in-place (without using extra space for another array) is to use the reversal method.
+
+Here is the step-by-step approach:
+
+## Adjust k:
+
+If k is greater than the length of the array, rotating by k steps is the same as rotating by k % n steps (where n is the length of the array). This is because rotating by the length of the array brings it back to the original position.
+Calculate k = k % n.
+
+## Reverse the entire array:
+
+By reversing the entire array, the last k elements (which we want to move to the front) will be at the beginning, but in reverse order.
+For example, reversing [1, 2, 3, 4, 5, 6, 7] gives [7, 6, 5, 4, 3, 2, 1].
+
+## Reverse the first k elements:
+
+The first k elements are now the elements that were originally at the end of the array. Reverse these to restore their original order.
+Continuing the example, reversing the first 3 elements of [7, 6, 5, 4, 3, 2, 1] gives [5, 6, 7, 4, 3, 2, 1].
+
+## Reverse the remaining n - k elements:
+
+Finally, reverse the rest of the array (from the k-th element to the end) to restore their order.
+In the example, reversing the elements from index 3 to 6 of [5, 6, 7, 4, 3, 2, 1] gives [5, 6, 7, 1, 2, 3, 4].
+
+
+
+#### Code in Different Languages
+
+<Tabs>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@mahek0620"/>
+   ```java
+
+    class Solution {
+    public void rotate(int[] nums, int k) {
+        // Ensure k is within the bounds of the array length
+        k = k % nums.length;
+
+        // Reverse the entire array
+        reverse(nums, 0, nums.length - 1);
+
+        // Reverse the first k elements
+        reverse(nums, 0, k - 1);
+
+        // Reverse the remaining elements
+        reverse(nums, k, nums.length - 1);
+    }
+
+    // Helper function to reverse elements in the array from start to end
+    private void reverse(int[] nums, int start, int end) {
+        while (start < end) {
+            int temp = nums[start];
+            nums[start] = nums[end];
+            nums[end] = temp;
+            start++;
+            end--;
+        }
+    }
+
+    ```
+  </TabItem>
+  <TabItem value="Python" label="Python">
+  <SolutionAuthor name="@mahek0620"/>
+   ```python
+
+    class Solution(object):
+    def rotate(self, nums, k):
+        k = k % len(nums)
+        self.reverse(nums, 0, len(nums) - 1)
+        self.reverse(nums, 0, k - 1)
+        self.reverse(nums, k, len(nums) - 1)
+    def reverse(self, nums, start, end):
+        while start < end:
+            nums[start], nums[end] = nums[end], nums[start]
+            start += 1
+            end -= 1
+    ```
+  </TabItem>
+
+ <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@mahek0620"/>
+  ```cpp
+
+    #include <vector>
+    using namespace std;
+
+    class Solution {
+    public:
+    void rotate(vector<int>& nums, int k) {
+        // Ensure k is within the bounds of the array length
+        k = k % nums.size();
+
+        // Reverse the entire array
+        reverse(nums, 0, nums.size() - 1);
+
+        // Reverse the first k elements
+        reverse(nums, 0, k - 1);
+
+        // Reverse the remaining elements
+        reverse(nums, k, nums.size() - 1);
+    }
+
+    private:
+    void reverse(vector<int>& nums, int start, int end) {
+        while (start < end) {
+            int temp = nums[start];
+            nums[start] = nums[end];
+            nums[end] = temp;
+            start++;
+            end--;
+        }
+    }
+};
+    ```
+  </TabItem>
+</Tabs>
+
+
+
+## References
+
+- **LeetCode Problem:** [House robber Problem](https://leetcode.com/problems/house-robber/)
+- **Solution Link:** [House Robber Solution on LeetCode](https://leetcode.com/problems/house-robber/solutions/5273312/house-robber-solution)
+- **Authors GeeksforGeeks Profile:** [Mahek Patel](https://leetcode.com/u/mahekrpatel611/)