Added : Search in a Rotated Sorted Array

Author: rohitkumar-rk

LeetCode/Array/33. Search in Rotated Sorted Array/Problem Statement.txt

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+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
+
+You are given a target value to search. If found in the array return its index, otherwise return -1.
+
+You may assume no duplicate exists in the array.
+
+Your algorithm's runtime complexity must be in the order of O(log n).
+
+Example 1:
+
+Input: nums = [4,5,6,7,0,1,2], target = 0
+Output: 4
+Example 2:
+
+Input: nums = [4,5,6,7,0,1,2], target = 3
+Output: -1

LeetCode/Array/33. Search in Rotated Sorted Array/Solution.java

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+/*
+@lc id : 33
+@problem : Search in Rotated Sorted Array
+@author : rohit
+@url : https://leetcode.com/problems/search-in-rotated-sorted-array/
+@difficulty : medium
+*/
+
+class Solution {
+    
+    public int binarySearch(int[] nums, int start, int end, int target){
+        //Base Case
+        if(start > end)
+            return -1;
+        
+        //Rec case
+        int mid = (end + start) / 2;
+        
+        if(nums[mid] == target)
+            return mid;
+        
+        //Checking where mid point lies
+        //On first part or second part
+        
+        //Mid lies in Left part or first line
+        if(nums[start] <= nums[mid]){
+            
+            if(target >= nums[start] && target <= nums[mid])
+                return binarySearch(nums, start, mid-1, target);
+            else return binarySearch(nums, mid+1, end, target);
+            
+        }
+        
+        //Mid lies in second line
+        
+        if(target >= nums[mid] && target <= nums[end])
+            return binarySearch(nums, mid+1, end, target);
+        
+        return binarySearch(nums, start, mid-1, target);
+        
+    }
+    
+    public int search(int[] nums, int target) {
+        return binarySearch(nums, 0, nums.length-1, target);
+        
+    }
+}