Binary Search: Search in Rotated Sorted Array in C++ #432

Author: Yashi11

Binary Search/searchInRotatedSortedArray.cpp

@@ -0,0 +1,88 @@
+/*
+	Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
+	You must write an algorithm with O(log n) runtime complexity.
+
+	Input: nums = [4,5,6,7,0,1,2], target = 0
+	Output: 4
+	
+	Input: nums = [4,5,6,7,0,1,2], target = 3
+	Output: -1
+	
+	Input: nums = [1], target = 0
+	Output: -1
+
+	Constraints:
+	>  1 <= nums.length <= 5000
+	>  -104 <= nums[i] <= 104
+	>  All values of nums are unique.
+	>  nums is an ascending array that is possibly rotated.
+	>  -104 <= target <= 104
+
+	APPROACH
+	1. Find the index at which the array has been rotated with the help of Binary Search.
+	2. Store the index of rotation as INDX and find the range where target might be found using the following comparision:
+			RANGE = {INDX+1, HIGH} if TARGET < NUMS[LOW]
+			RANGE = {LOW, INDX-1}  if TARGET > NUMS{HIGH]
+	3. Perform Binary Search for TARGET in the required range.
+	4. If target is not found return -1.
+
+	TIME COMPLEXITY : O(NlogN)  SPACE COMPLEXITY: O(1)
+*/
+class Solution {
+public:
+    int search(vector<int>& nums, int target) {
+        // Initialize Variable for later Usage
+        int n=nums.size();
+        int low=0;
+        int high=n-1;
+
+        //Find the Rotation Point in the Rotated Array
+        while(low<=high){
+            int mid=(low+high)/2;
+            if(mid==0 || mid==n-1){
+                low=mid;
+                break;
+            }
+            if(nums[mid-1]>nums[mid] && nums[mid+1]>nums[mid]){
+                low=mid;
+                break;
+            }
+            else if(nums[mid]>nums[low] && nums[mid]>nums[high]){
+                low=mid+1;
+            }
+            else{
+                high=mid-1;
+            }
+        }
+
+        // Re-initialize Variables Needed
+        int indx=low;
+        low=0, high=n-1;
+        if(target==nums[indx]){
+            return indx;
+        }
+        else if(target>nums[high]){
+            high=indx-1;
+        }
+        else if(target<nums[low]){
+            low=indx+1;
+        }
+
+        // Binary  Search for Target in range low-high
+        while(low<=high){
+            int mid=(low+high)/2;
+            if(nums[mid]==target){
+                return mid;
+            }
+            else if(target>nums[mid]){
+                low=mid+1;
+            }
+            else{
+                high=mid-1;
+            }
+        }
+
+        // If target not found return -1
+        return -1;
+    }
+};