Adding Efficient Solution for Problem - 33 - Search Rotated Sorted Array

Author: Vanditg

Search_Rotated_Sorted_Array/EfficientSolution.py

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+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 33
+## Problem Name: Search in Rotated Sorted Array    
+##===================================
+#
+#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
+class Solution:
+    def search(self, nums, target):
+        if not in nums:	#Condition-check: If nums is empty
+            return -1	#Return -1 
+        low = 0	#Initialize low 
+        high = len(nums) - 1	#Initialize high 
+        while low < high:	#Loop 
+            mid = (low + high) // 2	#Initialize mid point
+            if target == nums[mid]:	#Condition-check: If target is mid 
+                return mid 	#We return mid 
+            if nums[low] <= nums[mid]:	#Condition-check: If nums[low] <= nums[mid]
+                if nums[low] <= target <= nums[mid]:	#Condition-check: If target is in the range 
+                    high = mid - 1	#Update high by mid - 1
+                else:	#Condition-check: Else
+                    low = mid + 1	#Update low by mid + 1
+            else:
+                if nums[mid] <= target <= nums[high]	#Condition-check: If target is in the range
+                    low = mid + 1	#Update low by mid + 1
+                else:	#Condition-check: Else
+                    high = mid - 1	#Update high by mid - 1
+		return -1	#Otherwise we'll return -1