Complete binary search 1

Problem1.py

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+# SEARCH IN ROTATED SORTED ARRAY
+# https://leetcode.com/problems/search-in-rotated-sorted-array/description/
+class Solution(object):
+    def search(self, nums, target):
+        """
+        Since its a partially rotated array half of array will be sorted out and we will have to do the comparisons and accordingly do binarys search
+        :type nums: List[int]
+        :type target: int
+        :rtype: int
+        """
+        #initialize low and high boundary
+        low, high = 0, len(nums)-1
+        while (low<=high):
+            #get mid index
+            mid = (low + high)//2
+            if(nums[mid]==target):
+                return mid
+            ##Left side sorted, check if target lies between low index and mid and accordingly update high index else update low index if target lies on right side
+            if(nums[low] <= nums[mid]):
+                if(nums[low]<=target < nums[mid]):
+                    high = mid - 1
+                else:
+                    low = mid + 1
+            else:
+                if(nums[mid]<target<=nums[high]):
+                    low = mid + 1
+                else:
+                    high = mid - 1
+        return -1

Problem2.py

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+#Search in a sorted array of unknown size
+import ArrayReader
+
+class Solution: 
+    def search(self, reader: ArrayReader, target: int) -> int:
+        #Step 1 is to find the low and high boundaries
+        low, high = 0, 1
+        #Perform while the value at index high is less than target, initialize low to high 
+        #multiply high by 2, the reason why we do that is for balance between time complexity 
+        #and ensures we find time complexity of O(log n)
+        while(reader.get(high) < target):
+            low = high
+            high *= 2
+
+        while low <= high:
+            mid=(low + high)//2
+            value = reader.get(mid)
+            if(value==target):
+                return mid
+            elif value < target:
+                high = mid - 1
+            else:
+                low = mid + 1
+        return -1 

Problem3.py

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+# SEARCH 2D MATRIX
+# https://leetcode.com/problems/search-a-2d-matrix/description/
+
+class Solution(object):
+    def searchMatrix(self, matrix, target):
+        """
+        Goal is consider the 2d matrix as a 1d array and then do binary search
+        :type matrix: List[List[int]]
+        :type target: int
+        :rtype: bool
+        """
+        m, n = len(matrix), len(matrix[0])
+        low, high = 0, (m * n - 1)
+
+        while(low <= high):
+            mid = (low + high)//2
+            row = mid//n
+            col = mid % n
+            if(matrix[row][col] == target):
+                return True
+            elif(matrix[row][col] < target):
+                low = mid+1
+            else:
+                high = mid-1
+        return False
+
+