Solved Problem 1, 2 and 3

Problem-1.py

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+'''
+    Time Complexity: O(logn)
+    Space Complexity: O(1)
+    Ran successfully on leetcode
+    Difficulty: Was able to understand, but took some time to build the final logic
+'''
+
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        low, high = 0, len(nums) - 1
+
+        while low <= high:
+            mid = (low + high) // 2
+
+            if target == nums[mid]:
+                return mid
+
+            if nums[low] <= nums[mid]:
+                if target > nums[mid] or target < nums[low]:
+                    low = mid + 1
+                else:
+                    high = mid - 1 
+            else:
+                if target < nums[mid] or target > nums[high]:
+                    high = mid - 1
+                else:
+                    low = mid + 1
+
+        return -1
+

Problem-2.py

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+'''
+    Time Complexity: O(logn)
+    Space Complexity: O(1)
+    Ran successfully on leetcode
+    Difficulty: Solved quickly by implementing simple Binary search and handling the out of bounds situation, 
+    updated the code by using binary search for getting the range as well
+'''
+
+class Solution:
+    def search(self, reader: 'ArrayReader', target: int) -> int:
+        low, high = 0, 1
+
+        while reader.get(high) < target:
+            low = high
+            high = 2*high
+
+        while low <= high:
+            mid = (low + high) // 2
+            midValue = reader.get(mid)
+
+            if target == midValue:
+                return mid
+            elif target < midValue:
+                high = mid - 1
+            else:
+                low = mid + 1
+
+        return -1

Problem-3.py

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+'''
+    Time Complexity: O(log(mxn))
+    Space Complexity: O(1)
+    Ran successfully on leetcode
+    Difficulty: Used the mlogn solution before but updated it to log(mn) solution
+'''
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        m = len(matrix)
+        n = len(matrix[0])
+        low, high = 0, m*n - 1
+
+
+        while low <= high:
+            mid = low + (high - low)//2
+
+            r = mid // n
+            c = mid % n
+
+            if matrix[r][c] == target:
+                return True
+            elif matrix[r][c] < target:
+                low = mid + 1
+            else:
+                high = mid - 1
+
+        return False