Completed BinarySearch-1

Exercise_1.py

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+# Time Complexity: O(log(m*n))
+# Space Complexity: O(1)
+# Were you able to run the code on Leetcode: Yes
+# Any problem you faced while coding this: No
+
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+
+        m = len(matrix)
+        n = len(matrix[0])
+
+        # set low and high to 0 and last flattened index
+        low,high = 0, m*n -1
+
+        while low<=high:
+            mid = low + (high-low)//2
+
+            # get the row and col idx by bounding it by n
+            # since array is sorted row wise
+            row = mid//n
+            col = mid%n
+
+            if matrix[row][col] == target:
+                return True
+            elif matrix[row][col] < target:
+                # search in right side of mid element 
+                low = mid + 1
+            elif matrix[row][col] > target:
+                # search left side of mid element
+                high = mid - 1
+        return False

Exercise_2.py

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+# Time Complexity: O(log n)
+# Space Complexity: O(1)
+# Were you able to run the code on Leetcode: Yes
+# Any problem you faced while coding this: No
+
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+
+        if not nums or len(nums) == 0:
+            return -1
+
+        low = 0
+        high = len(nums)-1
+
+        while(low<=high):
+            mid = low + (high-low) //2
+
+            if nums[mid] == target:
+                return mid
+
+            # check which side is sorted
+            if nums[low] <= nums[mid]:
+                # left side sorted
+                if nums[low] <= target < nums[mid]:
+                    # if target is in the sorted side then search there by decrementing high to mid -1
+                    high = mid - 1
+                else:
+                    low = mid+1
+            else:
+                # right side sorted
+                if nums[mid] < target <= nums[high]:
+                    # if target is in the sorted side then search there by incrementing low to mid+1
+                    low = mid+1
+                else:
+                    high = mid -1
+
+        return -1 
+
+
+

Exercise_3.py

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+# Time Complexity: O(log n)
+# Space Complexity: O(1)
+# Were you able to run the code on Leetcode: No
+# Any problem you faced while coding this: No
+
+
+# This is ArrayReader's API interface.
+# You should not implement it, or speculate about its implementation
+# """
+# class ArrayReader:
+#    def get(self, index: int) -> int:
+
+
+class Solution:
+    def search(self, reader, target):
+        """
+        :type reader: ArrayReader
+        :type target: int
+        :rtype: int
+        """
+        low, high = 0, 1
+
+        while reader.get(high) <= target:
+            # start with small nos, keep increasing high by factor of 2 for binary search until high is greater or equal to target
+            low = high
+            high *= 2
+
+        return self.searchRange(reader, target, low, high)
+
+
+
+
+    def searchRange(self, reader, target, low, high)-> int:
+
+        while low<=high:
+
+            mid = low+(high-low) //2 
+
+            if reader.get(mid) == target:
+                return mid
+
+            # search in left window if mid value greater than target else right window
+            if reader.get(mid) > target:
+                high = mid - 1
+            else:
+                low = mid + 1
+
+        return -1
+
+