Completed BinarySearch -1

Rotated_Sorted_Array.py

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+# Time Complexity : O(log n) - binary search.
+#  Space Complexity : O(1) - storing only pointers.
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+#Your code here along with comments explaining your approach in three sentences only
+#Search using binary search check if middle is the target if yes return.
+#If not check if left is sorted or right is sorted. If left is sorted check if target is in the range of left else check in right. 
+class Solution(object):
+    def search(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: int
+        """
+
+        l , r = 0 , len(nums) - 1
+        while l <= r:
+            mid = l + (r - l) // 2
+            if target == nums[mid]:
+                return mid
+
+            #left sorted
+            if nums[l] <= nums[mid]:
+                #right side
+                if target > nums[mid] or target < nums[l]:
+                    l = mid + 1
+                else:
+                    r = mid - 1
+            # Right sorted
+            else:
+                #left side
+                if target < nums[mid] or target > nums[r]:
+                    r = mid - 1
+                else:
+                    l = mid + 1
+        return -1

Search_2D_Matrix.py

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+# Time Complexity : O(log(m*n)) - binary search on rows and columns.
+# Space Complexity : O(1) - storing only variables.
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : Trouble calculating rows and columns part.
+
+#Your code here along with comments explaining your approach in three sentences only
+#Converting 2d matric to imagneary 1d array.
+#calculating r,c using mid to apply binary search and get return value.
+
+class Solution(object):
+    def searchMatrix(self, matrix, target):
+        """
+        :type matrix: List[List[int]]
+        :type target: int
+        :rtype: bool
+        """
+
+        m = len(matrix)
+        n = len(matrix[0])
+        l = 0
+        h = m * n - 1
+
+        while l <= h:
+            mid = l + (h - l) // 2
+            r = mid // n
+            c = mid % n
+            if matrix[r][c] == target:
+                return True
+            elif matrix[r][c] > target:
+                h = mid - 1
+            else:
+                l = mid + 1
+        return False

Sorted_Array_Unknow_Size.py

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+# Time Complexity : O(log n)
+# Space Complexity : O(1) - storing only variables.
+# Did this code successfully run on Leetcode : No - premium subscription.
+# Any problem you faced while coding this : No
+
+#Your code here along with comments explaining your approach in three sentences only
+#Until high is less than target or get out of bounds keep incresing it 2 times of low to get a range of numbers.
+#Once range is found apply binary search to find the target.
+
+class ArrayReader(object):
+    def __init__(self, arr):
+        self.data = arr
+
+    def get(self, index):
+        if index < len(self.data):
+            return self.data[index]
+        else:
+            # LeetCode returns 2^31 - 1 for out-of-bounds access
+            return 2147483647
+
+class Solution(object):
+    def search(self, reader, target):
+        """
+        :type reader: ArrayReader
+        :type target: int
+        :rtype: int
+        """
+        low, high = 0, 1
+        while reader.get(high) < target:
+            low = high
+            high = high * 2
+        while low <= high:
+            mid = low + (high-low) // 2
+            if reader.get(mid) == target:
+                return mid
+            elif reader.get(mid) < target:
+                low = mid + 1
+            else:
+                high = mid - 1
+        return -1
+
+# --- Local Testing Block ---
+if __name__ == "__main__":
+    # Example usage:
+    test_data = [-1, 0, 3, 5, 9, 12]
+    target_val = 9
+
+    reader = ArrayReader(test_data)
+    sol = Solution()
+    print(f"Result: {sol.search(reader, target_val)}")