Done Binary Search 1

Problem-1.py

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+# Time Complexity : O(log(m * n))
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : NA
+
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        # Approach: Treat the 2D matrix as a 1D sorted array to apply binary search
+        m = len(matrix)
+        n = len(matrix[0])
+        low = 0  # Initialize the low pointer
+        high = m * n - 1  # High pointer is the last element index (total elements - 1)
+
+        while low <= high:
+            mid = low + (high - low) // 2  # Calculate mid index
+            row = mid // n  # Convert 1D mid index to row index
+            column = mid % n  # Convert 1D mid index to column index
+
+            # Check if the mid element matches the target
+            if matrix[row][column] == target:
+                return True
+            # If mid element is smaller than target, search the right half
+            elif matrix[row][column] < target:
+                low = mid + 1
+            # If mid element is greater than target, search the left half
+            else:
+                high = mid - 1
+
+        # Target not found
+        return False

Problem-2.py

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+# Time Complexity : O(log n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : NA
+
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: int
+        """
+
+        left = 0  # Initialize left pointer
+        right = len(nums) - 1  # Initialize right pointer (last index)
+
+        while left <= right:
+            mid = (left + right) // 2  # Calculate mid index
+
+            if nums[mid] == target:  # Check if the mid element matches the target, return index
+                return mid
+
+            # Check if the left half is sorted
+            if nums[left] <= nums[mid]:
+                # If target lies within the left sorted half
+                if nums[left] <= target < nums[mid]:
+                    right = mid - 1
+                else:
+                    left = mid + 1
+            else:
+                # Right half is sorted
+                if nums[mid] < target <= nums[right]:
+                    left = mid + 1
+                else:
+                    right = mid - 1
+
+        # Target not found
+        return -1

Problem-3.py

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+
+# Time Complexity : O(log(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : NA(Premium program)
+# Any problem you faced while coding this : NA
+def search( arr, target: int) -> int:
+        # Approach: Expand the search boundary exponentially until 'high' is greater than target
+        low, high = 0, 1 
+        while arr[high]< target:
+            low =high
+            high*=2
+        while(low<=high):
+                mid = (low+high)//2 # Calculate mid index
+                # Check if the mid element matches the target
+                if arr[mid]==target: 
+                    return mid
+                 # If mid element is smaller than target, search the right half
+                elif arr[mid]< target:
+                    low= mid+1
+
+                # If mid element is greater than target, search the left half
+
+                else:
+                    high = mid -1
+        # Target not found            
+        return -1
+
+
+
+
+
+print(search([1,3,4,5,6,7,8,34,67,89,100,237, 268, 300,344, 566,799,800], 11))