Binary Search 1

Problem_1.py

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+'''
+The rotated sorted array will always have one part that is fully sorted.
+
+Finding the mid and then see which side of the mid is sorted.
+In the sorted side try to see in the numbers are in the range of target or it contains target by checking the first 
+and the last item of the side.
+
+If the target is not in the sorted side, do further binary search in the other side.
+If target is in the sorted side, do the further binary search in the sorted side.
+'''
+
+class Solution:
+    def search_helper(self, nums: List[int], l: int, r: int, target: int):
+        if l > r:
+            return -1
+
+        m = l + (r-l)//2
+
+        if nums[m] == target:
+            return m
+
+        # Left of mid is sorted array
+        if nums[l] <= nums[m]:
+            # If target lies in the sorted array
+            if nums[l] <= target < nums[m]:
+                return self.search_helper(nums, l, m-1, target)
+            else:
+                return self.search_helper(nums, m+1, r, target)
+
+        # Right of mid is sorted array
+        else:
+            # If target lies in the sorted array
+            if nums[m] < target <= nums[r]:
+                return self.search_helper(nums, m+1, r, target)
+            else:
+                return self.search_helper(nums, l, m-1, target)
+
+
+    def search(self, nums: List[int], target: int) -> int:
+        return self.search_helper(nums, 0, len(nums)-1, target)

Problem_2.py

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+'''
+We start with a search size of 1 and see if the array of search space can contain the target.
+If not we move the search window to high and move the high to 2*high meaning increase the search window by
+2 times.
+
+Once we get the search window where the target lies, we then do a binary search in the window to get the index.
+'''
+
+class Solution:
+    def search(self, reader: 'ArrayReader', target: int) -> 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
+            if reader.get(mid) > target:
+                high = mid - 1
+            else:
+                low = mid + 1
+
+        return -1

Problem_3.py

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+'''
+First we search the first column to find appropriate row.
+If the mid row has numbers that can contain the target then we return the mid to do a second searh 
+to find the target in the row.
+
+Then we perform the binary search on the row to find the target and return True if present or False if absent.
+'''
+
+class Solution:
+
+    def searchRow(self, matrix, target):
+        l = 0
+        r = len(matrix)-1
+
+        while l <= r:
+            m = l + (r-l)//2
+
+            if matrix[m][0] <= target <= matrix[m][-1]:
+                return m
+
+            if matrix[m][0] > target:
+                r = m-1
+            else:
+                l = m+1
+
+        return -1
+
+    def searchCol(self, matrix, row, target):
+        l = 0
+        r = len(matrix[row]) - 1
+
+        while l <= r:
+            m = l + (r-l)//2
+
+            if matrix[row][m] == target:
+                return True
+
+            if matrix[row][m] > target:
+                r = m-1
+            else:
+                l = m+1
+
+        return False
+
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        row = self.searchRow(matrix, target)
+        print(row)
+        if row == -1:
+            return False
+
+        return self.searchCol(matrix, row, target)
+