Binary search done

Problem1.py

@@ -0,0 +1,75 @@
+#Search a 2D Matrix
+
+#Space complexity O(1), Time complexity O(log(m) + log(n))
+#Binary search on 1st column of rows will be used to identify which row target can be present
+#Binary search on that row will tell if target is present or not
+
+class Solution:
+#     def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+
+#         # edge case checks
+#         if matrix[0][0] > target:
+#             return False
+
+#         if matrix[len(matrix)-1][len(matrix[0])-1] < target:
+#             return False
+
+#         #1st for the input using binary search found which row does the number belongs using 1st column of each row
+#         rowLow = 0
+#         highRow = len(matrix) - 1
+
+#         while (rowLow<highRow):
+#             mid  = (rowLow+highRow)//2 + 1
+#             # if we found target return true
+#             if matrix[mid][0] == target:
+#                 return True
+
+#             if target < matrix[mid][0] :
+#                 # target is before the mid
+#                 highRow = mid - 1
+#             else:
+#                 rowLow  = mid
+
+#         # once and high are equals means we found the row in which the target can lie
+
+#         # now binary search on that row
+
+#         lowC  = 0
+#         highC = len(matrix[rowLow]) - 1
+
+#         while lowC <= highC:
+#             mid = (lowC + highC)//2
+#             if matrix[rowLow][mid] == target:
+#                 return True
+#             if target < matrix[rowLow][mid] :
+#                 highC = mid -1
+#             else:
+#                 lowC = mid + 1
+
+#         return False
+
+# This solution is more cleaner however, might be a bit slower due to the extra mathematical computations
+# The time complexity for both will be O(log(m*n))
+
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        # assuming it is a long array just place in 2d patrix
+        left = 0
+        right = len(matrix) * len(matrix[0]) - 1
+        rowLength = len(matrix[0])
+        while left<=right:
+            mid = (left+right)//2
+            #Now we need to find where in the 2d matrix this mid will be present
+            #if we take the length of 1st row, devision with that should give us the row number
+            #while modulus will give the column
+            row = mid // rowLength
+            column = mid % rowLength
+            element = matrix[row][column]
+            print("left is {}, right is {}, mid is {}, row is {}, column is {} and element at mid is {}".format(left, right, mid, row, column,element))
+            if element == target:
+                return True
+
+            if target >= matrix[row][column] :
+                left = mid + 1
+            else:
+                right = mid - 1
+        return False

Problem2.py

@@ -0,0 +1,25 @@
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        left = 0
+        right = len(nums) - 1
+
+        while left <= right:
+            mid = left+(right-left)//2
+            if nums[mid] == target:
+                return mid
+
+            if nums[mid] >= nums[left]:
+                # left is sorted, or both are sorted
+                if target >= nums[left] and target < nums[mid]:
+                    right = mid - 1
+                else:
+                    left = mid + 1
+
+            else:
+                # right is sorted
+                if target > nums[mid] and target <= nums[right]:
+                    left = mid  + 1
+                else:
+                    right = mid -1
+
+        return -1

Problem3.py

@@ -0,0 +1,33 @@
+# """
+# 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: 'ArrayReader', target: int) -> int:
+        #search the 1st occurence of the array out of bound index to get the length 
+        #of the arrya, then normal binary search in sorted array
+        left = 0
+        right = 1
+        valAtRight = reader.get(right)
+        while target >= valAtRight:
+            if valAtRight == target:
+                return right
+            left  = right
+            right = right*2
+            valAtRight = reader.get(right)
+
+        while left <= right:
+            mid = left + (right-left)//2
+            val = reader.get(mid)
+            if val == target:
+                return mid
+
+            if target > val:
+                left = mid+1
+            else:
+                right = mid -1
+
+        return -1