Sort modification

Author: deathlyhallows010

Binary Search/Median of Array.py

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+# https://www.interviewbit.com/problems/median-of-array/
+
+# Median of Array
+
+# There are two sorted arrays A and B of size m and n respectively.
+# Find the median of the two sorted arrays ( The median of the array formed by merging both the arrays ).
+# The overall run time complexity should be O(log (m+n)).
+
+
+def Med(A):
+ mid = len(A)//2
+ 
+ if mid%2==0:
+ return (A[mid-1]+A[mid])/2
+ else:
+ return A[mid]
+ 
+
+class Solution:
+ # @param A : tuple of integers
+ # @param B : tuple of integers
+ # @return a double
+ def findMedianSortedArrays(self, A, B):
+ 
+ A = list(A)
+ B = list(B)
+ 
+ if len(A)==0 or len(B)==0:
+ C = A+B
+ if len(C) == 1:
+ return C[0]
+ else:
+ return Med(C)
+ 
+ return (Med(A)+Med(B))/2

Binary Search/Rotated Sorted Array Search.py

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+# https://www.interviewbit.com/problems/rotated-sorted-array-search/
+
+# Rotated Sorted Array Search
+
+# Given an array of integers A of size N and an integer B.
+# array A is rotated at some pivot unknown to you beforehand.
+# You are given a target value B to search. If found in the array, return its index, otherwise return -1.
+# You may assume no duplicate exists in the array.
+
+def BinarySearch(A,target):
+ start = 0
+ end = len(A) - 1
+ 
+ while(start<=end):
+ mid = (start + end)//2
+ if A[mid] == target:
+ return mid
+ elif A[mid]>target:
+ end = mid -1
+ else:
+ start = mid + 1
+ 
+ return -1
+
+def findMin(A):
+ start = 0
+ end = len(A)-1
+ N = len(A)
+ while(start<=end):
+ if A[start]<=A[end]:
+ return start
+ mid = (start + end)//2
+ nxt = (mid+1)%N
+ prev = (mid-1+N)%N
+ if A[mid]<=A[nxt] and A[mid]<=A[prev]:
+ return mid
+ elif A[mid]<=A[end]:
+ end = mid -1
+ elif A[mid]>=A[end]:
+ start = mid +1
+ 
+ return -1
+
+class Solution:
+ # @param A : tuple of integers
+ # @return an integer
+ def search(self,A,B):
+ A = list(A)
+ n = findMin(A)
+ if n == -1:
+ return -1
+ d
+ 
+ A.sort()
+ return (BinarySearch(A,B)+n)%len(A)