I think I did everything correctly, but the base case return None, instead of False if the value does not exists. I cannot understand why.
def binary_search(lst, value):
if len(lst) == 1:
return lst[0] == value
mid = len(lst)/2
if lst[mid] < value:
binary_search(lst[:mid], value)
elif lst[mid] > value:
binary_search(lst[mid+1:], value)
else:
return True
print binary_search([1,2,4,5], 15)
You need to return the result of the recursive method invocation:
def binary_search(lst, value):
#base case here
if len(lst) == 1:
return lst[0] == value
mid = len(lst)/2
if lst[mid] < value:
return binary_search(lst[:mid], value)
elif lst[mid] > value:
return binary_search(lst[mid+1:], value)
else:
return True
And I think your if and elif condition are reversed. That should be:
if lst[mid] > value: # Should be `>` instead of `<`
# If value at `mid` is greater than `value`,
# then you should search before `mid`.
return binary_search(lst[:mid], value)
elif lst[mid] < value:
return binary_search(lst[mid+1:], value)
O(k) in python. A better solution will pass in the original list with start:end indices. See more: interactivepython.org/runestone/static/pythonds/SortSearch/…Because if return nothing!
if lst[mid] < value:
binary_search(lst[:mid], value)
# hidden return None
elif lst[mid] > value:
binary_search(lst[mid+1:], value)
# hidden return None
else:
return True
You need to return from if and elif too.
def binary_search(lst, value):
#base case here
if len(lst) == 1:
return lst[0] == value
mid = len(lst) / 2
if lst[mid] < value:
return binary_search(lst[:mid], value)
elif lst[mid] > value:
return binary_search(lst[mid+1:], value)
else:
return True
>>> print binary_search([1,2,4,5], 15)
False
def Binary_search(num,desired_value,left,right):
while left <= right:
mid = (left + right)//2
if desired_value == num[mid]:
return mid
elif desired_value > num[mid]:
left = mid + 1
else:
right = mid - 1
return -1
num =[12,15,19,20,22,29,38,41,44,90,106,397,399,635]
desired_value = 41
result = Binary_search(num,desired_value,0,len(num)-1)
if result != -1:
print("Number found at " + str(result),'th index')
else:
print("number not found")
def rBinarySearch(list,element):
if len(list) == 1:
return element == list[0]
mid = len(list)/2
if list[mid] > element:
return rBinarySearch( list[ : mid] , element )
if list[mid] < element:
return rBinarySearch( list[mid : ] , element)
return True
def binary_search(lists,x):
lists.sort()
mid = (len(lists) - 1)//2
if len(lists)>=1:
if x == lists[mid]:
return True
elif x < lists[mid]:
lists = lists[0:mid]
return binary_search(lists,x)
else:
lists = lists[mid+1:]
return binary_search(lists,x)
else:
return False
a = list(map(int,input('enter list :').strip().split()))
x = int(input('enter number for binary search : '))
(binary_search(a,x))
None? The code presented here lacks code comments. binary_search(sequence, key) desperately needs a doc string: it is expected to run in O(log(len(sequence))) time, but lists.sort() makes it o(len(sequence)).def binary_search(arr, elm):
low, high = 0, len(arr) - 1
while low <= high:
mid = (high + low) // 2
val = arr[mid]
if val == elm:
return mid
elif val <= elm:
low = mid + 1
else:
high = mid - 1
return -1
print(binary_search([2, 3, 4, 6, 12, 19, 20, 21], 12)) # 4
print(binary_search([2, 3, 4, 6, 12, 19, 20, 21], 3333)) # -1
def Binary_search(li, e, f, l):
mid = int((f+l)/2)
if li[mid] == e:
print("Found",li[mid] )
elif f == l-1 and li[mid] != e:
print("Not Found ")
elif e < li[mid]:
Binary_search(li, e, f,mid)
elif e > li[mid]:
Binary_search(li, e, mid,l)
elements = [1,2,4,6,8,9,20,30,40,50,60,80,90,100,120,130,666]
Binary_search(elements, 120, 0, len(elements))
None? Does Binary_search() return anything useful? There is the Style Guide for Python Code.
class binar_search:
def __init__(self,arr , element):
self.arr = arr
self.element = element
def search(self):
n = len(self.arr)
low = 0
high = n-1
while(low <= high):
mid = (low+high)//2
if self.arr[mid] == self.element:
return mid
elif self.arr[mid] < self.element:
low = mid+1
else:
high = mid -1
return 0
