"binary search in python" Code Answer's
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iterative binary search python
def binary_search(a, key):
low = 0
high = len(a) - 1
while low < high:
mid = (low + high) // 2
if key == a[mid]:
return True
elif key < mid:
high = mid - 1
else:
low = mid + 1
return False
binary search in python
def binary_search(item_list,item):
first = 0
last = len(item_list)-1
found = False
while( first<=last and not found):
mid = (first + last)//2
if item_list[mid] == item :
found = True
else:
if item < item_list[mid]:
last = mid - 1
else:
first = mid + 1
return found
Source: www.w3resource.com
binary search python
# This is real binary search
# this algorithm works very good because it is recursive
def binarySearch(arr, min, max, x):
if max >= min:
i = int(min + (max - min) / 2) # average
if arr[i] == x:
return i
elif arr[i] < x:
return binarySearch(arr, i + 1, max, x)
else:
return binarySearch(arr, min, i - 1, x)
binary search python
def binary_search(records:list, search_target, key=None):
"""Returns {'index':search_target_index, 'records':sorted_records_list} of
the search query else False.
Uses binary search algorithm. Searches both [1,2,3,...] & [{},{},{},...].
The later uses the "key" parameter else key=None.
For example:
r=binary_search(records=my_recs, search_target=4377017701570, key="orderId")
record=r['records'][r['index']]
~~~
l=binary_search(records=my_list, search_target='foo')
item=l['records'][l['index']]
Important Note:
Because the "records" parameter is a mutable of type list, the code in the
function will mutate and operate on the original argument object in memory
and not a copy of it. If you do not want the original object altered, pass
a copy and not the original object. The index returned corresponds to the
mutated/sorted object.
For example:
bs_list = [10,2,5,8,9,3,6,7,4,1]
bs_list_copy = bs_list.copy()
l = binary_search(records=bs_list_copy, search_target=5)
5 == l['records'][l['index']]
5 == bs_list_copy[[l['index']]
bs_list== [10,2,5,8,9,3,6,7,4,1]
bs_list_copy == [1,2,3,4,5,6,7,8,9,10]
Design choice:
Used "doesn't alter list" to eliminate mutation issue. Thus, "Important Note"
doesn't apply for this implementation.
Use "alters list" if you don't mind making a copy of the original list or if
you don't mind altering the original list; "Important Note" applies here.
"""
lower_bound = 0
upper_bound = len(records)-1
# List MUST be sorted in ascending order, due to the
# conditinal inequalities & the arithmitic.
# sort_records = f'records.sort(key=lambda record: record["{key}"], reverse=False)' # alters list
sort_records = f'sorted(records,key=lambda record: record["{key}"], reverse=False)' # doesn't alter list
reference_target = f'records[mid_index]["{key}"]'
if key == None:
sort_records = sort_records.replace(f'key=lambda record: record["{key}"], ','')
reference_target = reference_target.replace(f'["{key}"]','')
# eval(sort_records) # alters list
records = eval(sort_records) # doesn't alter list
while lower_bound <= upper_bound:
mid_index = (lower_bound+upper_bound) // 2
if search_target == eval(reference_target):
# Return records to remove sorting ambiguity;
# it has been sorted in ascending order.
return {'index':mid_index, 'records':records}
elif search_target < eval(reference_target):
upper_bound = mid_index - 1
else:
lower_bound = mid_index + 1
return False
Source: www.youtube.com
binary search in python
def binary_search(group, suspect):
group.sort()
midpoint = len(group)//2
while(True):
if(group[midpoint] == suspect):
return midpoint
if(suspect > group[midpoint]):
group = group[midpoint]
if(suspect < group[midpoint]):
group = group[0: midpoint]
midpoint = (len(group)//2)
code of binary search in python
def binary_search(mylist,low,k,key):
high = k - 1
mid = (low + high)//2
if mylist[mid]==key:
return mid
elif key > mylist[mid]:
return binary_search(mylist,mid + 1,k ,key)
else:
return binary_search(mylist,0,mid, key)
low = 0
k = int(input("Enter total amount of elements in k : "))
mylist = [int(input()) for x in range(k)]
key = int(input("Which element do we have to find: "))
print(binary_search(mylist,low,k,key))
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