'binary search implementation with python

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)
pythonalgorithmpython-2.x


Solution 1:[1]

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)

Solution 2:[2]

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

Solution 3:[3]

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

Solution 4:[4]

Binary Search:

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")

Solution 5:[5]

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

Solution 6:[6]

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))

Solution 7:[7]

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

Solution 8:[8]

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))

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 Rustam Safin
Solution 3 Vivek Jain
Solution 4 Nazmul Hassan Riaz
Solution 5 David Yachnis
Solution 6 vamshi madineni
Solution 7 Hamid Azimi
Solution 8 Sar ibra

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