Binary Search Implementation In Python

I am trying to implement a solution using binary search. I have a list of numbers list = [1, 2, 3, 4, 6] value to be searched = 2 I have written something like this def searchBina

Solution 1:

A few things.

1. Your naming is inconsistent. Also, do not use list as a variable name, you're shadowing the global builtin.

2. The stopping condition is while low <= high. This is important.

3. You do not break when you find a value. This will result in infinite recursion.

---

def searchBinary(l2s, sval): # do not use 'list'as a variable
    low = 0
    high = len(l2s) 

    while low <= high: # thisis the main issue. As long as low is not greater than high, the while loop must run
        mid = (high + low) // 2if l2s[mid] == sval:
            print("found : ", sval)
            return
        elif l2s[mid] > sval:
            high = mid - 1else:
            low = mid + 1

And now,

list_ = [1, 2, 3, 4, 6]
searchBinary(list_, 2)

Output:

found :2

Solution 2:

UPDATEhigh = len(lst) - 1 per comments below.

---

Three issues:

1. You used l2s instead of list (the actual name of the parameter).
2. Your while condition should be low <= high, not low < high.
3. You should presumably return the index when the value is found, or None (or perhaps -1?) if it's not found.

A couple other small changes I made:

• It's a bad idea to hide the built-in list. I renamed the parameter to lst, which is commonly used in Python in this situation.
• mid = (low + high) // 2 is a simpler form of finding the midpoint.
• Python convention is to use snake_case, not camelCase, so I renamed the function.

Fixed code:

defbinary_search(lst, sval):
    low = 0
    high = len(lst) - 1while low <= high:
        mid = (low + high) // 2if lst[mid] == sval:
            return mid
        elif lst[mid] > sval:
            high = mid - 1else:
            low = mid + 1returnNoneprint(binary_search([1, 2, 3, 4, 6], 2))  # 1