Need help with this code for binary search in python. Please tell me what's wrong in this?

Question

arr = [1,2,3,4,5]
n = int(input())
L = 0
H = (len(arr) - 1)

while L <= H:
M = (H+L)//2
if arr[M] == n:
print("Found")

elif arr[M] < n:
L = M+1

elif arr[M] > n:
H = M-1

else:
print("Not Found")

Answer 1

Edward Finkelstein why my code didn't work?

Answer 2

Because in the if-statement, you need a break or else you get an infinite loop. And the else does not correctly test for it it is not found. Actually, there is a bug in the code I just posted, it is better to check if a max number of iterations, which for binary search is given as log2(len(arr)).

arr = [1,2,3,4,5]
n = int(input())
L, count = 0, 0
H = (len(arr) - 1)

while L <= H:
M = (H+L)//2
if arr[M] == n:
print("Found")
break

elif len(arr) <= 2**count:
print("Not found")
break

elif arr[M] < n:
L = M+1

elif arr[M] > n:
H = M

count+=1

Answer 3

Better?

arr = [1,2,3,4,5]
n = int(input())
L = 0
H = (len(arr) - 1)
flag=False
while L <= H:
M = (H+L)//2
if arr[M] == n:
flag=True
break

elif arr[M] < n:
L = M+1

elif arr[M] > n:
H = M-1

print("found") if flag else print("not found")

Answer 4

Edward Finkelstein I don't want to use flag

Answer 5

Ok, then this maybe?

arr = [1,2,3,4,5]
n = int(input())
L = 0
H = (len(arr) - 1)

while L <= H:
M = (H+L)//2
if arr[M] == n:
print("Found")
break

elif arr[M] < n:
if M==len(arr)-1:
print(n,"Not found")
break
L = M+1

elif arr[M] > n:
if M==0:
print(n,"Not found")
break
H = M-1