计算二进制数中连续的 1

Question

This is a HackerRank Day 10 Code Problem (30 Days of Code Challenge)

I have figured out most of the part but I am stuck at the part where you have to count the number of consecutive 1s in a binary representation.

The code works just fine for consecutive 1s in the beginning or in the middle of the whole series, But for consecutive 1s at the end I can't figure out what to do

cons = 0
def bin_no():
    global rem, cons
    
    #Input
    n = int(input("Enter Number : ")) 
    rem=[]
    
    #Checking If The Number Is Valid Or Not.
    if n < 0:
        print("Please Enter A Valid Number : ")
        bin_no()
    elif n == 0:
        print('0000')
        
    #While Loop for Conversion
    while n >=1 :   
        rem.append(n%2)
        n = n//2
    rem.reverse()
    print("Binary Representation ",*rem, sep = '')

    #Finding Maximum Consecutive 1s.
    maxcon = 0
    for x in rem:
        if x == 1:
            cons += 1
        elif x != 1 and cons > maxcon:
            maxcon = cons
            cons = 0
                    
    print(maxcon)

    
bin_no()

You can see the Maximum Consecutive 1s code block near the end of the code.

I know the error is caused due to the ending of the list doesn't contain any element that is

not equal to 1

Thus the ending elif code block is ignored.

Answer 1

You need to setup a condition to check the maximality after the loop itself incase as you mentioned the maximum-consecutive-1s are towards the end:

#Finding Maximum Consecutive 1s.
    maxcon, cons = 0, 0
    for x in rem:
        if x == 1:
            cons += 1
        elif x != 1 and cons > maxcon:
            maxcon = cons
            cons = 0
    maxcon = max(maxcon, cons)

^^This should fix it

EDIT

Btw, I come up with a more elegant solution which doesnt require converting the number into its binary format:

N = int(14)

cons, maxcon = 0, 0
while N > 0:
    if N & 1:
        cons += 1
    else:
        maxcon = max(cons, maxcon)
        cons = 0
    N >>= 1

maxcon = max(cons, maxcon)
print(maxcon)

3

Answer 2

Serial Lazer's edit is definitely the way to go, but I just wanted to correct their first answer.

def consec_ones_a(rem):
    count, largest = 0, 0
    for x in rem:
        if x == 1:
            count += 1
        else:
            if count > largest:
                largest = count
            count = 0
    return max(largest, count)
    
def consec_ones_b(rem):
    maxcon, cons = 0, 0
    for x in rem:
        if x == 1:
            cons += 1
        elif x != 1 and cons > maxcon:
            maxcon = cons
            cons = 0
    return max(maxcon, cons)

# 4 consecutive 1s
binary = [1,1,1,0,1,0,1,1,1,1]
print(consec_ones_a(binary))
print(consec_ones_b(binary))

outputs

4
5