DEV Community: Bharathi R

DLC - 1

Bharathi R — Thu, 01 Jan 2026 03:54:31 +0000

Hello community! Happy New Year! Within two posts of practicing here, the universe decided to show some kindness, and I got an internship! I took the rest of the year off, so this year I'm hoping to be more productive and advance to the next level!

That being said, here's the solution i did for today's daily problem on leetcode.

Problem: 66. Plus One

Given an array representing a large integer, from left-right (msb-lsb), add 1 and return the new integer in a similar form.

My first approach:

def plusOne(self, digits: List[int]) -> List[int]: dig = int("".join(str(x) for x in digits)) dig+= 1 dig = str(dig) res = [int(i) for i in dig] return res

A second approach would be to check if dig[i]==9, if it is, make it zero, and the dig[i-1]+=1. and we return the dig. no str->int conversions here.

CSES-1 Weird Algorithm

Bharathi R — Thu, 02 Oct 2025 07:39:13 +0000

Attempt - 1 of the weird algorithm. Upon looking at it, it seems simple.

if the num is odd, do num*3 + 1
if it is even, divide by 2

keep doing this until your num is 1.

One thing I did struggle a bit was with the output formatting. At this point I'm so used to leetcode style that input-output formatting goes over my head.

Code:

n = int(input())
''' the logic: to see odd or even, we check the lsb. that way it is faster than the % function '''
import timeit # this will help in checking for time in local env.
def func(x):
    while x!=1:
        print(x, end= " ")
        if x&1==1:
            x = (x*3) + 1
        else:
            x//=2
    else:
        print(1)

    '''else:
        print("done")'''

start = timeit.default_timer()
func(n)
stop = timeit.default_timer()
#print("time taken: ",stop-start)

With this, I'd like to say, Hello, dev world!