Plus One Problem in Python | Leetcode 500 problem challenge in Python

[@Antel_007]

very insightful this.

[@codetodesign]

I'm glad you found it helpful.🙂

[@ravikumargangula]

sir pls explain this--> if we take list as [9,8,9]

[@codetodesign]

digits = [9, 8, 9] n = 3 First Iteration (i=2): digits[2] is incremented by 1: digits[2] = 9 + 1 = 10 Since digits[2] is 10 (which means there's a carry), digits[2] is set to 0. digits becomes [9, 8, 0]. Second Iteration (i=1): digits[1] is incremented by 1: digits[1] = 8 + 1 = 9 Since digits[1] is less than 10, no carry is generated, and the loop breaks. digits becomes [9, 9, 0]. Final State: The loop terminates because there is no further carry. The function returns [9, 9, 0]. I hope now its clear.

[@ravikumargangula]

@@codetodesign Tq sir

[@CodeWithKushwaha]

Plz do 500 leetcode problems in java, we really need it

[@codetodesign]

After python, will do it in Java.