# Plus One Problem on LeetCode in Python

## Introduction

In this article, we are going to discuss a relatively simple yet interesting problem called “Plus One” which is listed on LeetCode. This problem tests basic array manipulation skills and is a common interview question for software engineering roles. We will explore different methods to solve this problem in Python, analyze their time complexities, and understand how these methods can be applied in real-world scenarios.

1. Introduction
2. Understanding the Problem Statement
3. Using String Conversion
4. Iterative Approach
5. Recursive Approach
6. Time and Space Complexity Analysis
7. Conclusion

## Understanding the Problem Statement

The “Plus One” problem can be described as follows:

Given a non-empty array of decimal digits representing a non-negative integer, increment the integer by one. You may assume the integer does not contain any leading zeros, except the number 0 itself. The digits are stored such that the most significant digit is at the head of the list, and each element in the array contains a single digit. Return the array representation of the incremented integer.

For example:

Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123. Incremented by one, it becomes 124, which is returned as [1,2,4].

## Using String Conversion

One way to solve this problem is by converting the array into an integer, incrementing it by one, and then converting it back to an array form.

def plusOne(digits):
# Convert the array to an integer represented as a string
num_str = ''.join(map(str, digits))

# Convert the string representation to an integer and add one
num = int(num_str) + 1

# Convert the result back to an array of digits
return list(map(int, str(num)))

## Iterative Approach

Another approach to solving this problem is to simulate the way we perform addition manually, starting from the rightmost digit (least significant) and moving towards the left, carrying over if necessary.

def plusOne(digits):
n = len(digits)

# Iterate from the rightmost digit
for i in range(n - 1, -1, -1):
# If the current digit is less than 9, increment it and return the result
if digits[i] < 9:
digits[i] += 1
return digits

# Otherwise, set the current digit to 0 (carry over)
digits[i] = 0

# If we reach this point, it means there was a carry from the leftmost digit
# We need to insert 1 at the beginning
return  + digits

## Recursive Approach

We can also solve this problem using a recursive approach which is a bit more elegant but essentially follows the same logic as the iterative approach.

def plusOne(digits):
def helper(index):
# Base case: if index is negative, it means we have a carry from the leftmost digit
if index < 0:
return  + digits

# If the current digit is less than 9, increment it and return the result
if digits[index] < 9:
digits[index] += 1
return digits

# Otherwise, set the current digit to 0 and recur with the next left digit
digits[index] = 0
return helper(index - 1)

return helper(len(digits) - 1)

## Time and Space Complexity Analysis

1. Using String Conversion: The time complexity is O(N) where N is the number of digits in the input array. The space complexity is also O(N) because we create additional strings and arrays.
2. Iterative Approach: This approach has a time complexity of O(N) and a space complexity of O(1), making it more efficient in terms of space compared to the string conversion method.
3. Recursive Approach: The time complexity is O(N) and the space complexity is O(N) due to the recursive call stack.

## Conclusion

The “Plus One” problem is a simple yet powerful example that tests the understanding of array manipulation. Through different approaches including using string conversion, iterative and recursive methods, we understand different ways of thinking about the problem. Understanding the underlying concept is useful not just for interview preparation but also in understanding how arithmetic and counters work in low-level systems.