Plus One

The task here is to increment the given large integer (represented as an array of digits) by one. You can solve this by working your way from the rightmost digit, simulating a typical addition operation with carry.

Here’s a step-by-step process to implement the solution:

  1. Initialize a Carry: Start with a carry of 1, since you want to increment the number by one.

  2. Iterate from the Rightmost Digit: Traverse the digits from right to left. At each step, add the current digit with the carry and calculate the new digit and carry for that position.

  3. Handle Overflow: If after the final addition, there’s still a carry left (e.g., the number was 9999), you will need to append one more digit at the beginning of the result.

Here’s the code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        n = len(digits)
        carry = 1

        # Iterate from the rightmost digit
        for i in range(n-1, -1, -1):
            total = digits[i] + carry
            digits[i] = total % 10  # The new digit for that position
            carry = total // 10     # Carry for the next step

        # If there is still a carry, add a new digit at the beginning
        if carry:
            digits.insert(0, carry)

        return digits

This code will correctly handle the increment operation, even in cases with overflow. The time complexity of this solution is ( O(n) ), where ( n ) is the length of the input array, and the space complexity is ( O(1) ), since we are modifying the input array in place.

Identifying Problem Isomorphism

“Plus One” can be approximately mapped to the following problems:

  1. “Add Binary” (#67) on LeetCode
  2. “Multiply Strings” (#43) on LeetCode

Reasoning:

These problems involve operations on numerical digits, which is a commonality with the problem “Plus One”.

  1. “Add Binary” is a simpler problem compared to “Plus One”. The task in both problems involves adding numbers. However, in “Add Binary”, we are adding two binary numbers represented as strings. The carry-over concept in binary addition is similar to the one in “Plus One”.

  2. “Multiply Strings” is more complex compared to “Plus One”. Here, the problem is to multiply two numbers represented as strings. It’s an operation on numerical digits similar to “Plus One” but involves a more complicated operation (multiplication vs addition).

While these problems are not exact isomorphs of “Plus One”, they involve similar operations to be performed on numerical digits.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        if digits[-1] < 9:
            digits[-1] += 1
            return digits
        elif len(digits) == 1 and digits[0] == 9:
            return [1, 0]
        else:
            digits[-1] = 0
            digits[0:-1] = self.plusOne(digits[0:-1])
            return digits

Problem Classification

  1. Array Manipulation: The problem involves manipulating an array (or list) of integers.

  2. Arithmetic: The problem involves basic arithmetic operations - specifically, the concept of addition and carrying over in number systems.

  3. Recursion: The problem requires the application of a recursive approach to reach the solution.

This asks for incrementing the number represented by an array by one, which is a combination of the above categories - array manipulation and arithmetic operations (increment by one), and it’s solved by using a recursive approach.

Language Agnostic Coding Drills

1. List and Element Manipulation: Understanding how to manipulate elements in a list is a fundamental skill. Here, we are changing the values of the list in-place.

2. Recursion: This problem introduces a recursive solution to solve a problem. Recursion is the concept of a function calling itself to solve smaller instances of the same problem.

3. Conditional Statements: Applying conditional statements (if-elif-else) is another core concept used in this solution. Based on certain conditions, we decide what operation to perform on the list.

4. Array Indexing: Understanding how to work with array indices, including negative indexing, is crucial to this problem. In Python, negative indexing is used to access elements from the end of the list.

Approach to the problem:

The problem requires us to increment the last digit of a number represented as a list. If the last digit is less than 9, we simply increment it. If it’s 9, it becomes 0, and we need to carry over the increment to the previous digit.

The process follows these steps:

  1. If the last digit is less than 9, increment it and return the list.

  2. If the list length is 1 and the only digit is 9, return a new list [1, 0].

  3. If the last digit is 9, set it to 0 and call the function recursively for the list excluding the last digit (effectively carrying over the increment to the previous digit).

Note that recursion ends when it encounters a digit less than 9 (which is incremented and returned), or when all digits have been processed (case when the list length is 1, and the digit is 9). The modified list is returned at each level of the recursion.

Targeted Drills in Python

Drill 1: List and Element Manipulation

The first step is to understand how to manipulate elements in a list.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

# Initialize a list
digits = [1, 2, 3, 4, 5]

# Change an element in the list
digits[0] = 10
print(digits)  # Output: [10, 2, 3, 4, 5]

# Change the last element in the list
digits[-1] = 0
print(digits)  # Output: [10, 2, 3, 4, 0]

Drill 2: Array Indexing

Understanding how to work with array indices, including negative indexing, is important for this problem.

1
2
3
4
5

# Access the last element of the list
print(digits[-1])  # Output: 0

# Access the list excluding the last element
print(digits[:-1])  # Output: [10, 2, 3, 4]

Drill 3: Conditional Statements

Applying conditional statements (if-elif-else) is another fundamental concept used in this solution.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

# Initialize a variable
x = 5

# Check conditions using if-elif-else statements
if x < 5:
    print("Less than 5")
elif x == 5:
    print("Equal to 5")
else:
    print("Greater than 5")

Drill 4: Recursion

This problem introduces a recursive solution to solve a problem.

1
2
3
4
5
6
7
8
9

# Define a recursive function to calculate factorial of a number
def factorial(n):
    if n == 1 or n == 0:
        return 1
    else:
        return n * factorial(n-1)

# Test the function
print(factorial(5))  # Output: 120

Once we are comfortable with all these concepts, we can understand how they all fit together to solve the problem. We can increment the last digit of the list if it’s less than 9. If it’s 9, we set it to 0 and recursively call the function on the list excluding the last element, effectively carrying over the increment to the previous digit. We continue this until we encounter a digit less than 9 or have processed all digits.