Python - Given a list of integers that represents a decimal value, increment the last element by 1

When working with a list of integers that represents a decimal value, we sometimes need to increment it by 1. This is similar to adding 1 to a number, but we need to handle carry operations when digits are 9.

Problem Understanding

Given a list like [1, 2, 3] representing the number 123, we want to increment it to get [1, 2, 4] representing 124. If the last digit is 9, we need to handle carries properly.

Using Recursive Approach

Here's a recursive solution that handles the carry operation when incrementing ?

def increment_num(digits, index):
    # If the current digit is less than 9, simply increment it
    if digits[index] < 9:
        digits[index] += 1
        return
    
    # If digit is 9, set it to 0 and carry over
    digits[index] = 0
    
    # If we're at the first digit and it becomes 0, we need a new digit
    if index == 0:
        digits.insert(0, 1)
        return
    
    # Recursively handle the carry for the previous digit
    increment_num(digits, index - 1)

# Example 1: Normal increment
digits = [1, 2, 3]
print("Original list:", digits)
increment_num(digits, len(digits) - 1)
print("After increment:", digits)

Original list: [1, 2, 3]
After increment: [1, 2, 4]

Handling Carry Operations

When the last digit is 9, we need to handle the carry operation ?

def increment_num(digits, index):
    if digits[index] < 9:
        digits[index] += 1
        return
    
    digits[index] = 0
    
    if index == 0:
        digits.insert(0, 1)
        return
    
    increment_num(digits, index - 1)

# Example 2: With carry operation
digits = [1, 2, 9]
print("Original list:", digits)
increment_num(digits, len(digits) - 1)
print("After increment:", digits)

# Example 3: Multiple carries
digits = [9, 9, 9]
print("\nOriginal list:", digits)
increment_num(digits, len(digits) - 1)
print("After increment:", digits)

Original list: [1, 2, 9]
After increment: [1, 3, 0]

Original list: [9, 9, 9]
After increment: [1, 0, 0, 0]

Iterative Approach

We can also solve this problem using an iterative approach ?

def increment_iterative(digits):
    # Start from the last digit
    carry = 1
    
    for i in range(len(digits) - 1, -1, -1):
        total = digits[i] + carry
        digits[i] = total % 10
        carry = total // 10
        
        # If no carry, we're done
        if carry == 0:
            break
    
    # If there's still a carry, add it to the front
    if carry:
        digits.insert(0, carry)

# Test the iterative approach
digits = [9, 8, 7]
print("Original list:", digits)
increment_iterative(digits)
print("After increment:", digits)

Original list: [9, 8, 7]
After increment: [9, 8, 8]

Comparison

Conclusion

Both recursive and iterative approaches work well for incrementing a list representing a decimal number. The iterative approach is more efficient in terms of space complexity, while the recursive approach is easier to understand conceptually.