Problem Description

You are given a positive 4-digit integer num. Your task is to split this number into two new integers new1 and new2 using all four digits from the original number.

The key rules are:

• You must use all four digits from num to form the two new numbers
• Leading zeros are allowed in the new numbers
• Each digit from num must be used exactly once

For example, if num = 2932:

• The digits available are: 2, 9, 3, 2
• Valid splits include: [22, 93], [23, 92], [223, 9], [2, 329]
• Invalid would be using only some digits or using a digit more times than it appears

Your goal is to find the split that produces the minimum possible sum of new1 + new2.

The solution approach sorts the four digits in ascending order. To minimize the sum, the two smallest digits should be placed in the tens position of each number, while the two larger digits go in the ones position. This is why the solution returns 10 * (nums[0] + nums[1]) + nums[2] + nums[3], which effectively creates two 2-digit numbers: one with digits nums[0] and nums[2], and another with digits nums[1] and nums[3].

Intuition

To minimize the sum of two numbers formed from four digits, we need to think about place values. A digit in the tens position contributes 10 times its value to the sum, while a digit in the ones position contributes only its face value.

Consider what happens when we form two 2-digit numbers from four digits. The sum would be:

• (10 * digit1 + digit2) + (10 * digit3 + digit4)
• This simplifies to: 10 * (digit1 + digit3) + (digit2 + digit4)

From this formula, we can see that the digits placed in the tens positions get multiplied by 10, making them contribute much more to the final sum. Therefore, to minimize the sum, we should place the smallest digits in the tens positions and the larger digits in the ones positions.

Let's trace through an example with num = 2932:

• Digits sorted: [2, 2, 3, 9]
• If we place the two smallest (2, 2) in tens positions: 20 + 20 = 40 contribution from tens
• Then place the larger ones (3, 9) in ones positions: 3 + 9 = 12 contribution from ones
• Total: 40 + 12 = 52, which gives us numbers 23 and 29

Compare this with a poor choice like 92 + 23 = 115. The large digit 9 in the tens position contributes 90 alone, making the sum much larger.

This greedy approach of sorting the digits and assigning the smallest to the tens positions guarantees the minimum possible sum: 10 * (nums[0] + nums[1]) + nums[2] + nums[3].

Learn more about Greedy, Math and Sorting patterns.

Solution Approach

The implementation follows a straightforward approach to extract digits, sort them, and construct the minimum sum:

Step 1: Extract the digits from the number

nums = []
while num:
    nums.append(num % 10)
    num //= 10

• We use the modulo operator num % 10 to get the last digit
• Integer division num //= 10 removes the last digit
• This process continues until num becomes 0
• For example, with num = 2932: we get digits [2, 3, 9, 2] (in reverse order)

Step 2: Sort the digits in ascending order

nums.sort()

• After sorting, we have the digits arranged from smallest to largest
• For our example: [2, 2, 3, 9]

Step 3: Calculate the minimum sum

return 10 * (nums[0] + nums[1]) + nums[2] + nums[3]

• nums[0] and nums[1] are the two smallest digits - these go in the tens positions
• nums[2] and nums[3] are the two larger digits - these go in the ones positions
• The formula constructs two 2-digit numbers:
  • First number: 10 * nums[0] + nums[2]
  • Second number: 10 * nums[1] + nums[3]
  • Sum: 10 * (nums[0] + nums[1]) + nums[2] + nums[3]

Time Complexity: O(1) - We always process exactly 4 digits and sorting 4 elements is constant time.

Space Complexity: O(1) - We use a fixed-size array to store 4 digits.

The beauty of this solution lies in its simplicity - by recognizing that minimizing the tens positions is key, we can solve the problem with just sorting and a simple arithmetic formula.

Example Walkthrough

Let's walk through the solution with num = 4325:

Step 1: Extract digits Starting with 4325, we extract each digit using modulo and integer division:

• 4325 % 10 = 5, then 4325 // 10 = 432
• 432 % 10 = 2, then 432 // 10 = 43
• 43 % 10 = 3, then 43 // 10 = 4
• 4 % 10 = 4, then 4 // 10 = 0 (stop)

We get digits: [5, 2, 3, 4]

Step 2: Sort the digits After sorting in ascending order: [2, 3, 4, 5]

Step 3: Form the minimum sum To minimize the sum, we place the two smallest digits (2 and 3) in the tens positions:

• First number: 10 × 2 + 4 = 24
• Second number: 10 × 3 + 5 = 35
• Minimum sum: 24 + 35 = 59

Using the formula: 10 × (2 + 3) + 4 + 5 = 10 × 5 + 9 = 59

Why this works: If we had placed larger digits in the tens positions instead (like forming 52 + 43), we'd get a sum of 95, which is much larger. The key insight is that digits in tens positions contribute 10× their value, so we want the smallest digits there.

Solution Implementation

Time and Space Complexity

Time Complexity: O(1)

The algorithm performs the following operations:

• Extracting 4 digits from the number: O(4) = O(1) since we're always dealing with exactly 4 digits
• Sorting 4 elements: O(4 log 4) = O(1) since the array size is constant
• Computing the final sum: O(1)

Since all operations are performed on a fixed-size input (4-digit number), the overall time complexity is O(1).

Space Complexity: O(1)

The algorithm uses:

• A list nums to store exactly 4 digits: O(4) = O(1)
• A few variables for intermediate calculations: O(1)

The space usage is constant regardless of the input value (as long as it's a 4-digit number), so the space complexity is O(1).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Assuming Fixed Number Distribution

A common mistake is assuming that we always need to form two 2-digit numbers. While this works for most cases, beginners might not realize why this is optimal and might try to form numbers like a 3-digit and a 1-digit number.

Why it happens: The problem statement allows splits like [223, 9] or [2, 329], which might lead to trying different digit distributions.

The issue: Forming uneven splits (like 3-digit + 1-digit) will always result in a larger sum than forming two 2-digit numbers when trying to minimize the total.

Solution: Understand that for minimum sum, we want the most significant digits to be as small as possible. Having two 2-digit numbers ensures both numbers stay in the same magnitude range.

2. Incorrect Digit Extraction Order

When extracting digits using modulo and division, the digits come out in reverse order (right to left). Some might forget this and assume the array maintains the original digit order.

Example mistake:

# Wrong assumption about digit order
digits = []
while num:
    digits.append(num % 10)
    num //= 10
# If num = 2932, digits = [2, 3, 9, 2] ❌ Wrong!
# Actually, digits = [2, 9, 3, 2] ✓ (extracted right to left)

Solution: The order doesn't matter since we sort the array anyway, but being aware of this prevents confusion during debugging.

3. Overthinking the Pairing Logic

Some might implement complex logic to try different pairings of digits, using nested loops or recursion to find the minimum sum.

Overcomplicated approach:

# Unnecessary complexity
min_sum = float('inf')
for i in range(4):
    for j in range(4):
        if i != j:
            # Try different combinations...

Solution: Recognize that after sorting, the optimal pairing is always fixed: smallest with third-smallest, and second-smallest with largest. This gives us the formula 10 * (digits[0] + digits[1]) + digits[2] + digits[3].

4. Handling Leading Zeros Incorrectly

Some might worry about leading zeros and try to avoid them, thinking 02 is not a valid number.

The misconception: Trying to ensure no digit starts with 0, which adds unnecessary complexity.

Solution: The problem explicitly states "leading zeros are allowed," and mathematically, 02 equals 2. The formula naturally handles this - if digits[0] or digits[1] is 0, it simply contributes 0 to the tens place, which is correct.