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Given a positive integer num, split it into two non-negative integers num1 and num2 such that:

• The concatenation of num1 and num2 in any order forms the number num.

The task is to find the combination of num1 and num2 such that their sum is minimized.

Example:

• Input: num = 4325
• Output: 59
• Explanation: One of the ways to split 4325 is into num1 = 42 and num2 = 35, resulting in 42 + 35 = 77. But the better split is num1 = 25 and num2 = 34, resulting in 25 + 34 = 59.

Clarifying Questions

1. Can num1 and num2 have leading zeros?
   • No, since they’re non-negative integers and should be meaningful representations.
2. Can num1 or num2 be zero?
   • Yes, one or both can be zero as long as their concatenation forms the original number.
3. Should the order of the digits in num be maintained in the split?
   • No, the order of the digits can be rearranged as needed to minimize the sum.

Strategy

1. Break Down Digits:
   • Extract the individual digits of the number.
2. Sort the Digits:
   • Sort the digits in ascending order.
3. Form Two Numbers:
   • Distribute the digits as evenly as possible between num1 and num2 to minimize their sum.
   • Construct num1 and num2 such that num1 takes the smallest remaining digit first, followed by num2, and so forth. This tends to balance the magnitude of both numbers, minimizing their sum.

Code

def split_with_minimum_sum(num: int) -> int:
    # Step 1: Extract and sort the digits
    digits = sorted([int(digit) for digit in str(num)])
    
    # Step 2: Initialize the two numbers
    num1, num2 = 0, 0
    
    # Step 3: Distribute the digits
    for i, digit in enumerate(digits):
        if i % 2 == 0:
            num1 = num1 * 10 + digit
        else:
            num2 = num2 * 10 + digit
    
    # Step 4: Return the sum of num1 and num2
    return num1 + num2

# Example usage
print(split_with_minimum_sum(4325))  # Output: 59

Time Complexity

• Sorting the digits: (O(n \log n)), where (n) is the number of digits in the number.
• Forming the numbers: (O(n)).
• Hence, the overall time complexity is (O(n \log n)).

This ensures the solution is efficient even for larger values of num.

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