Split With Minimum Sum

This problem requires some observations to be solved optimally.

To minimize the sum of num1 and num2, we need to split num into two numbers such that their values are as close as possible. If the values are close, their sum will be minimal. For example, for num=1234, the optimal split is 123 and 4, not 1 and 234.
To get two numbers as close as possible, we should divide the digits of num into two groups with almost equal sums. The optimal way to do this is to start with the smallest digits. The reason is that using the smallest digits allows us to get a larger number of digits in each of num1 and num2, which helps to get the values of num1 and num2 closer.

Based on these observations, we can solve this problem using a greedy algorithm.

Here’s the Python solution:

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class Solution:
    def splitNum(self, num: int) -> int:
        digits = list(str(num))
        digits.sort() # sort the digits in non-decreasing order
        num1 = []
        num2 = []
        sum1 = sum2 = 0

        for digit in digits:
            if sum1 <= sum2:
                num1.append(digit)
                sum1 += int(digit)
            else:
                num2.append(digit)
                sum2 += int(digit)

        return (int(''.join(num1)) if num1 else 0) + (int(''.join(num2)) if num2 else 0)

The above solution first converts the integer num into a list of its digits. Then, it sorts the digits in non-decreasing order. After that, it iterates over the sorted digits and adds each digit to either num1 or num2, based on which number has a smaller sum. The algorithm ensures that num1 and num2 are as close in value as possible, thereby minimizing their sum.

The time complexity of this solution is O(d log d), where d is the number of digits in num. This is because we are sorting the digits, which takes O(d log d) time. The space complexity is O(d), because we are storing the digits in a list.

Abstract Representation of the Problem

We have a number, let’s call it ‘N’. We need to split ‘N’ into two parts, ‘A’ and ‘B’. The digits in ‘A’ and ‘B’ should be the same as in ‘N’, but they can be in any order.

Also, ‘A’ and ‘B’ can start with zero. The goal is to choose ‘A’ and ‘B’ so that when we add them together, we get the smallest possible result.

Q&A

How permutation works here? Somebody explain please.

The term “permutation” in this problem refers to the reordering of the digits of the original number num. The condition “The concatenation of num1 and num2 is a permutation of num” means that if we take the digits of num1 and num2 and combine them (not necessarily in the same order), we would be able to form the original number num.

For example, if num is 4321, then num1 could be 21 and num2 could be 43. Here, num1 and num2 are a permutation of num because they contain exactly the same digits (4, 3, 2, 1), although not necessarily in the same order.

So in the context of this problem, when it says “a permutation of num”, it does not mean we should preserve the order of the digits, rather it means we need to use all the digits from the original number num when we form num1 and num2.

Same vibe https://leetcode.com/problems/next-greater-element-iii/ . Re-arrange digits to achieve smallest number possible

Yes, the vibe is indeed similar to the “Next Greater Element III” problem, but with some differences. In the “Next Greater Element III” problem, we rearrange the digits to get the next larger number that uses the same digits.

In this problem, we are rearranging the digits and splitting them into two separate numbers, with the goal of minimizing the sum of these two numbers.

So the core task in both problems involves rearranging digits, but the specific goals and constraints differ, making each problem unique in its own right.

“Next Greater Element III” is simpler.

Here’s why:

In the “Next Greater Element III” problem, you’re looking for the next larger permutation of the same digits. The problem can be solved by following a standard approach to generating the next lexicographical permutation, which is well-documented in combinatorics.

In contrast, the “Split Minimum Sum” problem involves not only rearranging the digits but also splitting them into two separate numbers. This adds an additional layer of complexity because you need to consider different ways to split the digits as well as rearrange them. This problem also has more edge cases to consider because the split can occur at any position, and both resulting numbers can have any permutation of the original digits.

Therefore, while both problems require an understanding of permutations and combinatorics, the “Split Minimum Sum” problem introduces more complexity with the addition of splitting the number.