#2578

Easy

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Split With Minimum Sum

Split a positive integer into two parts to minimize their sum using a greedy approach and sorting.

Problem Statement

Given a positive integer num, you need to split its digits into two non-negative integers, num1 and num2. The objective is to minimize the sum of num1 and num2. The digits of num` should be split in such a way that both integers are formed from the digits, and no digit is left out.

For example, when splitting the number 4325, the correct split would give num1 = 24 and num2 = 35, resulting in a minimal sum of 59. You must return the minimum sum possible after splitting the number in this way.

Examples

Example 1

Input: num = 4325

Output: 59

We can split 4325 so that num1 is 24 and num2 is 35, giving a sum of 59. We can prove that 59 is indeed the minimal possible sum.

Example 2

Input: num = 687

Output: 75

We can split 687 so that num1 is 68 and num2 is 7, which would give an optimal sum of 75.

Constraints

10 <= num <= 109

Solution Approach

Greedy Approach with Sorting

The key to solving this problem efficiently is sorting the digits of the number in non-decreasing order. Once sorted, the digits are distributed between the two numbers in a way that minimizes their sum. This greedy approach ensures that the digits are as evenly distributed as possible, leading to the minimal possible sum.

Alternating Distribution

After sorting the digits, alternate the digits between the two numbers num1 and num2. This ensures that the largest and smallest digits are balanced between the two numbers, preventing one number from becoming significantly larger than the other.

Final Calculation

Once the digits have been assigned to num1 and num2, calculate their sum. This step involves simply adding the two numbers together, ensuring the smallest possible sum is returned.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is determined by the sorting step, which is O(d log d) where d is the number of digits in the number. The space complexity is O(d) due to the need to store the digits of the number.

What Interviewers Usually Probe

The candidate demonstrates understanding of greedy algorithms and sorting techniques.
The candidate shows the ability to break down a problem into manageable subproblems.
The candidate approaches the problem with the correct choice of sorting and distribution strategy.

Common Pitfalls or Variants

Common pitfalls

Not sorting the digits of the number, leading to suboptimal splits.
Failing to alternate the digits between num1 and num2 effectively.
Overcomplicating the solution by trying to use dynamic programming or other complex methods when sorting and a greedy approach are sufficient.

Follow-up variants

What if the number has repeating digits?
What if the number is large, near the upper constraint?
How would you modify the approach to handle smaller numbers with fewer digits?

FAQ

What is the best approach to solve the Split With Minimum Sum problem?

The best approach is to sort the digits of the number in non-decreasing order and then alternately distribute them between two numbers, minimizing their sum.

Why do we sort the digits of the number in the Split With Minimum Sum problem?

Sorting ensures that the largest and smallest digits are balanced, preventing one of the resulting numbers from becoming too large, leading to a minimal sum.

How can GhostInterview help with the Split With Minimum Sum problem?

GhostInterview helps by providing structured steps for solving the problem, including recognizing the greedy choice pattern and optimizing the solution using sorting.

What are some common pitfalls in the Split With Minimum Sum problem?

Common pitfalls include failing to sort the digits correctly, improperly distributing the digits, or overcomplicating the solution.

What is the time complexity of the Split With Minimum Sum problem?

The time complexity is O(d log d), where d is the number of digits in the number, due to the sorting step.

terminal

Solution

Solution 1: Counting + Greedy

First, we use a hash table or array $cnt$ to count the occurrences of each digit in $num$, and use a variable $n$ to record the number of digits in $num$.

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class Solution:
    def splitNum(self, num: int) -> int:
        cnt = Counter()
        n = 0
        while num:
            cnt[num % 10] += 1
            num //= 10
            n += 1
        ans = [0] * 2
        j = 0
        for i in range(n):
            while cnt[j] == 0:
                j += 1
            cnt[j] -= 1
            ans[i & 1] = ans[i & 1] * 10 + j
        return sum(ans)

Solution 2: Sorting + Greedy

We can convert $num$ to a string or character array, then sort it, and then alternately allocate the digits in the sorted array to $num1$ and $num2$ in ascending order. Finally, we return the sum of $num1$ and $num2$.

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class Solution:
    def splitNum(self, num: int) -> int:
        cnt = Counter()
        n = 0
        while num:
            cnt[num % 10] += 1
            num //= 10
            n += 1
        ans = [0] * 2
        j = 0
        for i in range(n):
            while cnt[j] == 0:
                j += 1
            cnt[j] -= 1
            ans[i & 1] = ans[i & 1] * 10 + j
        return sum(ans)

Continue Practicing

#2607 make k subarray sums equal #2790 maximum number of groups with increasing length #2818 apply operations to maximize score