Leetcode #2578: Split With Minimum Sum
In this guide, we solve Leetcode #2578 Split With Minimum Sum in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a positive integer num, split it into two non-negative integers num1 and num2 such that: The concatenation of num1 and num2 is a permutation of num. In other words, the sum of the number of occurrences of each digit in num1 and num2 is equal to the number of occurrences of that digit in num.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Greedy, Math, Sorting
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: num = 4325
Output: 59
Explanation: 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.
Python Solution
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)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.