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Leetcode 2578. Split With Minimum Sum

Problem Statement

You are given a positive integer num consisting of exactly n digits. You should split the number into two new integers num1 and num2 such that:

The task is to minimize the sum of num1 and num2.

Example:

Input: num = 4325
Output: 59
Explanation: num1 is 4 and num2 is 325, and their sum is 4 + 55 = 59. 

Input: num = 687
Output: 75
Explanation: num1 is 6 and num2 is 87, and their sum is 6 + 69 = 75.

Clarifying Questions

  1. What is the minimum number of digits num can have?
    • num will have at least 2 digits.
  2. Are there any constraints on the size of num?
    • The problem doesn’t specify, but in general, assume reasonable limits for typical coding challenge environments (e.g., within the range of a standard data type).
  3. Can num1 and num2 be equal to zero?
    • No, num1 and num2 cannot have leading zeros and must be valid non-negative integers when split.

Strategy

To minimize the sum of the resulting numbers from the split, the lower digits should be distributed into the smaller number and the higher digits should focus on creating a larger number. Sorting the digits will help in arranging them optimally.

  1. Convert the number to a string for easy manipulation of its digits.
  2. Sort the digits in ascending order to process from the smallest to largest.
  3. Distribute the digits alternately between two numbers, which can help in minimizing the sum.
  4. Convert the results back into integers and return their sum.

Code

Here is a concise implementation of the above strategy in C++:

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;

int splitWithMinimumSum(int num) {
    string numStr = to_string(num);
    sort(numStr.begin(), numStr.end());

    string num1, num2;
    for (int i = 0; i < numStr.size(); ++i) {
        if (i % 2 == 0)
            num1 += numStr[i];
        else
            num2 += numStr[i];
    }

    // Convert the resulting strings back to integers
    int num1Int = num1.empty() ? 0 : stoi(num1);
    int num2Int = num2.empty() ? 0 : stoi(num2);
    
    return num1Int + num2Int;
}

int main() {
    // Example usage:
    int num = 4325;
    cout << "Minimum sum after split: " << splitWithMinimumSum(num) << endl;  // Output: 59

    num = 687;
    cout << "Minimum sum after split: " << splitWithMinimumSum(num) << endl;  // Output: 75

    return 0;
}

Time Complexity

The main operations are:

  1. Conversion of integer to string - O(n), where n is the number of digits.
  2. Sorting the digits - O(n log n).
  3. Distributing the digits - O(n).

Thus, the overall time complexity is O(n log n).

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