LeetCode-in-Java

2578. Split With Minimum Sum

Easy

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.
• num1 and num2 can contain leading zeros.

Return the minimum possible sum of num1 and num2.

Notes:

• It is guaranteed that num does not contain any leading zeros.
• The order of occurrence of the digits in num1 and num2 may differ from the order of occurrence of num.

Example 1:

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.

Example 2:

Input: num = 687

Output: 75

Explanation: 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

public class Solution {
    public int splitNum(int number) {
        int num1 = 0;
        int num2 = 0;
        boolean addToOne = true;
        for (int i = 0; i <= 9; i++) {
            int tmpNumber = number;
            while (tmpNumber > 0) {
                int digit = tmpNumber % 10;
                if (digit == i) {
                    if (addToOne) {
                        num1 *= 10;
                        num1 += digit;
                        addToOne = false;
                    } else {
                        num2 *= 10;
                        num2 += digit;
                        addToOne = true;
                    }
                }
                tmpNumber /= 10;
            }
        }
        return num1 + num2;
    }
}

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