g2501_2600.s2578_split_with_minimum_sum.Solution Maven / Gradle / Ivy

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package g2501_2600.s2578_split_with_minimum_sum;

// #Easy #Math #Sorting #Greedy #2023_08_22_Time_0_ms_(100.00%)_Space_39.3_MB_(76.63%)

/**
 * 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
**/
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;
    }
}