2578. Split With Minimum Sum

Description

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

Solutions

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 \(num\).

Next, we enumerate all the digits \(i\) in \(nums\), and alternately allocate the digits in \(cnt\) to \(num1\) and \(num2\) in ascending order, recording them in an array \(ans\) of length \(2\). Finally, we return the sum of the two numbers in \(ans\).

The time complexity is \(O(n)\), and the space complexity is \(O(C)\). Where \(n\) is the number of digits in \(num\); and \(C\) is the number of different digits in \(num\), in this problem, \(C \leq 10\).

Python3JavaC++GoTypeScriptRust

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)

class Solution {
    public int splitNum(int num) {
        int[] cnt = new int[10];
        int n = 0;
        for (; num > 0; num /= 10) {
            ++cnt[num % 10];
            ++n;
        }
        int[] ans = new int[2];
        for (int i = 0, j = 0; i < n; ++i) {
            while (cnt[j] == 0) {
                ++j;
            }
            --cnt[j];
            ans[i & 1] = ans[i & 1] * 10 + j;
        }
        return ans[0] + ans[1];
    }
}

class Solution {
public:
    int splitNum(int num) {
        int cnt[10]{};
        int n = 0;
        for (; num; num /= 10) {
            ++cnt[num % 10];
            ++n;
        }
        int ans[2]{};
        for (int i = 0, j = 0; i < n; ++i) {
            while (cnt[j] == 0) {
                ++j;
            }
            --cnt[j];
            ans[i & 1] = ans[i & 1] * 10 + j;
        }
        return ans[0] + ans[1];
    }
};

func splitNum(num int) int {
    cnt := [10]int{}
    n := 0
    for ; num > 0; num /= 10 {
        cnt[num%10]++
        n++
    }
    ans := [2]int{}
    for i, j := 0, 0; i < n; i++ {
        for cnt[j] == 0 {
            j++
        }
        cnt[j]--
        ans[i&1] = ans[i&1]*10 + j
    }
    return ans[0] + ans[1]
}

function splitNum(num: number): number {
    const cnt: number[] = Array(10).fill(0);
    let n = 0;
    for (; num > 0; num = Math.floor(num / 10)) {
        ++cnt[num % 10];
        ++n;
    }
    const ans: number[] = Array(2).fill(0);
    for (let i = 0, j = 0; i < n; ++i) {
        while (cnt[j] === 0) {
            ++j;
        }
        --cnt[j];
        ans[i & 1] = ans[i & 1] * 10 + j;
    }
    return ans[0] + ans[1];
}

impl Solution {
    pub fn split_num(mut num: i32) -> i32 {
        let mut cnt = vec![0; 10];
        let mut n = 0;

        while num != 0 {
            cnt[(num as usize) % 10] += 1;
            num /= 10;
            n += 1;
        }

        let mut ans = vec![0; 2];
        let mut j = 0;
        for i in 0..n {
            while cnt[j] == 0 {
                j += 1;
            }
            cnt[j] -= 1;

            ans[i & 1] = ans[i & 1] * 10 + (j as i32);
        }

        ans[0] + ans[1]
    }
}

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 \(num2\).

The time complexity is \(O(n \times \log n)\), and the space complexity is \(O(n)\). Where \(n\) is the number of digits in \(num\).

Python3JavaC++GoTypeScriptRust

class Solution:
    def splitNum(self, num: int) -> int:
        s = sorted(str(num))
        return int(''.join(s[::2])) + int(''.join(s[1::2]))

class Solution {
    public int splitNum(int num) {
        char[] s = (num + "").toCharArray();
        Arrays.sort(s);
        int[] ans = new int[2];
        for (int i = 0; i < s.length; ++i) {
            ans[i & 1] = ans[i & 1] * 10 + s[i] - '0';
        }
        return ans[0] + ans[1];
    }
}

class Solution {
public:
    int splitNum(int num) {
        string s = to_string(num);
        sort(s.begin(), s.end());
        int ans[2]{};
        for (int i = 0; i < s.size(); ++i) {
            ans[i & 1] = ans[i & 1] * 10 + s[i] - '0';
        }
        return ans[0] + ans[1];
    }
};

func splitNum(num int) int {
    s := []byte(strconv.Itoa(num))
    sort.Slice(s, func(i, j int) bool { return s[i] < s[j] })
    ans := [2]int{}
    for i, c := range s {
        ans[i&1] = ans[i&1]*10 + int(c-'0')
    }
    return ans[0] + ans[1]
}

function splitNum(num: number): number {
    const s: string[] = String(num).split('');
    s.sort();
    const ans: number[] = Array(2).fill(0);
    for (let i = 0; i < s.length; ++i) {
        ans[i & 1] = ans[i & 1] * 10 + Number(s[i]);
    }
    return ans[0] + ans[1];
}

impl Solution {
    pub fn split_num(num: i32) -> i32 {
        let mut s = num.to_string().into_bytes();
        s.sort_unstable();

        let mut ans = vec![0; 2];
        for (i, c) in s.iter().enumerate() {
            ans[i & 1] = ans[i & 1] * 10 + ((c - b'0') as i32);
        }

        ans[0] + ans[1]
    }
}

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