README_EN.md

2578. Split With Minimum Sum

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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.

  <ul>
  	<li>In other words, the sum of the number of occurrences of each digit in <code>num1</code> and <code>num2</code> is equal to the number of occurrences of that digit in <code>num</code>.</li>
  </ul>
  </li>
  <li><code>num1</code> and <code>num2</code> can contain leading zeros.</li>
  

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, $C \leq 10$.

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 $num$.

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]
    }
}