README_EN.md

1694. Reformat Phone Number

中文文档

Description

You are given a phone number as a string number. number consists of digits, spaces ' ', and/or dashes '-'.

You would like to reformat the phone number in a certain manner. Firstly, remove all spaces and dashes. Then, group the digits from left to right into blocks of length 3 until there are 4 or fewer digits. The final digits are then grouped as follows:

• 2 digits: A single block of length 2.
• 3 digits: A single block of length 3.
• 4 digits: Two blocks of length 2 each.

The blocks are then joined by dashes. Notice that the reformatting process should never produce any blocks of length 1 and produce at most two blocks of length 2.

Return the phone number after formatting.

 

Example 1:

Input: number = "1-23-45 6"
Output: "123-456"
Explanation: The digits are "123456".
Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
Step 2: There are 3 digits remaining, so put them in a single block of length 3. The 2nd block is "456".
Joining the blocks gives "123-456".

Example 2:

Input: number = "123 4-567"
Output: "123-45-67"
Explanation: The digits are "1234567".
Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
Step 2: There are 4 digits left, so split them into two blocks of length 2. The blocks are "45" and "67".
Joining the blocks gives "123-45-67".

Example 3:

Input: number = "123 4-5678"
Output: "123-456-78"
Explanation: The digits are "12345678".
Step 1: The 1st block is "123".
Step 2: The 2nd block is "456".
Step 3: There are 2 digits left, so put them in a single block of length 2. The 3rd block is "78".
Joining the blocks gives "123-456-78".

 

Constraints:

• 2 <= number.length <= 100
• number consists of digits and the characters '-' and ' '.
• There are at least two digits in number.

Solutions

Solution 1: Simple Simulation

First, according to the problem description, we remove all spaces and hyphens from the string.

Let the current string length be $n$. Then we traverse the string from the beginning, grouping every $3$ characters together and adding them to the result string. We take a total of $n / 3$ groups.

If there is $1$ character left in the end, we form a new group of two characters with the last character of the last group and this character, and add it to the result string. If there are $2$ characters left, we directly form a new group with these two characters and add it to the result string.

Finally, we add hyphens between all groups and return the result string.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.

class Solution:
    def reformatNumber(self, number: str) -> str:
        number = number.replace("-", "").replace(" ", "")
        n = len(number)
        ans = [number[i * 3 : i * 3 + 3] for i in range(n // 3)]
        if n % 3 == 1:
            ans[-1] = ans[-1][:2]
            ans.append(number[-2:])
        elif n % 3 == 2:
            ans.append(number[-2:])
        return "-".join(ans)

class Solution {
    public String reformatNumber(String number) {
        number = number.replace("-", "").replace(" ", "");
        int n = number.length();
        List<String> ans = new ArrayList<>();
        for (int i = 0; i < n / 3; ++i) {
            ans.add(number.substring(i * 3, i * 3 + 3));
        }
        if (n % 3 == 1) {
            ans.set(ans.size() - 1, ans.get(ans.size() - 1).substring(0, 2));
            ans.add(number.substring(n - 2));
        } else if (n % 3 == 2) {
            ans.add(number.substring(n - 2));
        }
        return String.join("-", ans);
    }
}

class Solution {
public:
    string reformatNumber(string number) {
        string s;
        for (char c : number) {
            if (c != ' ' && c != '-') {
                s.push_back(c);
            }
        }
        int n = s.size();
        vector<string> res;
        for (int i = 0; i < n / 3; ++i) {
            res.push_back(s.substr(i * 3, 3));
        }
        if (n % 3 == 1) {
            res.back() = res.back().substr(0, 2);
            res.push_back(s.substr(n - 2));
        } else if (n % 3 == 2) {
            res.push_back(s.substr(n - 2));
        }
        string ans;
        for (auto& v : res) {
            ans += v;
            ans += "-";
        }
        ans.pop_back();
        return ans;
    }
};

func reformatNumber(number string) string {
	number = strings.ReplaceAll(number, " ", "")
	number = strings.ReplaceAll(number, "-", "")
	n := len(number)
	ans := []string{}
	for i := 0; i < n/3; i++ {
		ans = append(ans, number[i*3:i*3+3])
	}
	if n%3 == 1 {
		ans[len(ans)-1] = ans[len(ans)-1][:2]
		ans = append(ans, number[n-2:])
	} else if n%3 == 2 {
		ans = append(ans, number[n-2:])
	}
	return strings.Join(ans, "-")
}

function reformatNumber(number: string): string {
    const cs = [...number].filter(c => c !== ' ' && c !== '-');
    const n = cs.length;
    return cs
        .map((v, i) => {
            if (((i + 1) % 3 === 0 && i < n - 2) || (n % 3 === 1 && n - 3 === i)) {
                return v + '-';
            }
            return v;
        })
        .join('');
}

impl Solution {
    pub fn reformat_number(number: String) -> String {
        let cs: Vec<char> = number
            .chars()
            .filter(|&c| c != ' ' && c != '-')
            .collect();
        let n = cs.len();
        cs.iter()
            .enumerate()
            .map(|(i, c)| {
                if ((i + 1) % 3 == 0 && i < n - 2) || (n % 3 == 1 && i == n - 3) {
                    return c.to_string() + &"-";
                }
                c.to_string()
            })
            .collect()
    }
}