1417. Reformat The String

Description

You are given an alphanumeric string s. (Alphanumeric string is a string consisting of lowercase English letters and digits).

You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type.

Return the reformatted string or return an empty string if it is impossible to reformat the string.

Example 1:

Input: s = "a0b1c2"
Output: "0a1b2c"
Explanation: No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.

Example 2:

Input: s = "leetcode"
Output: ""
Explanation: "leetcode" has only characters so we cannot separate them by digits.

Example 3:

Input: s = "1229857369"
Output: ""
Explanation: "1229857369" has only digits so we cannot separate them by characters.

Constraints:

1 <= s.length <= 500
s consists of only lowercase English letters and/or digits.

Solutions

Solution 1: Simulation

We classify all characters in string \(s\) into two categories: "digits" and "letters", and put them into arrays \(a\) and \(b\) respectively.

Compare the lengths of \(a\) and \(b\). If the length of \(a\) is less than \(b\), swap \(a\) and \(b\). Then check the difference in lengths; if it exceeds \(1\), return an empty string.

Next, iterate through both arrays simultaneously, appending characters from \(a\) and \(b\) alternately to the answer. After the iteration, if \(a\) is longer than \(b\), append the last character of \(a\) to the answer.

The time complexity is \(O(n)\) and the space complexity is \(O(n)\), where \(n\) is the length of string \(s\).

Python3JavaC++GoTypeScriptRust

class Solution:
    def reformat(self, s: str) -> str:
        a = [c for c in s if c.islower()]
        b = [c for c in s if c.isdigit()]
        if abs(len(a) - len(b)) > 1:
            return ''
        if len(a) < len(b):
            a, b = b, a
        ans = []
        for x, y in zip(a, b):
            ans.append(x + y)
        if len(a) > len(b):
            ans.append(a[-1])
        return ''.join(ans)

class Solution {
    public String reformat(String s) {
        StringBuilder a = new StringBuilder();
        StringBuilder b = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (Character.isDigit(c)) {
                a.append(c);
            } else {
                b.append(c);
            }
        }
        int m = a.length(), n = b.length();
        if (Math.abs(m - n) > 1) {
            return "";
        }
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < Math.min(m, n); ++i) {
            if (m > n) {
                ans.append(a.charAt(i));
                ans.append(b.charAt(i));
            } else {
                ans.append(b.charAt(i));
                ans.append(a.charAt(i));
            }
        }
        if (m > n) {
            ans.append(a.charAt(m - 1));
        }
        if (m < n) {
            ans.append(b.charAt(n - 1));
        }
        return ans.toString();
    }
}

class Solution {
public:
    string reformat(string s) {
        string a = "", b = "";
        for (char c : s) {
            if (isdigit(c))
                a += c;
            else
                b += c;
        }
        int m = a.size(), n = b.size();
        if (abs(m - n) > 1) return "";
        string ans = "";
        for (int i = 0; i < min(m, n); ++i) {
            if (m > n) {
                ans += a[i];
                ans += b[i];
            } else {
                ans += b[i];
                ans += a[i];
            }
        }
        if (m > n) ans += a[m - 1];
        if (m < n) ans += b[n - 1];
        return ans;
    }
};

func reformat(s string) string {
    a := []byte{}
    b := []byte{}
    for _, c := range s {
        if unicode.IsLetter(c) {
            a = append(a, byte(c))
        } else {
            b = append(b, byte(c))
        }
    }
    if len(a) < len(b) {
        a, b = b, a
    }
    if len(a)-len(b) > 1 {
        return ""
    }
    var ans strings.Builder
    for i := range b {
        ans.WriteByte(a[i])
        ans.WriteByte(b[i])
    }
    if len(a) > len(b) {
        ans.WriteByte(a[len(a)-1])
    }
    return ans.String()
}

function reformat(s: string): string {
    let a: string[] = [];
    let b: string[] = [];

    for (const c of s) {
        if (c >= 'a' && c <= 'z') a.push(c);
        else if (c >= '0' && c <= '9') b.push(c);
    }

    if (Math.abs(a.length - b.length) > 1) {
        return '';
    }

    if (a.length < b.length) {
        [a, b] = [b, a];
    }

    const ans: string[] = [];

    for (let i = 0; i < b.length; i++) {
        ans.push(a[i] + b[i]);
    }

    if (a.length > b.length) {
        ans.push(a[a.length - 1]);
    }

    return ans.join('');
}

impl Solution {
    pub fn reformat(s: String) -> String {
        let mut a: Vec<char> = Vec::new();
        let mut b: Vec<char> = Vec::new();

        for c in s.chars() {
            if c.is_ascii_lowercase() {
                a.push(c);
            } else if c.is_ascii_digit() {
                b.push(c);
            }
        }

        if (a.len() as i32 - b.len() as i32).abs() > 1 {
            return String::new();
        }

        if a.len() < b.len() {
            std::mem::swap(&mut a, &mut b);
        }

        let mut ans = String::new();

        for i in 0..b.len() {
            ans.push(a[i]);
            ans.push(b[i]);
        }

        if a.len() > b.len() {
            ans.push(a[a.len() - 1]);
        }

        ans
    }
}

1417. Reformat The String

Description

Solutions

Solution 1: Simulation

Comments