README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1400-1499/1417.Reformat%20The%20String/README_EN.md	1241	Weekly Contest 185 Q1	String

1417. Reformat The String

中文文档

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

Python3

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)

Java

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();
    }
}

C++

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;
    }
};

Go

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()
}