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2904. Shortest and Lexicographically Smallest Beautiful String

Description

You are given a binary string s and a positive integer k.

A substring of s is beautiful if the number of 1's in it is exactly k.

Let len be the length of the shortest beautiful substring.

Return the lexicographically smallest beautiful substring of string s with length equal to len. If s doesn't contain a beautiful substring, return an empty string.

A string a is lexicographically larger than a string b (of the same length) if in the first position where a and b differ, a has a character strictly larger than the corresponding character in b.

  • For example, "abcd" is lexicographically larger than "abcc" because the first position they differ is at the fourth character, and d is greater than c.

 

Example 1:

Input: s = "100011001", k = 3
Output: "11001"
Explanation: There are 7 beautiful substrings in this example:
1. The substring "100011001".
2. The substring "100011001".
3. The substring "100011001".
4. The substring "100011001".
5. The substring "100011001".
6. The substring "100011001".
7. The substring "100011001".
The length of the shortest beautiful substring is 5.
The lexicographically smallest beautiful substring with length 5 is the substring "11001".

Example 2:

Input: s = "1011", k = 2
Output: "11"
Explanation: There are 3 beautiful substrings in this example:
1. The substring "1011".
2. The substring "1011".
3. The substring "1011".
The length of the shortest beautiful substring is 2.
The lexicographically smallest beautiful substring with length 2 is the substring "11".

Example 3:

Input: s = "000", k = 1
Output: ""
Explanation: There are no beautiful substrings in this example.

 

Constraints:

  • 1 <= s.length <= 100
  • 1 <= k <= s.length

Solutions

Solution 1: Enumeration

We can enumerate all substrings $s[i: j]$, where $i \lt j$, and check if they are beautiful substrings. If so, we update the answer.

The time complexity is $O(n^3)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

Solution 2: Two Pointers

We can also use two pointers to maintain a sliding window, where pointer $i$ points to the left boundary of the window, and pointer $j$ points to the right boundary of the window. Initially, $i$ and $j$ both point to $0$. In addition, we use a variable $cnt$ to record the number of $1$s in the sliding window.

We first move pointer $j$ to the right, add $s[j]$ to the sliding window, and update $cnt$. If $cnt$ is greater than $k$, or if $i$ is less than $j$ and $s[i]$ is $0$, we move pointer $i$ to the right and update $cnt$.

When $cnt$ equals $k$, we have found a beautiful substring. We compare it with the current answer and update the answer if necessary.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

  • class Solution {
    public String shortestBeautifulSubstring(String s, int k) {
        int i = 0, j = 0, cnt = 0;
        int n = s.length();
        String ans = "";
        while (j < n) {
            cnt += s.charAt(j) - '0';
            while (cnt > k || (i < j && s.charAt(i) == '0')) {
                cnt -= s.charAt(i) - '0';
                ++i;
            }
            ++j;
            String t = s.substring(i, j);
            if (cnt == k
                && ("".equals(ans) || j - i < ans.length()
                    || (j - i == ans.length() && t.compareTo(ans) < 0))) {
                ans = t;
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    string shortestBeautifulSubstring(string s, int k) {
        int i = 0, j = 0, cnt = 0;
        int n = s.size();
        string ans = "";
        while (j < n) {
            cnt += s[j] == '1';
            while (cnt > k || (i < j && s[i] == '0')) {
                cnt -= s[i++] == '1';
            }
            ++j;
            string t = s.substr(i, j - i);
            if (cnt == k && (ans == "" || j - i < ans.size() || (j - i == ans.size() && t < ans))) {
                ans = t;
            }
        }
        return ans;
    }
};

  • class Solution:
    def shortestBeautifulSubstring(self, s: str, k: int) -> str:
        i = j = cnt = 0
        n = len(s)
        ans = ""
        while j < n:
            cnt += s[j] == "1"
            while cnt > k or (i < j and s[i] == "0"):
                cnt -= s[i] == "1"
                i += 1
            j += 1
            if cnt == k and (
                not ans or j - i < len(ans) or (j - i == len(ans) and s[i:j] < ans)
            ):
                ans = s[i:j]
        return ans


  • func shortestBeautifulSubstring(s string, k int) (ans string) {
	i, j, cnt := 0, 0, 0
	n := len(s)
	for j < n {
		cnt += int(s[j] - '0')
		for cnt > k || (i < j && s[i] == '0') {
			cnt -= int(s[i] - '0')
			i++
		}
		j++
		t := s[i:j]
		if cnt == k && (ans == "" || j-i < len(ans) || (j-i == len(ans) && t < ans)) {
			ans = t
		}
	}
	return
}

  • function shortestBeautifulSubstring(s: string, k: number): string {
    let [i, j, cnt] = [0, 0, 0];
    const n = s.length;
    let ans: string = '';
    while (j < n) {
        cnt += s[j] === '1' ? 1 : 0;
        while (cnt > k || (i < j && s[i] === '0')) {
            cnt -= s[i++] === '1' ? 1 : 0;
        }
        ++j;
        const t = s.slice(i, j);
        if (cnt === k && (ans === '' || j - i < ans.length || (j - i === ans.length && t < ans))) {
            ans = t;
        }
    }
    return ans;
}


  • impl Solution {
    pub fn shortest_beautiful_substring(s: String, k: i32) -> String {
        let s_chars: Vec<char> = s.chars().collect();
        let mut i = 0;
        let mut j = 0;
        let mut cnt = 0;
        let mut ans = String::new();
        let n = s.len();

        while j < n {
            if s_chars[j] == '1' {
                cnt += 1;
            }

            while cnt > k || (i < j && s_chars[i] == '0') {
                if s_chars[i] == '1' {
                    cnt -= 1;
                }
                i += 1;
            }

            j += 1;

            if cnt == k && (ans.is_empty() || j - i < ans.len() || (j - i == ans.len() && &s[i..j] < &ans)) {
                ans = s_chars[i..j].iter().collect();
            }
        }

        ans
    }
}

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