2609. Find the Longest Balanced Substring of a Binary String

Description

You are given a binary string s consisting only of zeroes and ones.

A substring of s is considered balanced if all zeroes are before ones and the number of zeroes is equal to the number of ones inside the substring. Notice that the empty substring is considered a balanced substring.

Return the length of the longest balanced substring of s.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "01000111"
Output: 6
Explanation: The longest balanced substring is "000111", which has length 6.

Example 2:

Input: s = "00111"
Output: 4
Explanation: The longest balanced substring is "0011", which has length 4.

Example 3:

Input: s = "111"
Output: 0
Explanation: There is no balanced substring except the empty substring, so the answer is 0.

Constraints:

1 <= s.length <= 50
'0' <= s[i] <= '1'

Solutions

Solution 1: Brute force

Since the range of $n$ is small, we can enumerate all substrings $s[i..j]$ to check if it is a balanced string. If so, update the answer.

The time complexity is $O(n^3)$, and the space complexity is $O(1)$. Where $n$ is the length of string $s$.

Python3JavaC++GoTypeScriptRust

class Solution:
    def findTheLongestBalancedSubstring(self, s: str) -> int:
        def check(i, j):
            cnt = 0
            for k in range(i, j + 1):
                if s[k] == '1':
                    cnt += 1
                elif cnt:
                    return False
            return cnt * 2 == (j - i + 1)

        n = len(s)
        ans = 0
        for i in range(n):
            for j in range(i + 1, n):
                if check(i, j):
                    ans = max(ans, j - i + 1)
        return ans

class Solution {
    public int findTheLongestBalancedSubstring(String s) {
        int n = s.length();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (check(s, i, j)) {
                    ans = Math.max(ans, j - i + 1);
                }
            }
        }
        return ans;
    }

    private boolean check(String s, int i, int j) {
        int cnt = 0;
        for (int k = i; k <= j; ++k) {
            if (s.charAt(k) == '1') {
                ++cnt;
            } else if (cnt > 0) {
                return false;
            }
        }
        return cnt * 2 == j - i + 1;
    }
}

class Solution {
public:
    int findTheLongestBalancedSubstring(string s) {
        int n = s.size();
        int ans = 0;
        auto check = [&](int i, int j) -> bool {
            int cnt = 0;
            for (int k = i; k <= j; ++k) {
                if (s[k] == '1') {
                    ++cnt;
                } else if (cnt) {
                    return false;
                }
            }
            return cnt * 2 == j - i + 1;
        };
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (check(i, j)) {
                    ans = max(ans, j - i + 1);
                }
            }
        }
        return ans;
    }
};

func findTheLongestBalancedSubstring(s string) (ans int) {
    n := len(s)
    check := func(i, j int) bool {
        cnt := 0
        for k := i; k <= j; k++ {
            if s[k] == '1' {
                cnt++
            } else if cnt > 0 {
                return false
            }
        }
        return cnt*2 == j-i+1
    }
    for i := 0; i < n; i++ {
        for j := i + 1; j < n; j++ {
            if check(i, j) {
                ans = max(ans, j-i+1)
            }
        }
    }
    return
}

function findTheLongestBalancedSubstring(s: string): number {
    const n = s.length;
    let ans = 0;
    const check = (i: number, j: number): boolean => {
        let cnt = 0;
        for (let k = i; k <= j; ++k) {
            if (s[k] === '1') {
                ++cnt;
            } else if (cnt > 0) {
                return false;
            }
        }
        return cnt * 2 === j - i + 1;
    };
    for (let i = 0; i < n; ++i) {
        for (let j = i + 1; j < n; j += 2) {
            if (check(i, j)) {
                ans = Math.max(ans, j - i + 1);
            }
        }
    }
    return ans;
}

impl Solution {
    pub fn find_the_longest_balanced_substring(s: String) -> i32 {
        let check = |i: usize, j: usize| -> bool {
            let mut cnt = 0;

            for k in i..=j {
                if s.as_bytes()[k] == b'1' {
                    cnt += 1;
                } else if cnt > 0 {
                    return false;
                }
            }

            cnt * 2 == j - i + 1
        };

        let mut ans = 0;
        let n = s.len();
        for i in 0..n - 1 {
            for j in (i + 1..n).rev() {
                if j - i + 1 < ans {
                    break;
                }

                if check(i, j) {
                    ans = std::cmp::max(ans, j - i + 1);
                    break;
                }
            }
        }

        ans as i32
    }
}

Solution 2: Enumeration optimization

We use variables $zero$ and $one$ to record the number of continuous $0$ and $1$.

Traverse the string $s$, for the current character $c$:

If the current character is '0', we check if $one$ is greater than $0$, if so, we reset $zero$ and $one$ to $0$, and then add $1$ to $zero$.
If the current character is '1', we add $1$ to $one$, and update the answer to $ans = max(ans, 2 \times min(one, zero))$.

After the traversal is complete, we can get the length of the longest balanced substring.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Where $n$ is the length of string $s$.