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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

Thinking

A balanced substring is zeros followed by the same number of ones. \(n \le 50\) allows enumerating \(O(n^2)\) substrings and scanning each in \(O(n)\).

For every interval we count ones, require every zero to precede the first one, and require the one-count to be exactly half the length.

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 \(s\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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

Thinking

Solution 1 scans every interval in cubic time. A valid string is only a run of zeros followed by a run of ones, so we keep those two run lengths.

A zero after ones starts a new pair of runs; a one updates the answer with \(2\times\min(\textit{zero},\textit{one})\). One pass suffices.

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 \(s\).

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class Solution:
    def findTheLongestBalancedSubstring(self, s: str) -> int:
        ans = zero = one = 0
        for c in s:
            if c == '0':
                if one:
                    zero = one = 0
                zero += 1
            else:
                one += 1
                ans = max(ans, 2 * min(one, zero))
        return ans

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class Solution {
    public int findTheLongestBalancedSubstring(String s) {
        int zero = 0, one = 0;
        int ans = 0, n = s.length();
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) == '0') {
                if (one > 0) {
                    zero = 0;
                    one = 0;
                }
                ++zero;
            } else {
                ans = Math.max(ans, 2 * Math.min(zero, ++one));
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int findTheLongestBalancedSubstring(string s) {
        int zero = 0, one = 0;
        int ans = 0;
        for (char& c : s) {
            if (c == '0') {
                if (one > 0) {
                    zero = 0;
                    one = 0;
                }
                ++zero;
            } else {
                ans = max(ans, 2 * min(zero, ++one));
            }
        }
        return ans;
    }
};

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func findTheLongestBalancedSubstring(s string) (ans int) {
    zero, one := 0, 0
    for _, c := range s {
        if c == '0' {
            if one > 0 {
                zero, one = 0, 0
            }
            zero++
        } else {
            one++
            ans = max(ans, 2*min(zero, one))
        }
    }
    return
}

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function findTheLongestBalancedSubstring(s: string): number {
    let zero = 0;
    let one = 0;
    let ans = 0;
    for (const c of s) {
        if (c === '0') {
            if (one > 0) {
                zero = 0;
                one = 0;
            }
            ++zero;
        } else {
            ans = Math.max(ans, 2 * Math.min(zero, ++one));
        }
    }
    return ans;
}

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impl Solution {
    pub fn find_the_longest_balanced_substring(s: String) -> i32 {
        let mut zero = 0;
        let mut one = 0;
        let mut ans = 0;

        for &c in s.as_bytes().iter() {
            if c == b'0' {
                if one > 0 {
                    zero = 0;
                    one = 0;
                }
                zero += 1;
            } else {
                one += 1;
                ans = std::cmp::max(ans, std::cmp::min(zero, one) * 2);
            }
        }

        ans
    }
}

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