LeetCode #2609 — EASY

Find the Longest Balanced Substring of a Binary String

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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'

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"01000111"

Example 2

"00111"

Example 3

"111"

Step 02

Core Insight

What unlocks the optimal approach

Consider iterating over each subarray and checking if it’s balanced or not.
Among all balanced subarrays, the answer is the longest one of them.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2609: Find the Longest Balanced Substring of a Binary String
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;
    }
}

// Accepted solution for LeetCode #2609: Find the Longest Balanced Substring of a Binary String
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
}

# Accepted solution for LeetCode #2609: Find the Longest Balanced Substring of a Binary String
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

// Accepted solution for LeetCode #2609: Find the Longest Balanced Substring of a Binary String
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
    }
}

// Accepted solution for LeetCode #2609: Find the Longest Balanced Substring of a Binary String
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^3)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.