LeetCode #761 — HARD

Special Binary String

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Special binary strings are binary strings with the following two properties:

The number of 0's is equal to the number of 1's.
Every prefix of the binary string has at least as many 1's as 0's.

You are given a special binary string s.

A move consists of choosing two consecutive, non-empty, special substrings of s, and swapping them. Two strings are consecutive if the last character of the first string is exactly one index before the first character of the second string.

Return the lexicographically largest resulting string possible after applying the mentioned operations on the string.

Example 1:

Input: s = "11011000"
Output: "11100100"
Explanation: The strings "10" [occuring at s[1]] and "1100" [at s[3]] are swapped.
This is the lexicographically largest string possible after some number of swaps.

Example 2:

Input: s = "10"
Output: "10"

Constraints:

1 <= s.length <= 50
s[i] is either '0' or '1'.
s is a special binary string.

Step 01

Brute Force Baseline

Problem summary: Special binary strings are binary strings with the following two properties: The number of 0's is equal to the number of 1's. Every prefix of the binary string has at least as many 1's as 0's. You are given a special binary string s. A move consists of choosing two consecutive, non-empty, special substrings of s, and swapping them. Two strings are consecutive if the last character of the first string is exactly one index before the first character of the second string. Return the lexicographically largest resulting string possible after applying the mentioned operations on the string.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"11011000"

Example 2

"10"

Core Insight

What unlocks the optimal approach

Draw a line from (x, y) to (x+1, y+1) if we see a "1", else to (x+1, y-1). A special substring is just a line that starts and ends at the same y-coordinate, and that is the lowest y-coordinate reached. Call a mountain a special substring with no special prefixes - ie. only at the beginning and end is the lowest y-coordinate reached. If F is the answer function, and S has mountain decomposition M1,M2,M3,...,Mk, then the answer is: reverse_sorted(F(M1), F(M2), ..., F(Mk)). However, you'll also need to deal with the case that S is a mountain, such as 11011000 -> 11100100.

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

Largest constraint values

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 #761: Special Binary String
class Solution {
    public String makeLargestSpecial(String s) {
        if ("".equals(s)) {
            return "";
        }
        List<String> ans = new ArrayList<>();
        int cnt = 0;
        for (int i = 0, j = 0; i < s.length(); ++i) {
            cnt += s.charAt(i) == '1' ? 1 : -1;
            if (cnt == 0) {
                String t = "1" + makeLargestSpecial(s.substring(j + 1, i)) + "0";
                ans.add(t);
                j = i + 1;
            }
        }
        ans.sort(Comparator.reverseOrder());
        return String.join("", ans);
    }
}

// Accepted solution for LeetCode #761: Special Binary String
func makeLargestSpecial(s string) string {
	if s == "" {
		return ""
	}
	ans := sort.StringSlice{}
	cnt := 0
	for i, j := 0, 0; i < len(s); i++ {
		if s[i] == '1' {
			cnt++
		} else {
			cnt--
		}
		if cnt == 0 {
			ans = append(ans, "1"+makeLargestSpecial(s[j+1:i])+"0")
			j = i + 1
		}
	}
	sort.Sort(sort.Reverse(ans))
	return strings.Join(ans, "")
}

# Accepted solution for LeetCode #761: Special Binary String
class Solution:
    def makeLargestSpecial(self, s: str) -> str:
        if s == '':
            return ''
        ans = []
        cnt = 0
        i = j = 0
        while i < len(s):
            cnt += 1 if s[i] == '1' else -1
            if cnt == 0:
                ans.append('1' + self.makeLargestSpecial(s[j + 1 : i]) + '0')
                j = i + 1
            i += 1
        ans.sort(reverse=True)
        return ''.join(ans)

// Accepted solution for LeetCode #761: Special Binary String
struct Solution;

use std::cmp::Reverse;

impl Solution {
    fn make_largest_special(s: String) -> String {
        let mut count = 0;
        let mut v = vec![];
        let mut ss = "".to_string();
        for c in s.chars() {
            ss.push(c);
            if c == '1' {
                count += 1;
            } else {
                count -= 1;
            }
            if count == 0 {
                let n = ss.len();
                let t = format!("1{}0", Self::make_largest_special(ss[1..n - 1].to_string()));
                v.push(t);
                ss = "".to_string();
            }
        }
        v.sort_unstable_by_key(|s| Reverse(s.to_string()));
        v.join("")
    }
}

#[test]
fn test() {
    let s = "11011000".to_string();
    let res = "11100100".to_string();
    assert_eq!(Solution::make_largest_special(s), res);
}

// Accepted solution for LeetCode #761: Special Binary String
function makeLargestSpecial(s: string): string {
    if (s.length === 0) {
        return '';
    }

    const ans: string[] = [];
    let cnt = 0;

    for (let i = 0, j = 0; i < s.length; ++i) {
        cnt += s[i] === '1' ? 1 : -1;
        if (cnt === 0) {
            const t = '1' + makeLargestSpecial(s.substring(j + 1, i)) + '0';
            ans.push(t);
            j = i + 1;
        }
    }

    ans.sort((a, b) => b.localeCompare(a));
    return ans.join('');
}

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^2)

Space

O(n)

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.