Leetcode #761: Special Binary String
In this guide, we solve Leetcode #761 Special Binary String in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Recursion, String
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Example
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.
Python Solution
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)
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.