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Problem solution · Python

Longest Balanced Substring II

Longest Balanced Substring II: a Python solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Hash-based lookup
Source
Kamyu LeetCode Solutions
Length
61 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For Longest Balanced Substring II, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 61 lines of Python from the credited upstream file longest-balanced-substring-ii.py.
  • The implementation visibly relies on hash lookup.
  • No explicit loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeLongest Balanced Substring II · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(n) import collections  # hash table, prefix sumclass Solution(object):    def longestBalanced(self, s):        """        :type s: str        :rtype: int        """        def count1():            result = cnt = 0            for i in xrange(len(s)):                cnt += 1                if i+1 == len(s) or s[i+1] != s[i]:                    result = max(result, cnt)                    cnt = 0            return result         def count2(a, b):            result = cnt = 0            lookup = collections.defaultdict(int, {cnt:-1})            for i, x in enumerate(s):                if x == a:                    cnt += 1                elif x == b:                    cnt -= 1                else:                    cnt = 0                    lookup = collections.defaultdict(int, {cnt:i})                    continue                if cnt in lookup:                    result = max(result, i-lookup[cnt])                else:                    lookup[cnt] = i            return result            def count3():            result = a = b = 0            lookup = collections.defaultdict(int, {(a, b):-1})            for i, x in enumerate(s):                if x == 'a':                    a += 1                elif x == 'b':                    b += 1                else:                    a -= 1                    b -= 1                if (a, b) in lookup:                    result = max(result, i-lookup[a, b])                else:                    lookup[a, b] = i            return result                return max(count1(),                    count2('a', 'b'), count2('b', 'c'), count2('c', 'a'),                   count3())

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