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Problem solution · Python

Maximum Length of a Concatenated String with Unique Characters

Maximum Length of a Concatenated String with Unique Characters: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Dynamic programming
Source
Kamyu LeetCode Solutions
Length
76 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

Dynamic programming

For Maximum Length of a Concatenated String with Unique Characters, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 76 lines of Python from the credited upstream file maximum-length-of-a-concatenated-string-with-unique-characters.py.
  • The implementation visibly relies on sequence storage, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeMaximum Length of a Concatenated String with Unique Characters · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n) ~ O(2^n)# Space: O(1) ~ O(2^n) power = [1]log2 = {1:0}for i in xrange(1, 26):    power.append(power[-1]<<1)    log2[power[i]] = i  class Solution(object):    def maxLength(self, arr):        """        :type arr: List[str]        :rtype: int        """        def bitset(s):            result = 0            for c in s:                if result & power[ord(c)-ord('a')]:                    return 0                result |= power[ord(c)-ord('a')]            return result                def number_of_one(n):            result = 0            while n:                n &= n-1                result += 1            return result         dp = [0]        for x in arr:            x_set = bitset(x)            if not x_set:                continue            curr_len = len(dp)            for i in xrange(curr_len):                if dp[i] & x_set:                    continue                dp.append(dp[i] | x_set)        return max(number_of_one(s_set) for s_set in dp)  # Time:  O(2^n)# Space: O(1)class Solution2(object):    def maxLength(self, arr):        """        :type arr: List[str]        :rtype: int        """         def bitset(s):            result = 0            for c in s:                if result & power[ord(c)-ord('a')]:                    return 0                result |= power[ord(c)-ord('a')]            return result            bitsets = [bitset(x) for x in arr]        result = 0        for i in xrange(power[len(arr)]):            curr_bitset, curr_len = 0, 0            while i:                j = i & -i  # rightmost bit                i ^= j                j = log2[j]  # log2(j)                if not bitsets[j] or (curr_bitset & bitsets[j]):                    break                curr_bitset |= bitsets[j]                curr_len += len(arr[j])            else:                result = max(result, curr_len)        return result

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