O(2^n) Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/o2n/ Tue, 29 Oct 2019 07:04:08 +0000 en hourly 1 https://wordpress.org/?v=6.0.8 https://zxi.mytechroad.com/blog/wp-content/uploads/2017/09/cropped-photo-32x32.jpg O(2^n) Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/o2n/ 32 32 花花酱 LeetCode 1239. Maximum Length of a Concatenated String with Unique Characters https://zxi.mytechroad.com/blog/searching/leetcode-1239-maximum-length-of-a-concatenated-string-with-unique-characters/ https://zxi.mytechroad.com/blog/searching/leetcode-1239-maximum-length-of-a-concatenated-string-with-unique-characters/#respond Mon, 28 Oct 2019 03:56:41 +0000 https://zxi.mytechroad.com/blog/?p=5792 Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s. Example 1: Input: arr = ["un","iq","ue"] Output:…

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]]> Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters.

Return the maximum possible length of s.

Example 1:

Input: arr = ["un","iq","ue"]
Output: 4
Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique".
Maximum length is 4.

Example 2:

Input: arr = ["cha","r","act","ers"]
Output: 6
Explanation: Possible solutions are "chaers" and "acters".

Example 3:

Input: arr = ["abcdefghijklmnopqrstuvwxyz"]
Output: 26

Constraints:

  • 1 <= arr.length <= 16
  • 1 <= arr[i].length <= 26
  • arr[i] contains only lower case English letters.

Solution: Combination + Bit

Time complexity: O(2^n)
Space complexity: O(n)

C++

// Author: Huahua
class Solution {
public:
  int maxLength(vector<string>& arr) {
    vector<int> a;
    
    for (const string& x : arr) {
      set<char> s(begin(x), end(x));
      if (s.size() != x.length()) continue;
      a.push_back(0);      
      for (char c : x) a.back() |= 1 << (c - 'a');      
    }
    
    int ans = 0;
    
    function<void(int, int)> dfs = [&](int s, int cur) {
      ans = max(ans, __builtin_popcount(cur));
      for (int i = s; i < a.size(); ++i)
        if ((cur & a[i]) == 0)
          dfs(i + 1, cur | a[i]);      
    };
    
    dfs(0, 0);
    return ans;
  }
};

Solution 2: DP

Time complexity: O(2^n)
Space complexity: O(2^n)

C++

// Author: Huahua
class Solution {
public:
  int maxLength(vector<string>& arr) {
    vector<int> a;
    
    for (const string& x : arr) {
      int mask = 0;      
      for (char c : x) mask |= 1 << (c - 'a');
      if (__builtin_popcount(mask) != x.length()) continue;
      a.push_back(mask);
    }
    
    int ans = 0;
    
    vector<int> dp{0};
    for (int i = 0; i < a.size(); ++i) {
      int size = dp.size();
      for (int j = 0; j < size; ++j) {
        if (dp[j] & a[i]) continue;
        int t = dp[j] | a[i];        
        dp.push_back(t);
        ans = max(ans, __builtin_popcount(t));
      }
    }
    
    return ans;
  }
};

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