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1181. Before and After Puzzle

Description

Given a list of phrases, generate a list of Before and After puzzles.

A phrase is a string that consists of lowercase English letters and spaces only. No space appears in the start or the end of a phrase. There are no consecutive spaces in a phrase.

Before and After puzzles are phrases that are formed by merging two phrases where the last word of the first phrase is the same as the first word of the second phrase.

Return the Before and After puzzles that can be formed by every two phrases phrases[i] and phrases[j] where i != j. Note that the order of matching two phrases matters, we want to consider both orders.

You should return a list of distinct strings sorted lexicographically.

 

Example 1:

Input: phrases = ["writing code","code rocks"]
Output: ["writing code rocks"]

Example 2:

Input: phrases = ["mission statement",
                  "a quick bite to eat",
                  "a chip off the old block",
                  "chocolate bar",
                  "mission impossible",
                  "a man on a mission",
                  "block party",
                  "eat my words",
                  "bar of soap"]
Output: ["a chip off the old block party",
         "a man on a mission impossible",
         "a man on a mission statement",
         "a quick bite to eat my words",
         "chocolate bar of soap"]

Example 3:

Input: phrases = ["a","b","a"]
Output: ["a"]

 

Constraints:

• 1 <= phrases.length <= 100
• 1 <= phrases[i].length <= 100

Solutions

Solution 1: Hash Table + Sorting

First, we traverse the phrases list, storing the first and last words of each phrase in the array $ps$, where $ps[i][0]$ and $ps[i][1]$ represent the first and last words of the $i$th phrase, respectively.

Next, we enumerate all $(i, j)$, where $i, j \in [0, n)$ and $i \neq j$. If $ps[i][1] = ps[j][0]$, then we can concatenate the $i$th phrase and the $j$th phrase to get a new phrase $phrases[i] + phrases[j][len(ps[j][0]):]$, and add the new phrase to the hash table $s$.

Finally, we convert the hash table $s$ into an array and sort it to get the answer.

The time complexity is $O(n^2 \times m \times (\log n + \log m))$, and the space complexity is $O(n^2 \times m)$. Here, $n$ and $m$ represent the length of the phrases array and the average length of each phrase, respectively.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public List<String> beforeAndAfterPuzzles(String[] phrases) {
          int n = phrases.length;
          var ps = new String[n][];
          for (int i = 0; i < n; ++i) {
              var ws = phrases[i].split(" ");
              ps[i] = new String[] {ws[0], ws[ws.length - 1]};
          }
          Set<String> s = new HashSet<>();
          for (int i = 0; i < n; ++i) {
              for (int j = 0; j < n; ++j) {
                  if (i != j && ps[i][1].equals(ps[j][0])) {
                      s.add(phrases[i] + phrases[j].substring(ps[j][0].length()));
                  }
              }
          }
          var ans = new ArrayList<>(s);
          Collections.sort(ans);
          return ans;
      }
  }
  

• class Solution {
  public:
      vector<string> beforeAndAfterPuzzles(vector<string>& phrases) {
          int n = phrases.size();
          pair<string, string> ps[n];
          for (int i = 0; i < n; ++i) {
              int j = phrases[i].find(' ');
              if (j == string::npos) {
                  ps[i] = {phrases[i], phrases[i]};
              } else {
                  int k = phrases[i].rfind(' ');
                  ps[i] = {phrases[i].substr(0, j), phrases[i].substr(k + 1)};
              }
          }
          unordered_set<string> s;
          for (int i = 0; i < n; ++i) {
              for (int j = 0; j < n; ++j) {
                  if (i != j && ps[i].second == ps[j].first) {
                      s.insert(phrases[i] + phrases[j].substr(ps[i].second.size()));
                  }
              }
          }
          vector<string> ans(s.begin(), s.end());
          sort(ans.begin(), ans.end());
          return ans;
      }
  };
  

• class Solution:
      def beforeAndAfterPuzzles(self, phrases: List[str]) -> List[str]:
          ps = []
          for p in phrases:
              ws = p.split()
              ps.append((ws[0], ws[-1]))
          n = len(ps)
          ans = []
          for i in range(n):
              for j in range(n):
                  if i != j and ps[i][1] == ps[j][0]:
                      ans.append(phrases[i] + phrases[j][len(ps[j][0]) :])
          return sorted(set(ans))
  
  

• func beforeAndAfterPuzzles(phrases []string) []string {
  	n := len(phrases)
  	ps := make([][2]string, n)
  	for i, p := range phrases {
  		ws := strings.Split(p, " ")
  		ps[i] = [2]string{ws[0], ws[len(ws)-1]}
  	}
  	s := map[string]bool{}
  	for i := 0; i < n; i++ {
  		for j := 0; j < n; j++ {
  			if i != j && ps[i][1] == ps[j][0] {
  				s[phrases[i]+phrases[j][len(ps[j][0]):]] = true
  			}
  		}
  	}
  	ans := make([]string, 0, len(s))
  	for k := range s {
  		ans = append(ans, k)
  	}
  	sort.Strings(ans)
  	return ans
  }
  

• function beforeAndAfterPuzzles(phrases: string[]): string[] {
      const ps: string[][] = [];
      for (const p of phrases) {
          const ws = p.split(' ');
          ps.push([ws[0], ws[ws.length - 1]]);
      }
      const n = ps.length;
      const s: Set<string> = new Set();
      for (let i = 0; i < n; ++i) {
          for (let j = 0; j < n; ++j) {
              if (i !== j && ps[i][1] === ps[j][0]) {
                  s.add(`${phrases[i]}${phrases[j].substring(ps[j][0].length)}`);
              }
          }
      }
      return [...s].sort();
  }
  
  

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