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3045. Count Prefix and Suffix Pairs II

Description

You are given a 0-indexed string array words.

Let's define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2:

• isPrefixAndSuffix(str1, str2) returns true if str1 is both a prefix and a suffix of str2, and false otherwise.

For example, isPrefixAndSuffix("aba", "ababa") is true because "aba" is a prefix of "ababa" and also a suffix, but isPrefixAndSuffix("abc", "abcd") is false.

Return an integer denoting the number of index pairs (i, j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.

 

Example 1:

Input: words = ["a","aba","ababa","aa"]
Output: 4
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("a", "aba") is true.
i = 0 and j = 2 because isPrefixAndSuffix("a", "ababa") is true.
i = 0 and j = 3 because isPrefixAndSuffix("a", "aa") is true.
i = 1 and j = 2 because isPrefixAndSuffix("aba", "ababa") is true.
Therefore, the answer is 4.

Example 2:

Input: words = ["pa","papa","ma","mama"]
Output: 2
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("pa", "papa") is true.
i = 2 and j = 3 because isPrefixAndSuffix("ma", "mama") is true.
Therefore, the answer is 2.  

Example 3:

Input: words = ["abab","ab"]
Output: 0
Explanation: In this example, the only valid index pair is i = 0 and j = 1, and isPrefixAndSuffix("abab", "ab") is false.
Therefore, the answer is 0.

 

Constraints:

• 1 <= words.length <= 105
• 1 <= words[i].length <= 105
• words[i] consists only of lowercase English letters.
• The sum of the lengths of all words[i] does not exceed 5 * 105.

Solutions

Solution 1: Trie

We can treat each string $s$ in the string array as a list of character pairs, where each character pair $(s[i], s[m - i - 1])$ represents the $i$th character pair of the prefix and suffix of string $s$.

We can use a trie to store all the character pairs, and then for each string $s$, we search for all the character pairs $(s[i], s[m - i - 1])$ in the trie, and add their counts to the answer.

The time complexity is $O(n \times m)$, and the space complexity is $O(n \times m)$. Here, $n$ and $m$ are the lengths of words and the maximum length of the strings, respectively.

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

• class Node {
      Map<Integer, Node> children = new HashMap<>();
      int cnt;
  }
  
  class Solution {
      public long countPrefixSuffixPairs(String[] words) {
          long ans = 0;
          Node trie = new Node();
          for (String s : words) {
              Node node = trie;
              int m = s.length();
              for (int i = 0; i < m; ++i) {
                  int p = s.charAt(i) * 32 + s.charAt(m - i - 1);
                  node.children.putIfAbsent(p, new Node());
                  node = node.children.get(p);
                  ans += node.cnt;
              }
              ++node.cnt;
          }
          return ans;
      }
  }
  

• class Node {
  public:
      unordered_map<int, Node*> children;
      int cnt;
  
      Node()
          : cnt(0) {}
  };
  
  class Solution {
  public:
      long long countPrefixSuffixPairs(vector<string>& words) {
          long long ans = 0;
          Node* trie = new Node();
          for (const string& s : words) {
              Node* node = trie;
              int m = s.length();
              for (int i = 0; i < m; ++i) {
                  int p = s[i] * 32 + s[m - i - 1];
                  if (node->children.find(p) == node->children.end()) {
                      node->children[p] = new Node();
                  }
                  node = node->children[p];
                  ans += node->cnt;
              }
              ++node->cnt;
          }
          return ans;
      }
  };
  

• class Node:
      __slots__ = ["children", "cnt"]
  
      def __init__(self):
          self.children = {}
          self.cnt = 0
  
  
  class Solution:
      def countPrefixSuffixPairs(self, words: List[str]) -> int:
          ans = 0
          trie = Node()
          for s in words:
              node = trie
              for p in zip(s, reversed(s)):
                  if p not in node.children:
                      node.children[p] = Node()
                  node = node.children[p]
                  ans += node.cnt
              node.cnt += 1
          return ans
  
  

• type Node struct {
  	children map[int]*Node
  	cnt      int
  }
  
  func countPrefixSuffixPairs(words []string) (ans int64) {
  	trie := &Node{children: make(map[int]*Node)}
  	for _, s := range words {
  		node := trie
  		m := len(s)
  		for i := 0; i < m; i++ {
  			p := int(s[i])*32 + int(s[m-i-1])
  			if _, ok := node.children[p]; !ok {
  				node.children[p] = &Node{children: make(map[int]*Node)}
  			}
  			node = node.children[p]
  			ans += int64(node.cnt)
  		}
  		node.cnt++
  	}
  	return
  }
  

• class Node {
      children: Map<number, Node> = new Map<number, Node>();
      cnt: number = 0;
  }
  
  function countPrefixSuffixPairs(words: string[]): number {
      let ans: number = 0;
      const trie: Node = new Node();
      for (const s of words) {
          let node: Node = trie;
          const m: number = s.length;
          for (let i: number = 0; i < m; ++i) {
              const p: number = s.charCodeAt(i) * 32 + s.charCodeAt(m - i - 1);
              if (!node.children.has(p)) {
                  node.children.set(p, new Node());
              }
              node = node.children.get(p)!;
              ans += node.cnt;
          }
          ++node.cnt;
      }
      return ans;
  }
  
  

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