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3042. Count Prefix and Suffix Pairs I

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 <= 50
  • 1 <= words[i].length <= 10
  • words[i] consists only of lowercase English letters.

Solutions

Solution 1: Enumeration

We can enumerate all index pairs $(i, j)$, where $i < j$, and then determine whether words[i] is a prefix or suffix of words[j]. If it is, we increment the count.

The time complexity is $O(n^2 \times m)$, where $n$ and $m$ are the length of words and the maximum length of the strings, respectively.

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class Solution:
    def countPrefixSuffixPairs(self, words: List[str]) -> int:
        ans = 0
        for i, s in enumerate(words):
            for t in words[i + 1 :]:
                ans += t.endswith(s) and t.startswith(s)
        return ans

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class Solution {
    public int countPrefixSuffixPairs(String[] words) {
        int ans = 0;
        int n = words.length;
        for (int i = 0; i < n; ++i) {
            String s = words[i];
            for (int j = i + 1; j < n; ++j) {
                String t = words[j];
                if (t.startsWith(s) && t.endsWith(s)) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int countPrefixSuffixPairs(vector<string>& words) {
        int ans = 0;
        int n = words.size();
        for (int i = 0; i < n; ++i) {
            string s = words[i];
            for (int j = i + 1; j < n; ++j) {
                string t = words[j];
                if (t.find(s) == 0 && t.rfind(s) == t.length() - s.length()) {
                    ++ans;
                }
            }
        }
        return ans;
    }
};

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func countPrefixSuffixPairs(words []string) (ans int) {
    for i, s := range words {
        for _, t := range words[i+1:] {
            if strings.HasPrefix(t, s) && strings.HasSuffix(t, s) {
                ans++
            }
        }
    }
    return
}

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function countPrefixSuffixPairs(words: string[]): number {
    let ans = 0;
    for (let i = 0; i < words.length; ++i) {
        const s = words[i];
        for (const t of words.slice(i + 1)) {
            if (t.startsWith(s) && t.endsWith(s)) {
                ++ans;
            }
        }
    }
    return ans;
}

Solution 2: 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.

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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

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class Node {
    Map<Integer, Node> children = new HashMap<>();
    int cnt;
}

class Solution {
    public int countPrefixSuffixPairs(String[] words) {
        int 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;
    }
}

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class Node {
public:
    unordered_map<int, Node*> children;
    int cnt;

    Node()
        : cnt(0) {}
};

class Solution {
public:
    int countPrefixSuffixPairs(vector<string>& words) {
        int 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;
    }
};

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type Node struct {
    children map[int]*Node
    cnt      int
}

func countPrefixSuffixPairs(words []string) (ans int) {
    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 += node.cnt
        }
        node.cnt++
    }
    return
}

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