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2416. Sum of Prefix Scores of Strings

Description

You are given an array words of size n consisting of non-empty strings.

We define the score of a string word as the number of strings words[i] such that word is a prefix of words[i].

  • For example, if words = ["a", "ab", "abc", "cab"], then the score of "ab" is 2, since "ab" is a prefix of both "ab" and "abc".

Return an array answer of size n where answer[i] is the sum of scores of every non-empty prefix of words[i].

Note that a string is considered as a prefix of itself.

 

Example 1:

Input: words = ["abc","ab","bc","b"]
Output: [5,4,3,2]
Explanation: The answer for each string is the following:
- "abc" has 3 prefixes: "a", "ab", and "abc".
- There are 2 strings with the prefix "a", 2 strings with the prefix "ab", and 1 string with the prefix "abc".
The total is answer[0] = 2 + 2 + 1 = 5.
- "ab" has 2 prefixes: "a" and "ab".
- There are 2 strings with the prefix "a", and 2 strings with the prefix "ab".
The total is answer[1] = 2 + 2 = 4.
- "bc" has 2 prefixes: "b" and "bc".
- There are 2 strings with the prefix "b", and 1 string with the prefix "bc".
The total is answer[2] = 2 + 1 = 3.
- "b" has 1 prefix: "b".
- There are 2 strings with the prefix "b".
The total is answer[3] = 2.

Example 2:

Input: words = ["abcd"]
Output: [4]
Explanation:
"abcd" has 4 prefixes: "a", "ab", "abc", and "abcd".
Each prefix has a score of one, so the total is answer[0] = 1 + 1 + 1 + 1 = 4.

 

Constraints:

  • 1 <= words.length <= 1000
  • 1 <= words[i].length <= 1000
  • words[i] consists of lowercase English letters.

Solutions

Solution 1: Trie

Use a trie to maintain the prefixes of all strings and the occurrence count of each prefix.

Then, traverse each string, accumulating the occurrence count of each prefix.

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

  • class Trie {
    private Trie[] children = new Trie[26];
    private int cnt;

    public void insert(String w) {
        Trie node = this;
        for (char c : w.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
            ++node.cnt;
        }
    }

    public int search(String w) {
        Trie node = this;
        int ans = 0;
        for (char c : w.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                return ans;
            }
            node = node.children[c];
            ans += node.cnt;
        }
        return ans;
    }
}

class Solution {
    public int[] sumPrefixScores(String[] words) {
        Trie trie = new Trie();
        for (String w : words) {
            trie.insert(w);
        }
        int[] ans = new int[words.length];
        for (int i = 0; i < words.length; ++i) {
            ans[i] = trie.search(words[i]);
        }
        return ans;
    }
}

  • class Trie {
private:
    vector<Trie*> children;
    int cnt;

public:
    Trie()
        : children(26)
        , cnt(0) {}

    void insert(string& w) {
        Trie* node = this;
        for (char c : w) {
            int idx = c - 'a';
            if (!node->children[idx]) node->children[idx] = new Trie();
            node = node->children[idx];
            ++node->cnt;
        }
    }

    int search(string& w) {
        Trie* node = this;
        int ans = 0;
        for (char c : w) {
            int idx = c - 'a';
            if (!node->children[idx]) return ans;
            node = node->children[idx];
            ans += node->cnt;
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> sumPrefixScores(vector<string>& words) {
        Trie* trie = new Trie();
        for (auto& w : words) {
            trie->insert(w);
        }
        vector<int> ans;
        for (auto& w : words) {
            ans.push_back(trie->search(w));
        }
        return ans;
    }
};

  • class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.cnt = 0

    def insert(self, w):
        node = self
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
            node.cnt += 1

    def search(self, w):
        node = self
        ans = 0
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                return ans
            node = node.children[idx]
            ans += node.cnt
        return ans


class Solution:
    def sumPrefixScores(self, words: List[str]) -> List[int]:
        trie = Trie()
        for w in words:
            trie.insert(w)
        return [trie.search(w) for w in words]


  • type Trie struct {
	children [26]*Trie
	cnt      int
}

func newTrie() *Trie {
	return &Trie{}
}
func (this *Trie) insert(w string) {
	node := this
	for _, c := range w {
		c -= 'a'
		if node.children[c] == nil {
			node.children[c] = newTrie()
		}
		node = node.children[c]
		node.cnt++
	}
}

func (this *Trie) search(word string) int {
	node := this
	ans := 0
	for _, c := range word {
		c -= 'a'
		if node.children[c] == nil {
			return ans
		}
		node = node.children[c]
		ans += node.cnt
	}
	return ans
}

func sumPrefixScores(words []string) []int {
	trie := newTrie()
	for _, w := range words {
		trie.insert(w)
	}
	ans := make([]int, len(words))
	for i, w := range words {
		ans[i] = trie.search(w)
	}
	return ans
}

  • function sumPrefixScores(words: string[]): number[] {
    const map = new Map();

    for (const word of words) {
        const n = word.length;
        for (let i = 1; i <= n; i++) {
            const s = word.slice(0, i);
            map.set(s, (map.get(s) ?? 0) + 1);
        }
    }

    return words.map(word => {
        const n = word.length;
        let count = 0;
        for (let i = 1; i <= n; i++) {
            const s = word.slice(0, i);
            count += map.get(s);
        }
        return count;
    });
}

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