3093. Longest Common Suffix Queries

Description

You are given two arrays of strings wordsContainer and wordsQuery.

For each wordsQuery[i], you need to find a string from wordsContainer that has the longest common suffix with wordsQuery[i]. If there are two or more strings in wordsContainer that share the longest common suffix, find the string that is the smallest in length. If there are two or more such strings that have the same smallest length, find the one that occurred earlier in wordsContainer.

Return an array of integers ans, where ans[i] is the index of the string in wordsContainer that has the longest common suffix with wordsQuery[i].

Example 1:

Input: wordsContainer = ["abcd","bcd","xbcd"], wordsQuery = ["cd","bcd","xyz"]

Output: [1,1,1]

Explanation:

Let's look at each wordsQuery[i] separately:

For wordsQuery[0] = "cd", strings from wordsContainer that share the longest common suffix "cd" are at indices 0, 1, and 2. Among these, the answer is the string at index 1 because it has the shortest length of 3.
For wordsQuery[1] = "bcd", strings from wordsContainer that share the longest common suffix "bcd" are at indices 0, 1, and 2. Among these, the answer is the string at index 1 because it has the shortest length of 3.
For wordsQuery[2] = "xyz", there is no string from wordsContainer that shares a common suffix. Hence the longest common suffix is "", that is shared with strings at index 0, 1, and 2. Among these, the answer is the string at index 1 because it has the shortest length of 3.

Example 2:

Input: wordsContainer = ["abcdefgh","poiuygh","ghghgh"], wordsQuery = ["gh","acbfgh","acbfegh"]

Output: [2,0,2]

Explanation:

Let's look at each wordsQuery[i] separately:

For wordsQuery[0] = "gh", strings from wordsContainer that share the longest common suffix "gh" are at indices 0, 1, and 2. Among these, the answer is the string at index 2 because it has the shortest length of 6.
For wordsQuery[1] = "acbfgh", only the string at index 0 shares the longest common suffix "fgh". Hence it is the answer, even though the string at index 2 is shorter.
For wordsQuery[2] = "acbfegh", strings from wordsContainer that share the longest common suffix "gh" are at indices 0, 1, and 2. Among these, the answer is the string at index 2 because it has the shortest length of 6.

Constraints:

1 <= wordsContainer.length, wordsQuery.length <= 10⁴
1 <= wordsContainer[i].length <= 5 * 10³
1 <= wordsQuery[i].length <= 5 * 10³
wordsContainer[i] consists only of lowercase English letters.
wordsQuery[i] consists only of lowercase English letters.
Sum of wordsContainer[i].length is at most 5 * 10⁵.
Sum of wordsQuery[i].length is at most 5 * 10⁵.

Solutions

Solution 1: Trie

The problem requires us to find the longest common suffix, so we can consider using a Trie.

We define the structure of the Trie node as follows:

children: An array of length 26, used to store child nodes.
length: The length of the shortest string at the current node.
idx: The index of the string at the current node.

We traverse the string array wordsContainer, and insert each string in reverse order into the Trie. During the insertion process, we update the length and idx of each node.

Next, we traverse the string array wordsQuery. For each string, we search for the index of the longest common suffix string from the Trie. During the search process, if we encounter a null node, it means that there is no common suffix afterwards, and we can directly return the idx of the current node.

The time complexity is $(L_1 \times |\Sigma| + L_2)$, and the space complexity is $O(L_1 \times |\Sigma|)$. Here, $L_1$ and $L_2$ are the sum of the lengths of the strings in wordsContainer and wordsQuery respectively; and $\Sigma$ is the size of the character set, in this problem $\Sigma = 26$.

class Trie {
    private final int inf = 1 << 30;
    private Trie[] children = new Trie[26];
    private int length = inf;
    private int idx = inf;

    public void insert(String w, int i) {
        Trie node = this;
        if (node.length > w.length()) {
            node.length = w.length();
            node.idx = i;
        }
        for (int k = w.length() - 1; k >= 0; --k) {
            int idx = w.charAt(k) - 'a';
            if (node.children[idx] == null) {
                node.children[idx] = new Trie();
            }
            node = node.children[idx];
            if (node.length > w.length()) {
                node.length = w.length();
                node.idx = i;
            }
        }
    }

    public int query(String w) {
        Trie node = this;
        for (int k = w.length() - 1; k >= 0; --k) {
            int idx = w.charAt(k) - 'a';
            if (node.children[idx] == null) {
                break;
            }
            node = node.children[idx];
        }
        return node.idx;
    }
}

class Solution {
    public int[] stringIndices(String[] wordsContainer, String[] wordsQuery) {
        Trie trie = new Trie();
        for (int i = 0; i < wordsContainer.length; ++i) {
            trie.insert(wordsContainer[i], i);
        }
        int n = wordsQuery.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = trie.query(wordsQuery[i]);
        }
        return ans;
    }
}

class Trie {
private:
    const int inf = 1 << 30;
    Trie* children[26];
    int length = inf;
    int idx = inf;

public:
    Trie() {
        for (int i = 0; i < 26; ++i) {
            children[i] = nullptr;
        }
    }

    void insert(string w, int i) {
        Trie* node = this;
        if (node->length > w.length()) {
            node->length = w.length();
            node->idx = i;
        }
        for (int k = w.length() - 1; k >= 0; --k) {
            int idx = w[k] - 'a';
            if (node->children[idx] == nullptr) {
                node->children[idx] = new Trie();
            }
            node = node->children[idx];
            if (node->length > w.length()) {
                node->length = w.length();
                node->idx = i;
            }
        }
    }

    int query(string w) {
        Trie* node = this;
        for (int k = w.length() - 1; k >= 0; --k) {
            int idx = w[k] - 'a';
            if (node->children[idx] == nullptr) {
                break;
            }
            node = node->children[idx];
        }
        return node->idx;
    }
};

class Solution {
public:
    vector<int> stringIndices(vector<string>& wordsContainer, vector<string>& wordsQuery) {
        Trie* trie = new Trie();
        for (int i = 0; i < wordsContainer.size(); ++i) {
            trie->insert(wordsContainer[i], i);
        }
        int n = wordsQuery.size();
        vector<int> ans(n);
        for (int i = 0; i < n; ++i) {
            ans[i] = trie->query(wordsQuery[i]);
        }
        return ans;
    }
};

class Trie:
    __slots__ = ("children", "length", "idx")

    def __init__(self):
        self.children = [None] * 26
        self.length = inf
        self.idx = inf

    def insert(self, w: str, i: int):
        node = self
        if node.length > len(w):
            node.length = len(w)
            node.idx = i
        for c in w[::-1]:
            idx = ord(c) - ord("a")
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
            if node.length > len(w):
                node.length = len(w)
                node.idx = i

    def query(self, w: str) -> int:
        node = self
        for c in w[::-1]:
            idx = ord(c) - ord("a")
            if node.children[idx] is None:
                break
            node = node.children[idx]
        return node.idx


class Solution:
    def stringIndices(
        self, wordsContainer: List[str], wordsQuery: List[str]
    ) -> List[int]:
        trie = Trie()
        for i, w in enumerate(wordsContainer):
            trie.insert(w, i)
        return [trie.query(w) for w in wordsQuery]

const inf = 1 << 30

type Trie struct {
	children [26]*Trie
	length   int
	idx      int
}

func newTrie() *Trie {
	return &Trie{length: inf, idx: inf}
}

func (t *Trie) insert(w string, i int) {
	node := t
	if node.length > len(w) {
		node.length = len(w)
		node.idx = i
	}
	for k := len(w) - 1; k >= 0; k-- {
		idx := int(w[k] - 'a')
		if node.children[idx] == nil {
			node.children[idx] = newTrie()
		}
		node = node.children[idx]
		if node.length > len(w) {
			node.length = len(w)
			node.idx = i
		}
	}
}

func (t *Trie) query(w string) int {
	node := t
	for k := len(w) - 1; k >= 0; k-- {
		idx := int(w[k] - 'a')
		if node.children[idx] == nil {
			break
		}
		node = node.children[idx]
	}
	return node.idx
}

func stringIndices(wordsContainer []string, wordsQuery []string) (ans []int) {
	trie := newTrie()
	for i, w := range wordsContainer {
		trie.insert(w, i)
	}
	for _, w := range wordsQuery {
		ans = append(ans, trie.query(w))
	}
	return
}

class Trie {
    private children: Trie[] = new Array<Trie>(26);
    private length: number = Infinity;
    private idx: number = Infinity;

    public insert(w: string, i: number): void {
        let node: Trie = this;
        if (node.length > w.length) {
            node.length = w.length;
            node.idx = i;
        }
        for (let k: number = w.length - 1; k >= 0; --k) {
            let idx: number = w.charCodeAt(k) - 'a'.charCodeAt(0);
            if (node.children[idx] == null) {
                node.children[idx] = new Trie();
            }
            node = node.children[idx];
            if (node.length > w.length) {
                node.length = w.length;
                node.idx = i;
            }
        }
    }

    public query(w: string): number {
        let node: Trie = this;
        for (let k: number = w.length - 1; k >= 0; --k) {
            let idx: number = w.charCodeAt(k) - 'a'.charCodeAt(0);
            if (node.children[idx] == null) {
                break;
            }
            node = node.children[idx];
        }
        return node.idx;
    }
}

function stringIndices(wordsContainer: string[], wordsQuery: string[]): number[] {
    const trie: Trie = new Trie();
    for (let i: number = 0; i < wordsContainer.length; ++i) {
        trie.insert(wordsContainer[i], i);
    }
    const n: number = wordsQuery.length;
    const ans: number[] = new Array<number>(n);
    for (let i: number = 0; i < n; ++i) {
        ans[i] = trie.query(wordsQuery[i]);
    }
    return ans;
}

3093 - Longest Common Suffix Queries

3093. Longest Common Suffix Queries

Description

Solutions

Solution 1: Trie

All Problems

All Solutions

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3093. Longest Common Suffix Queries

Description

Solutions

Solution 1: Trie

All Problems

All Solutions