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g3001_3100.s3093_longest_common_suffix_queries.readme.md Maven / Gradle / Ivy

3093\. Longest Common Suffix Queries

Hard

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.