3076. Shortest Uncommon Substring in an Array

Description

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

Find a string array answer of size n such that:

• answer[i] is the shortest substring of arr[i] that does not occur as a substring in any other string in arr. If multiple such substrings exist, answer[i] should be the lexicographically smallest. And if no such substring exists, answer[i] should be an empty string.

Return the array answer.

 

Example 1:

Input: arr = ["cab","ad","bad","c"]
Output: ["ab","","ba",""]
Explanation: We have the following:
- For the string "cab", the shortest substring that does not occur in any other string is either "ca" or "ab", we choose the lexicographically smaller substring, which is "ab".
- For the string "ad", there is no substring that does not occur in any other string.
- For the string "bad", the shortest substring that does not occur in any other string is "ba".
- For the string "c", there is no substring that does not occur in any other string.

Example 2:

Input: arr = ["abc","bcd","abcd"]
Output: ["","","abcd"]
Explanation: We have the following:
- For the string "abc", there is no substring that does not occur in any other string.
- For the string "bcd", there is no substring that does not occur in any other string.
- For the string "abcd", the shortest substring that does not occur in any other string is "abcd".

 

Constraints:

• n == arr.length
• 2 <= n <= 100
• 1 <= arr[i].length <= 20
• arr[i] consists only of lowercase English letters.

Solutions

Solution 1: Enumeration

Given the small data scale, we can directly enumerate all substrings of each string and then determine whether it is a substring of other strings.

Specifically, we first enumerate each string arr[i], then enumerate the length \(j\) of each substring from small to large, and then enumerate the starting position \(l\) of each substring. We can get the current substring as sub = arr[i][l:l+j]. Then we determine whether sub is a substring of other strings. If it is, we skip the current substring; otherwise, we update the answer.

The time complexity is \(O(n^2 \times m^4)\), and the space complexity is \(O(m)\). Where \(n\) is the length of the string array arr, and \(m\) is the maximum length of the string. In this problem, \(m \le 20\).

Python3JavaC++GoTypeScript

class Solution:
    def shortestSubstrings(self, arr: List[str]) -> List[str]:
        ans = [""] * len(arr)
        for i, s in enumerate(arr):
            m = len(s)
            for j in range(1, m + 1):
                for l in range(m - j + 1):
                    sub = s[l : l + j]
                    if not ans[i] or ans[i] > sub:
                        if all(k == i or sub not in t for k, t in enumerate(arr)):
                            ans[i] = sub
                if ans[i]:
                    break
        return ans

class Solution {
    public String[] shortestSubstrings(String[] arr) {
        int n = arr.length;
        String[] ans = new String[n];
        Arrays.fill(ans, "");
        for (int i = 0; i < n; ++i) {
            int m = arr[i].length();
            for (int j = 1; j <= m && ans[i].isEmpty(); ++j) {
                for (int l = 0; l <= m - j; ++l) {
                    String sub = arr[i].substring(l, l + j);
                    if (ans[i].isEmpty() || sub.compareTo(ans[i]) < 0) {
                        boolean ok = true;
                        for (int k = 0; k < n && ok; ++k) {
                            if (k != i && arr[k].contains(sub)) {
                                ok = false;
                            }
                        }
                        if (ok) {
                            ans[i] = sub;
                        }
                    }
                }
            }
        }
        return ans;
    }
}

class Solution {
public:
    vector<string> shortestSubstrings(vector<string>& arr) {
        int n = arr.size();
        vector<string> ans(n);
        for (int i = 0; i < n; ++i) {
            int m = arr[i].size();
            for (int j = 1; j <= m && ans[i].empty(); ++j) {
                for (int l = 0; l <= m - j; ++l) {
                    string sub = arr[i].substr(l, j);
                    if (ans[i].empty() || sub < ans[i]) {
                        bool ok = true;
                        for (int k = 0; k < n && ok; ++k) {
                            if (k != i && arr[k].find(sub) != string::npos) {
                                ok = false;
                            }
                        }
                        if (ok) {
                            ans[i] = sub;
                        }
                    }
                }
            }
        }
        return ans;
    }
};

func shortestSubstrings(arr []string) []string {
    ans := make([]string, len(arr))
    for i, s := range arr {
        m := len(s)
        for j := 1; j <= m && len(ans[i]) == 0; j++ {
            for l := 0; l <= m-j; l++ {
                sub := s[l : l+j]
                if len(ans[i]) == 0 || ans[i] > sub {
                    ok := true
                    for k, t := range arr {
                        if k != i && strings.Contains(t, sub) {
                            ok = false
                            break
                        }
                    }
                    if ok {
                        ans[i] = sub
                    }
                }
            }
        }
    }
    return ans
}

function shortestSubstrings(arr: string[]): string[] {
    const n: number = arr.length;
    const ans: string[] = Array(n).fill('');
    for (let i = 0; i < n; ++i) {
        const m: number = arr[i].length;
        for (let j = 1; j <= m && ans[i] === ''; ++j) {
            for (let l = 0; l <= m - j; ++l) {
                const sub: string = arr[i].slice(l, l + j);
                if (ans[i] === '' || sub.localeCompare(ans[i]) < 0) {
                    let ok: boolean = true;
                    for (let k = 0; k < n && ok; ++k) {
                        if (k !== i && arr[k].includes(sub)) {
                            ok = false;
                        }
                    }
                    if (ok) {
                        ans[i] = sub;
                    }
                }
            }
        }
    }
    return ans;
}

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