LeetCode #3076 — MEDIUM

Shortest Uncommon Substring in an Array

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Trie

Example 1

["cab","ad","bad","c"]

Example 2

["abc","bcd","abcd"]

Step 02

Core Insight

What unlocks the optimal approach

Try a brute force solution where you check every substring.
Use a Hash map to keep track of the substrings.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3076: Shortest Uncommon Substring in an Array
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;
    }
}

// Accepted solution for LeetCode #3076: Shortest Uncommon Substring in an Array
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
}

# Accepted solution for LeetCode #3076: Shortest Uncommon Substring in an Array
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

// Accepted solution for LeetCode #3076: Shortest Uncommon Substring in an Array
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3076: Shortest Uncommon Substring in an Array
// 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;
//     }
// }

// Accepted solution for LeetCode #3076: Shortest Uncommon Substring in an Array
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2 × m^4)

Space

O(m)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.