LeetCode #2800 — MEDIUM

Shortest String That Contains Three Strings

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

The Problem

Problem Statement

Given three strings a, b, and c, your task is to find a string that has the minimum length and contains all three strings as substrings.

If there are multiple such strings, return the lexicographically smallest one.

Return a string denoting the answer to the problem.

Notes

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
A substring is a contiguous sequence of characters within a string.

Example 1:

Input: a = "abc", b = "bca", c = "aaa"
Output: "aaabca"
Explanation:  We show that "aaabca" contains all the given strings: a = ans[2...4], b = ans[3..5], c = ans[0..2]. It can be shown that the length of the resulting string would be at least 6 and "aaabca" is the lexicographically smallest one.

Example 2:

Input: a = "ab", b = "ba", c = "aba"
Output: "aba"
Explanation: We show that the string "aba" contains all the given strings: a = ans[0..1], b = ans[1..2], c = ans[0..2]. Since the length of c is 3, the length of the resulting string would be at least 3. It can be shown that "aba" is the lexicographically smallest one.

Constraints:

1 <= a.length, b.length, c.length <= 100
a, b, c consist only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given three strings a, b, and c, your task is to find a string that has the minimum length and contains all three strings as substrings. If there are multiple such strings, return the lexicographically smallest one. Return a string denoting the answer to the problem. Notes A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. A substring is a contiguous sequence of characters within a string.

Baseline thinking

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

Pattern signal: Greedy

Example 1

"abc"
"bca"
"aaa"

Example 2

"ab"
"ba"
"aba"

Core Insight

What unlocks the optimal approach

Think about how you can generate all possible strings that contain all three input strings as substrings. Can you come up with an efficient algorithm to do this?
Check all permutations of the words a, b, and c. For each permutation, begin by appending some letters to the end of the first word to form the second word. Then, proceed to add more letters to generate the third word.

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 #2800: Shortest String That Contains Three Strings
class Solution {
    public String minimumString(String a, String b, String c) {
        String[] s = {a, b, c};
        int[][] perm = {{0, 1, 2}, {0, 2, 1}, {1, 0, 2}, {1, 2, 0}, {2, 1, 0}, {2, 0, 1}};
        String ans = "";
        for (var p : perm) {
            int i = p[0], j = p[1], k = p[2];
            String t = f(f(s[i], s[j]), s[k]);
            if ("".equals(ans) || t.length() < ans.length()
                || (t.length() == ans.length() && t.compareTo(ans) < 0)) {
                ans = t;
            }
        }
        return ans;
    }

    private String f(String s, String t) {
        if (s.contains(t)) {
            return s;
        }
        if (t.contains(s)) {
            return t;
        }
        int m = s.length(), n = t.length();
        for (int i = Math.min(m, n); i > 0; --i) {
            if (s.substring(m - i).equals(t.substring(0, i))) {
                return s + t.substring(i);
            }
        }
        return s + t;
    }
}

// Accepted solution for LeetCode #2800: Shortest String That Contains Three Strings
func minimumString(a string, b string, c string) string {
	f := func(s, t string) string {
		if strings.Contains(s, t) {
			return s
		}
		if strings.Contains(t, s) {
			return t
		}
		m, n := len(s), len(t)
		for i := min(m, n); i > 0; i-- {
			if s[m-i:] == t[:i] {
				return s + t[i:]
			}
		}
		return s + t
	}
	s := [3]string{a, b, c}
	ans := ""
	for _, p := range [][]int{{0, 1, 2}, {0, 2, 1}, {1, 0, 2}, {1, 2, 0}, {2, 0, 1}, {2, 1, 0}} {
		i, j, k := p[0], p[1], p[2]
		t := f(f(s[i], s[j]), s[k])
		if ans == "" || len(t) < len(ans) || (len(t) == len(ans) && t < ans) {
			ans = t
		}
	}
	return ans
}

# Accepted solution for LeetCode #2800: Shortest String That Contains Three Strings
class Solution:
    def minimumString(self, a: str, b: str, c: str) -> str:
        def f(s: str, t: str) -> str:
            if s in t:
                return t
            if t in s:
                return s
            m, n = len(s), len(t)
            for i in range(min(m, n), 0, -1):
                if s[-i:] == t[:i]:
                    return s + t[i:]
            return s + t

        ans = ""
        for a, b, c in permutations((a, b, c)):
            s = f(f(a, b), c)
            if ans == "" or len(s) < len(ans) or (len(s) == len(ans) and s < ans):
                ans = s
        return ans

// Accepted solution for LeetCode #2800: Shortest String That Contains Three Strings
impl Solution {
    fn f(s1: String, s2: String) -> String {
        if s1.contains(&s2) {
            return s1;
        }
        if s2.contains(&s1) {
            return s2;
        }
        for i in 0..s1.len() {
            let s = &s1[i..];
            if s2.starts_with(s) {
                let n = s.len();
                return s1 + &s2[n..];
            }
        }
        s1 + s2.as_str()
    }

    pub fn minimum_string(a: String, b: String, c: String) -> String {
        let s = [&a, &b, &c];
        let perm = [
            [0, 1, 2],
            [0, 2, 1],
            [1, 0, 2],
            [1, 2, 0],
            [2, 0, 1],
            [2, 1, 0],
        ];
        let mut ans = String::new();
        for [i, j, k] in perm.iter() {
            let r = Self::f(Self::f(s[*i].clone(), s[*j].clone()), s[*k].clone());
            if ans == "" || r.len() < ans.len() || (r.len() == ans.len() && r < ans) {
                ans = r;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2800: Shortest String That Contains Three Strings
function minimumString(a: string, b: string, c: string): string {
    const f = (s: string, t: string): string => {
        if (s.includes(t)) {
            return s;
        }
        if (t.includes(s)) {
            return t;
        }
        const m = s.length;
        const n = t.length;
        for (let i = Math.min(m, n); i > 0; --i) {
            if (s.slice(-i) === t.slice(0, i)) {
                return s + t.slice(i);
            }
        }
        return s + t;
    };
    const s: string[] = [a, b, c];
    const perm: number[][] = [
        [0, 1, 2],
        [0, 2, 1],
        [1, 0, 2],
        [1, 2, 0],
        [2, 0, 1],
        [2, 1, 0],
    ];
    let ans = '';
    for (const [i, j, k] of perm) {
        const t = f(f(s[i], s[j]), s[k]);
        if (ans === '' || t.length < ans.length || (t.length === ans.length && t < ans)) {
            ans = t;
        }
    }
    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)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.