LeetCode #748 — EASY

Shortest Completing Word

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given a string licensePlate and an array of strings words, find the shortest completing word in words.

A completing word is a word that contains all the letters in licensePlate. Ignore numbers and spaces in licensePlate, and treat letters as case insensitive. If a letter appears more than once in licensePlate, then it must appear in the word the same number of times or more.

For example, if licensePlate = "aBc 12c", then it contains letters 'a', 'b' (ignoring case), and 'c' twice. Possible completing words are "abccdef", "caaacab", and "cbca".

Return the shortest completing word in words. It is guaranteed an answer exists. If there are multiple shortest completing words, return the first one that occurs in words.

Example 1:

Input: licensePlate = "1s3 PSt", words = ["step","steps","stripe","stepple"]
Output: "steps"
Explanation: licensePlate contains letters 's', 'p', 's' (ignoring case), and 't'.
"step" contains 't' and 'p', but only contains 1 's'.
"steps" contains 't', 'p', and both 's' characters.
"stripe" is missing an 's'.
"stepple" is missing an 's'.
Since "steps" is the only word containing all the letters, that is the answer.

Example 2:

Input: licensePlate = "1s3 456", words = ["looks","pest","stew","show"]
Output: "pest"
Explanation: licensePlate only contains the letter 's'. All the words contain 's', but among these "pest", "stew", and "show" are shortest. The answer is "pest" because it is the word that appears earliest of the 3.

Constraints:

1 <= licensePlate.length <= 7
licensePlate contains digits, letters (uppercase or lowercase), or space ' '.
1 <= words.length <= 1000
1 <= words[i].length <= 15
words[i] consists of lower case English letters.

Step 01

Brute Force Baseline

Problem summary: Given a string licensePlate and an array of strings words, find the shortest completing word in words. A completing word is a word that contains all the letters in licensePlate. Ignore numbers and spaces in licensePlate, and treat letters as case insensitive. If a letter appears more than once in licensePlate, then it must appear in the word the same number of times or more. For example, if licensePlate = "aBc 12c", then it contains letters 'a', 'b' (ignoring case), and 'c' twice. Possible completing words are "abccdef", "caaacab", and "cbca". Return the shortest completing word in words. It is guaranteed an answer exists. If there are multiple shortest completing words, return the first one that occurs in words.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

"1s3 PSt"
["step","steps","stripe","stepple"]

Example 2

"1s3 456"
["looks","pest","stew","show"]

Step 02

Core Insight

What unlocks the optimal approach

Count only the letters (possibly converted to lowercase) of each word. If a word is shorter and the count of each letter is at least the count of that letter in the licensePlate, it is the best answer we've seen yet.

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 #748: Shortest Completing Word
class Solution {
    public String shortestCompletingWord(String licensePlate, String[] words) {
        int[] cnt = new int[26];
        for (int i = 0; i < licensePlate.length(); ++i) {
            char c = licensePlate.charAt(i);
            if (Character.isLetter(c)) {
                cnt[Character.toLowerCase(c) - 'a']++;
            }
        }
        String ans = "";
        for (String w : words) {
            if (!ans.isEmpty() && w.length() >= ans.length()) {
                continue;
            }
            int[] t = new int[26];
            for (int i = 0; i < w.length(); ++i) {
                t[w.charAt(i) - 'a']++;
            }
            boolean ok = true;
            for (int i = 0; i < 26; ++i) {
                if (t[i] < cnt[i]) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                ans = w;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #748: Shortest Completing Word
func shortestCompletingWord(licensePlate string, words []string) (ans string) {
	cnt := [26]int{}
	for _, c := range licensePlate {
		if unicode.IsLetter(c) {
			cnt[unicode.ToLower(c)-'a']++
		}
	}
	for _, w := range words {
		if len(ans) > 0 && len(ans) <= len(w) {
			continue
		}
		t := [26]int{}
		for _, c := range w {
			t[c-'a']++
		}
		ok := true
		for i, v := range cnt {
			if t[i] < v {
				ok = false
				break
			}
		}
		if ok {
			ans = w
		}
	}
	return
}

# Accepted solution for LeetCode #748: Shortest Completing Word
class Solution:
    def shortestCompletingWord(self, licensePlate: str, words: List[str]) -> str:
        cnt = Counter(c.lower() for c in licensePlate if c.isalpha())
        ans = None
        for w in words:
            if ans and len(w) >= len(ans):
                continue
            t = Counter(w)
            if all(v <= t[c] for c, v in cnt.items()):
                ans = w
        return ans

// Accepted solution for LeetCode #748: Shortest Completing Word
impl Solution {
    pub fn shortest_completing_word(license_plate: String, words: Vec<String>) -> String {
        let mut cnt = vec![0; 26];
        for c in license_plate.chars() {
            if c.is_ascii_alphabetic() {
                cnt[((c.to_ascii_lowercase() as u8) - b'a') as usize] += 1;
            }
        }
        let mut ans = String::new();
        for w in words {
            if !ans.is_empty() && w.len() >= ans.len() {
                continue;
            }
            let mut t = vec![0; 26];
            for c in w.chars() {
                t[((c as u8) - b'a') as usize] += 1;
            }
            let mut ok = true;
            for i in 0..26 {
                if t[i] < cnt[i] {
                    ok = false;
                    break;
                }
            }
            if ok {
                ans = w;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #748: Shortest Completing Word
function shortestCompletingWord(licensePlate: string, words: string[]): string {
    const cnt: number[] = Array(26).fill(0);
    for (const c of licensePlate) {
        const i = c.toLowerCase().charCodeAt(0) - 97;
        if (0 <= i && i < 26) {
            ++cnt[i];
        }
    }
    let ans = '';
    for (const w of words) {
        if (ans.length && ans.length <= w.length) {
            continue;
        }
        const t = Array(26).fill(0);
        for (const c of w) {
            ++t[c.charCodeAt(0) - 97];
        }
        let ok = true;
        for (let i = 0; i < 26; ++i) {
            if (t[i] < cnt[i]) {
                ok = false;
                break;
            }
        }
        if (ok) {
            ans = w;
        }
    }
    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 × |\Sigma|)

Space

O(|\Sigma|)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.