LeetCode #916 — MEDIUM

Word Subsets

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

Solve on LeetCode

The Problem

Problem Statement

You are given two string arrays words1 and words2.

A string b is a subset of string a if every letter in b occurs in a including multiplicity.

For example, "wrr" is a subset of "warrior" but is not a subset of "world".

A string a from words1 is universal if for every string b in words2, b is a subset of a.

Return an array of all the universal strings in words1. You may return the answer in any order.

Example 1:

Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"]

Output: ["facebook","google","leetcode"]

Example 2:

Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["lc","eo"]

Output: ["leetcode"]

Example 3:

Input: words1 = ["acaac","cccbb","aacbb","caacc","bcbbb"], words2 = ["c","cc","b"]

Output: ["cccbb"]

Constraints:

1 <= words1.length, words2.length <= 10⁴
1 <= words1[i].length, words2[i].length <= 10
words1[i] and words2[i] consist only of lowercase English letters.
All the strings of words1 are unique.

Step 01

Brute Force Baseline

Problem summary: You are given two string arrays words1 and words2. A string b is a subset of string a if every letter in b occurs in a including multiplicity. For example, "wrr" is a subset of "warrior" but is not a subset of "world". A string a from words1 is universal if for every string b in words2, b is a subset of a. Return an array of all the universal strings in words1. You may return the answer in any order.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

["amazon","apple","facebook","google","leetcode"]
["e","o"]

Example 2

["amazon","apple","facebook","google","leetcode"]
["lc","eo"]

Example 3

["acaac","cccbb","aacbb","caacc","bcbbb"]
["c","cc","b"]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #916: Word Subsets
class Solution {
    public List<String> wordSubsets(String[] words1, String[] words2) {
        int[] cnt = new int[26];
        for (var b : words2) {
            int[] t = new int[26];
            for (int i = 0; i < b.length(); ++i) {
                t[b.charAt(i) - 'a']++;
            }
            for (int i = 0; i < 26; ++i) {
                cnt[i] = Math.max(cnt[i], t[i]);
            }
        }
        List<String> ans = new ArrayList<>();
        for (var a : words1) {
            int[] t = new int[26];
            for (int i = 0; i < a.length(); ++i) {
                t[a.charAt(i) - 'a']++;
            }
            boolean ok = true;
            for (int i = 0; i < 26; ++i) {
                if (cnt[i] > t[i]) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                ans.add(a);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #916: Word Subsets
func wordSubsets(words1 []string, words2 []string) (ans []string) {
	cnt := [26]int{}
	for _, b := range words2 {
		t := [26]int{}
		for _, c := range b {
			t[c-'a']++
		}
		for i := range cnt {
			cnt[i] = max(cnt[i], t[i])
		}
	}
	for _, a := range words1 {
		t := [26]int{}
		for _, c := range a {
			t[c-'a']++
		}
		ok := true
		for i, v := range cnt {
			if v > t[i] {
				ok = false
				break
			}
		}
		if ok {
			ans = append(ans, a)
		}
	}
	return
}

# Accepted solution for LeetCode #916: Word Subsets
class Solution:
    def wordSubsets(self, words1: List[str], words2: List[str]) -> List[str]:
        cnt = Counter()
        for b in words2:
            t = Counter(b)
            for c, v in t.items():
                cnt[c] = max(cnt[c], v)
        ans = []
        for a in words1:
            t = Counter(a)
            if all(v <= t[c] for c, v in cnt.items()):
                ans.append(a)
        return ans

// Accepted solution for LeetCode #916: Word Subsets
struct Solution;

impl Solution {
    fn word_subsets(a: Vec<String>, b: Vec<String>) -> Vec<String> {
        let mut max_count = [0; 26];
        for s in b {
            let mut count = [0; 26];
            for b in s.bytes() {
                count[(b - b'a') as usize] += 1;
            }
            for i in 0..26 {
                max_count[i] = max_count[i].max(count[i]);
            }
        }

        let mut res = vec![];
        'a: for s in a {
            let mut count = [0; 26];
            for b in s.bytes() {
                count[(b - b'a') as usize] += 1;
            }
            for i in 0..26 {
                if count[i] < max_count[i] {
                    continue 'a;
                };
            }
            res.push(s)
        }

        res
    }
}

#[test]
fn test() {
    let a = vec_string!["amazon", "apple", "facebook", "google", "leetcode"];
    let b = vec_string!["e", "o"];
    let res = vec_string!["facebook", "google", "leetcode"];
    assert_eq!(Solution::word_subsets(a, b), res);
    let a = vec_string!["amazon", "apple", "facebook", "google", "leetcode"];
    let b = vec_string!["l", "e"];
    let res = vec_string!["apple", "google", "leetcode"];
    assert_eq!(Solution::word_subsets(a, b), res);
    let a = vec_string!["amazon", "apple", "facebook", "google", "leetcode"];
    let b = vec_string!["e", "oo"];
    let res = vec_string!["facebook", "google"];
    assert_eq!(Solution::word_subsets(a, b), res);
    let a = vec_string!["amazon", "apple", "facebook", "google", "leetcode"];
    let b = vec_string!["lo", "eo"];
    let res = vec_string!["google", "leetcode"];
    assert_eq!(Solution::word_subsets(a, b), res);
    let a = vec_string!["amazon", "apple", "facebook", "google", "leetcode"];
    let b = vec_string!["ec", "oc", "ceo"];
    let res = vec_string!["facebook", "leetcode"];
    assert_eq!(Solution::word_subsets(a, b), res);
}

// Accepted solution for LeetCode #916: Word Subsets
function wordSubsets(words1: string[], words2: string[]): string[] {
    const cnt: number[] = Array(26).fill(0);
    for (const b of words2) {
        const t: number[] = Array(26).fill(0);
        for (const c of b) {
            t[c.charCodeAt(0) - 97]++;
        }
        for (let i = 0; i < 26; i++) {
            cnt[i] = Math.max(cnt[i], t[i]);
        }
    }

    const ans: string[] = [];
    for (const a of words1) {
        const t: number[] = Array(26).fill(0);
        for (const c of a) {
            t[c.charCodeAt(0) - 97]++;
        }

        let ok = true;
        for (let i = 0; i < 26; i++) {
            if (cnt[i] > t[i]) {
                ok = false;
                break;
            }
        }

        if (ok) {
            ans.push(a);
        }
    }

    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)

Space

O(1)

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.