LeetCode #1935 — EASY

Maximum Number of Words You Can Type

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

The Problem

Problem Statement

There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.

Given a string text of words separated by a single space (no leading or trailing spaces) and a string brokenLetters of all distinct letter keys that are broken, return the number of words in text you can fully type using this keyboard.

Example 1:

Input: text = "hello world", brokenLetters = "ad"
Output: 1
Explanation: We cannot type "world" because the 'd' key is broken.

Example 2:

Input: text = "leet code", brokenLetters = "lt"
Output: 1
Explanation: We cannot type "leet" because the 'l' and 't' keys are broken.

Example 3:

Input: text = "leet code", brokenLetters = "e"
Output: 0
Explanation: We cannot type either word because the 'e' key is broken.

Constraints:

1 <= text.length <= 10⁴
0 <= brokenLetters.length <= 26
text consists of words separated by a single space without any leading or trailing spaces.
Each word only consists of lowercase English letters.
brokenLetters consists of distinct lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly. Given a string text of words separated by a single space (no leading or trailing spaces) and a string brokenLetters of all distinct letter keys that are broken, return the number of words in text you can fully type using this keyboard.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"hello world"
"ad"

Example 2

"leet code"
"lt"

Example 3

"leet code"
"e"

Step 02

Core Insight

What unlocks the optimal approach

Check each word separately if it can be typed.
A word can be typed if all its letters are not broken.

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 #1935: Maximum Number of Words You Can Type
class Solution {
    public int canBeTypedWords(String text, String brokenLetters) {
        boolean[] s = new boolean[26];
        for (char c : brokenLetters.toCharArray()) {
            s[c - 'a'] = true;
        }
        int ans = 0;
        for (String w : text.split(" ")) {
            for (char c : w.toCharArray()) {
                if (s[c - 'a']) {
                    --ans;
                    break;
                }
            }
            ++ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1935: Maximum Number of Words You Can Type
func canBeTypedWords(text string, brokenLetters string) (ans int) {
	s := [26]bool{}
	for _, c := range brokenLetters {
		s[c-'a'] = true
	}
	for _, w := range strings.Split(text, " ") {
		for _, c := range w {
			if s[c-'a'] {
				ans--
				break
			}
		}
		ans++
	}
	return
}

# Accepted solution for LeetCode #1935: Maximum Number of Words You Can Type
class Solution:
    def canBeTypedWords(self, text: str, brokenLetters: str) -> int:
        s = set(brokenLetters)
        return sum(all(c not in s for c in w) for w in text.split())

// Accepted solution for LeetCode #1935: Maximum Number of Words You Can Type
impl Solution {
    pub fn can_be_typed_words(text: String, broken_letters: String) -> i32 {
        let mut s = vec![false; 26];
        for c in broken_letters.chars() {
            s[(c as usize) - ('a' as usize)] = true;
        }
        let mut ans = 0;
        let words = text.split_whitespace();
        for w in words {
            for c in w.chars() {
                if s[(c as usize) - ('a' as usize)] {
                    ans -= 1;
                    break;
                }
            }
            ans += 1;
        }
        ans
    }
}

// Accepted solution for LeetCode #1935: Maximum Number of Words You Can Type
function canBeTypedWords(text: string, brokenLetters: string): number {
    const s: boolean[] = Array(26).fill(false);
    for (const c of brokenLetters) {
        s[c.charCodeAt(0) - 'a'.charCodeAt(0)] = true;
    }
    let ans = 0;
    for (const w of text.split(' ')) {
        for (const c of w) {
            if (s[c.charCodeAt(0) - 'a'.charCodeAt(0)]) {
                --ans;
                break;
            }
        }
        ++ans;
    }
    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(|\Sigma|)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.