Count Strings With Required Characters

Count Strings With Required Characters - Problem

You are given a string required containing distinct lowercase English letters and an array of strings words.

A word is considered matching if it contains at least one character that also appears in required.

Return the count of matching words in the array words.

Input & Output

Example 1 — Some Match

$ Input: required = "ae", words = ["apple","box","cat","dog","egg"]

› Output: 3

💡 Note: "apple" contains 'a' and 'e', "cat" contains 'a', "egg" contains 'e'. "box" and "dog" contain neither 'a' nor 'e'.

Example 2 — All Match

$ Input: required = "xyz", words = ["fox","yell","buzz","ax","lazy"]

› Output: 5

💡 Note: Every word contains at least one of 'x', 'y', or 'z': fox(x), yell(y), buzz(z), ax(x), lazy(y,z).

Example 3 — None Match

$ Input: required = "q", words = ["hello","world","abc","def"]

› Output: 0

💡 Note: No word contains the character 'q'.

Constraints

1 ≤ words.length ≤ 10⁴
1 ≤ required.length ≤ 26
1 ≤ words[i].length ≤ 10
The characters in required are distinct
words[i] and required contain only lowercase English letters

Visualization

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Asked in

a Amazon 15 G Google 10 M Microsoft 8

The key insight is to use a hash set for O(1) character lookups. Build a set from required, then for each word check if any of its characters is in the set. Break early on first match. Time: O(n×m), Space: O(r) where r ≤ 26.

Common Approaches

✓ Hash Set Lookup

⏱️ Time: O(n × m) Space: O(r)

Build a hash set (or bitmask) from the required characters for O(1) lookup. Then for each word, check if any character exists in the set. Break early on the first match.

Nested Loop Check

⏱️ Time: O(n × m × r) Space: O(1)

For each word, iterate through its characters and for each character, scan the required string to check for a match. Count the word as matching if any character is found in required.

Hash Set Lookup — Algorithm Steps

Step 1: Build a set from the required characters
Step 2: For each word, iterate its characters
Step 3: If any character is in the set, count the word as matching and break
Step 4: Return the total count

Visualization

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Step-by-Step Walkthrough

Build Set

Build set from "ae" → {'a', 'e'}

O(1) Lookup Per Char

"apple": 'a' in set? O(1) YES → count++. "box": 'b'? NO, 'o'? NO, 'x'? NO → skip.

Result

3 matching words found

Code -

solution.c — C

#include <stdio.h>

int main() {
    char required[30];
    int reqMask = 0;
    int count = 0;
    int ch, i, inWord, found;
    
    i = 0;
    while ((ch = getchar()) != EOF && ch != '\n' && ch != '\r' && ch != ' ') {
        required[i++] = ch;
    }
    required[i] = '\0';
    
    for (i = 0; required[i]; i++) {
        reqMask |= (1 << (required[i] - 'a'));
    }
    
    inWord = 0;
    found = 0;
    while ((ch = getchar()) != EOF) {
        if (ch == '"') {
            if (inWord) {
                if (found) count++;
                inWord = 0;
                found = 0;
            } else {
                inWord = 1;
                found = 0;
            }
        } else if (inWord && ch >= 'a' && ch <= 'z') {
            if (reqMask & (1 << (ch - 'a'))) {
                found = 1;
            }
        }
    }
    
    printf("%d\n", count);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

For each password (n), check each character (m) with O(1) set lookup; early break on first match

✓ Linear Growth

Space Complexity

O(r)

Hash set or bitmask storing the required characters, where r ≤ 26

✓ Linear Space

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Medium Frequency

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Input & Output

Constraints

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Related Problems

Common Approaches

Hash Set Lookup — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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