Count and return the number of characters of str1 that makes appearances in str2 using JavaScript

We need to write a JavaScript function that takes two strings as parameters and counts how many characters from the first string appear in the second string, including duplicate occurrences.

Problem Statement

Given two strings str1 and str2, count how many characters from str1 also appear in str2. If a character appears multiple times in str2, count each occurrence separately.

Example:

• str1 = 'Kk' contains characters 'K' and 'k'
• str2 = 'klKKkKsl' contains: k(1), l(1), K(2), K(3), k(4), K(5), s(6), l(7)
• Characters from str1 found in str2: k, K, K, k, K = 5 matches

Solution

const str1 = 'Kk';
const str2 = 'klKKkKsl';

var countAppearances = (str1 = '', str2 = '') => {
    const map = {};
    
    // Create a map of characters in str1
    for(let c of str1) {
        map[c] = true;
    }
    
    let count = 0;
    
    // Check each character in str2
    for(let c of str2) {
        if(map[c]) {
            count += 1;
        }
    }
    
    return count;
};

console.log(countAppearances(str1, str2));

5

How It Works

The algorithm works in two steps:

1. Create a character map: Store all unique characters from str1 in an object for quick lookup
2. Count matches: Iterate through str2 and increment the counter whenever we find a character that exists in our map

Alternative Approach Using includes()

const countAppearancesSimple = (str1, str2) => {
    let count = 0;
    
    for(let char of str2) {
        if(str1.includes(char)) {
            count++;
        }
    }
    
    return count;
};

console.log(countAppearancesSimple('Kk', 'klKKkKsl'));

5

Comparison

Where n = length of str2, m = length of str1, k = unique characters in str1

Conclusion

The object map approach is more efficient for larger strings as it provides O(1) character lookup. For smaller strings, the simpler includes() method offers cleaner code with acceptable performance.