Find Second most frequent character in array - JavaScript

We are required to write a JavaScript function that takes in an array and returns the element that appears for the second most number of times.

Let's say the following is our array:

const arr = [1, 34, 4, 3, 2, 1, 4, 6, 4, 6, 5, 3, 6, 6];

The most frequent element is 6 (appears 4 times), but we want the second most frequent element, which is 4 (appears 3 times).

Example

const arr = [1, 34, 4, 3, 2, 1, 4, 6, 4, 6, 5, 3, 6, 6];

const secondMostFrequent = arr => {
    const map = arr.reduce((acc, val) => {
        if(acc.has(val)){
            acc.set(val, acc.get(val) + 1);
        } else {
            acc.set(val, 1);
        }
        return acc;
    }, new Map());
    
    const frequencyArray = Array.from(map);
    return frequencyArray.sort((a, b) => {
        return b[1] - a[1];
    })[1][0];
};

console.log(secondMostFrequent(arr));

Output

4

How It Works

The function works in three steps:

1. Count frequencies: Uses reduce() with a Map to count how many times each element appears
2. Convert to array: Transforms the Map into an array of [element, frequency] pairs
3. Sort and select: Sorts by frequency in descending order and returns the second element (index 1)

Alternative Approach Using Object

const arr = [1, 34, 4, 3, 2, 1, 4, 6, 4, 6, 5, 3, 6, 6];

const secondMostFrequentAlt = arr => {
    const frequency = {};
    
    // Count frequencies
    arr.forEach(element => {
        frequency[element] = (frequency[element] || 0) + 1;
    });
    
    // Sort by frequency
    const sorted = Object.entries(frequency).sort((a, b) => b[1] - a[1]);
    
    // Return second most frequent (convert back to number)
    return Number(sorted[1][0]);
};

console.log(secondMostFrequentAlt(arr));

Output

4

Key Points

• The function assumes there are at least two distinct elements with different frequencies
• If multiple elements tie for second place, it returns one of them
• Both Map and Object approaches work effectively for frequency counting
• The sorting step arranges elements by frequency in descending order

Conclusion

Finding the second most frequent element involves counting frequencies, sorting by count, and selecting the second-highest. Both Map and Object approaches provide clean solutions for this common programming problem.