What’s your alternative solution? Challenge #57

About this series

This is series of daily JavaScript coding challenges... for both beginners and advanced users.

Each day I’m gone present you a very simple coding challenge, together with the solution. The solution is intentionally written in a didactic way using classic JavaScript syntax in order to be accessible to coders of all levels.

Solutions are designed with increase level of complexity.

Today’s coding challenge

Find the frequency of characters inside a string. Return the result as an array of objects. Each object has 2 fields: character and number of occurrences.

(scroll down for solution)

Code newbies

If you are a code newbie, try to work on the solution on your own. After you finish it, or if you need help, please consult the provided solution.

Advanced developers

Please provide alternative solutions in the comments below.

You can solve it using functional concepts or solve it using a different algorithm... or just solve it using the latest ES innovations.

By providing a new solution you can show code newbies different ways to solve the same problem.

Solution

// Solution for challenge51

var ar = getCharFrequency("Find the frequency of characters inside a string");
println(JSON.stringify(ar));

function getCharFrequency(text)
{
    var ar = [];

    for(var chr of text)
    {
        updateFrequency(ar, chr);
    }

    return ar;
}

function updateFrequency(ar, chr)
{
    for(var el of ar)
    {
        if (el.chr === chr)
        {
            el.count++;
        }
    }

    ar.push({ chr : chr, count : 1 });
}

To quickly verify this solution, copy the code above in this coding editor and press "Run".

Note: The solution was originally designed for codeguppy.com environment, and therefore is making use of println. This is the almost equivalent of console.log in other environments. Please feel free to use your preferred coding playground / environment when implementing your solution.

Comments

[Mike Talbot ⭐]

   const count = str=>Object.entries(str.split('').reduce((c,a)=>(c[a] = (c[a] || 0) + 1, c), {})).map(([chr,count])=>({chr,count}))