Kenny

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# Character Frequency

Hi everyone Today, we're going to go over the Character Frequency Toy Problem!
This toy problem's input will be string and output an array of arrays that is sorted in descending order by frequency and then by ascending order by character for example:

``````characterFrequency('mississippi');
// =>
// [
//   ['i', 4],
//   ['s', 4],
//   ['p', 2],
//   ['m', 1],
// ]
``````

To begin we know our

• Input: String

• Output: An array of arrays Where the subarrays' first element is the character and second is a number of how many times it had appeared in the provided string.
• So lets break down the problem a little bit We know our output should be an array of arrays, so lets create an array variable, followed by an empty object, that will be our letter counter. When we're iterating through the string to see how many character out string has and next we're going to add them into our object as we're adding them into our string we're going to give each character a number value of how many occurrences the character appears

Our current output should look like this so far:

``````{ m: 1,
i: 5,
s: 2,
p: 4 }
``````
``````const characterFrequency = (string) => {
let result = [];
// create an object to keep store the value as key value pairs
let letterCount = {};
// loop through string
for (let i = 0; i < string.length; i++) {
// if the letter only appears once keep its count at one
//else increment the count
!letterCount[string[i]] ? letterCount[string[i]] = 1 : letterCount[string[i]]++;
}
``````

Next we're going to initialize a new variable grab the keys of our letter count object using Object.Keys now that now new variable is an array we could sort it accordingly

Afterwards we'll iterate one last time for our inner array
within our loop we'll create another array variable representing our inner array we're going to add the character in the first index of our inner array and the number in the second index of inner array and push the last part of our inner array into our final result array.

`````` for (let j = 0; j < arrayMaker.length; j++) {
let innerArray = [];
innerArray[0] = arrayMaker[j];
innerArray[1] = letterCount[arrayMaker[j]];
result.push(innerArray);
}
``````

Our final solution should look at little something like this:

``````const characterFrequency = (string) => {
let result = [];
// create an object to keep store the value as key value pairs
let letterCount = {};
// loop through string
for (let i = 0; i < string.length; i++) {
// if the letter only appears once keep its count at one
if (!letterCount[string[i]]) {
letterCount[string[i]] = 1;
} else {
//else increment the count
letterCount[string[i]]++;
}
}
// turn the object into an array
let arrayMaker = Object.keys(letterCount);
// console.log(arrayMaker);
// sort the array
arrayMaker.sort(function(a, b) {
if (letterCount[b] === letterCount[a]) {
if (a > b) {
return 1;
} else if (a < b) {
return -1;
} else {
return 0;
}
}
return letterCount[b] - letterCount[a];
});
// loop through the sorted array
for (let j = 0; j < arrayMaker.length; j++) {
let innerArray = [];
innerArray[0] = arrayMaker[j];
innerArray[1] = letterCount[arrayMaker[j]];
result.push(innerArray);
}
// console.log(result);
return result;

};
``````

Thank you for taking the time looking over this blog post! This problem was fun to solve out. I hope it helps those who had troubles with this problem!
Til next time Thank you!

``````const freqMap = str => [...str].reduce( (acc, val) => {acc[val] = acc[val] + 1 || 1; return acc}, new Map());