查找字符串中的唯一字符及其出现次数

Question

I need to count the occurrence of characters in a given string and print out the unique characters and the number of how many times they appeared. So, for example, if I receive a string of 'HELLO' it should print out:

H: 1, E: 1, L: 2, O: 1

This is a much-simplified version of a problem, but the answer should put me in the right direction. How can I approach this problem?

Thank you in advance.

Answer 1

This is more or less what it should look like in order to make it easier it prints it in JSON you can already convert it to String yourself if you want.

function count_occurrence(text = "") {
    const array_from_text = text.split("");
    const result = {};
    Array.from(new Set(array_from_text)).forEach(word => {
        const { length } = array_from_text.filter(w => w === word);
        result[word] = length;
    });
    return result;
};
const occurences = count_occurence("HELLO");
console.log(occurences); // {H: 1, E: 1, L: 2, O: 1}

Answer 2

const countChars = (str) => {
  const charCount = {} ;
  for (const c of [...str]) {
    charCount[c] = (charCount[c] || 0) + 1 ;
  }
  return charCount ;
}
console.log(countChars('HELLO')) ; // {H: 1, E: 1, L: 2, O: 1}

Answer 3

You could use Array.reduce() to get a count of each occurrence in the input word.

We convert the word to an array using the ... operator, then .reduce() to create an object with a property for each unique letter in the word.

 const input = 'HELLO'; const result = [...input].reduce((acc, chr) => { acc[chr] = (acc[chr] || 0) + 1; return acc; }, {}); console.log('Result:', result)

 .as-console-wrapper { max-height: 100% !important; }

Answer 4

My approach to this problem is:

 let str = "HELLO"; // An object for the final result {character:count} let counts = {}; // Loop through the str... for (let index = 0; index < str.length; ++index) { // Get each char let ch = str.charAt(index); // Get the count for that char let count = counts[ch]; // If we have one, store that count plus one; if (count) { counts[ch] += 1; } else { // if not, store one counts[ch] = 1; } // or more simply with ternary operator // counts[ch] = count ? count + 1 : 1;. } console.log(counts);

Answer 5

Maybe the easiest answer is just split to char and put it into the map.

const count={}
"HELLO".split("").forEach(e=>{
count[e]??=0;
    count[e]++;
})

count is what you want.

Answer 6

Use a dictionary like datastructure that gives you O(1) access and update times. In JS you can use an Object literat (not recommended) or a Map .

Iterate over the characters of your string and update the dictionary by incrementing the character count of the current character. If it isn't in your dictionary add it and set the count to one.

When done with the iteration, iterate over the keys of your dictionary, where the values are the the number of occurence of that specific character, and output them in any format of your liking.

 const myStr = "Hello" const myMap = new Map() for (let c of myStr) { if (myMap.has(c)) { myMap.set(c, myMap.get(c)+1) } else { myMap.set(c, 1) } } for (let k of myMap.keys()) { console.log(`${k} occures ${myMap.get(k)} times`) }