Discussion on: Algorithm 202 (My Interview Question): Grouping Anagrams in 3 Ways

[Wojciech Rebis]

Great article, and good interview question.
Good work, keep it up! :)

Just wanted to share my approach to this:

function groupAnagrams(anagrams) {
  const wordGroups = anagrams.reduce((groups, anagram) => {
    const word = anagram.split('').sort().join('');
    if (groups[word]) {
      groups[word].push(anagram);
    } else {
      groups[word] = [anagram];
    }
    return groups;
  }, {});

  return Object.values(wordGroups).map(group => group);
} 

What do you think about it?

[NJOKU SAMSON EBERE]

Thank you Wojciech. I love your method. It's very clever and coincise. Works perfectly too.

dev-to-uploads.s3.amazonaws.com/i/...

Do you mind explaining what is going on in the reduce() function probably by commenting the code?

[Wojciech Rebis]

I thought of reducing array of anagrams to an object which keys will be initial words and values arrays of anagrams from these words.

[Ganesh Shetty]

Tried pretty much in the same way

function groupAnagrams(array) {
var reducedObject = array.reduce((acc, cur) => {
var newcur=cur.split('').sort().join('');
if (!acc[newcur]) {
acc[newcur] = [cur]
} else {
acc[newcur] = [...acc[newcur], cur]
}
return acc;
}, {});
return Object.values(reducedObject);
}
console.log(groupAnagrams(["eat", "tea", "tan", "ate", "nat", "bat"]));