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

NJOKU SAMSON EBERE on March 23, 2020

Last year, I had a technical interview and one of the questions was on Anagrams. I solved the problem in 3 ways today and I want to share it with y...
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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);
}
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What do you think about it?

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NJOKU SAMSON EBERE •

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

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

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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.

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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"]));

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Bruno Oliveira •

Nice! :)

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NJOKU SAMSON EBERE •

Bruno, Thank you for taking your time to read through!