如何在javascript中返回O（n）时间复杂度相同的字母元素？

Question

I am trying to solve a problem with lower time complexity which in this case O(n) .

Here is the problem:

let arr = ['dog', 'come', 'ogd', 'something', 'emoc'];

It should return

[['dog', 'ogd], ['come', 'emoc']]; // which same using letters

So far I solved this problem using two functions it is working great but mine is nested loop will give me O(n2)

Here is my code

const isSameChars = (str1, str2) => {
  let str1sorted = str1.split("").sort().join("");
  let str2sorted = str2.split("").sort().join("");
  if (str1.length !== str2.length) {
    return false;
  }
  for (var i = 0; i < str1sorted.length; i++) {
    let char1 = str1sorted[i];
    let char2 = str2sorted[i];
    if (char1 !== char2) {
      return false;
    }
  }
  return true;
}

const isSameCharElements = (arr) => {
  let result = [];
  for(var i = 0; i < arr.length; i++) {
    for(var j = 0; j < i; j++) {
      if (isSameChars(arr[i], arr[j]) === true) {
        result.push(arr[j])
      }
    }
  }
  return result; 
}

console.log(isSameCharElements(['dog', 'come', 'ogd', 'something', 
'emoc'])) //  [['dog', 'ogd], ['come', 'emoc']]

Is there any way to solve this with O(n) time complexity?

Thank you an advance!

Answer 1

You can have a 'bag of letters' representation of any String by sorting the letters:

function sortLetters(word){
  return word.split('').sort().join('');
}

You can then iterate over your input, grouping words that have the same bag of letters representation into an object:

const grouped = arr.reduce(function (m, word) {
  var bagRepr = sortLetters(word);
  var withSameLetters =  m[bagRepr] || [];
  withSameLetters.push(word);
  m[bagRepr] = withSameLetters;
  return m;
}, {});

const result = Object.values(grouped)
  .filter(function (arr) {
    return arr.length > 1;  
  });

This is O(n), provided that sortLetters() is O(1), which is the case if the words lengths are bounded by a constant.

Disclaimer: note that we're only talking about asymptotic complexity here - this does not mean at all that this approach is the most efficient from a practical standpoint!