group-anagram

Link: https://leetcode.com/problems/group-anagrams/description/

• The concept here is make unique key from each string that represent the string even if the character are shuffle.
• store the key with string in map or object i.e. hashmap

function groupAnagrams(strs) {
  let ans = strs.reduce((acc, val) => {
    let count = Array(26).fill(0); // making array of 26 length fill with 0. a[0] -> a, a[1] -> b
    for (const char of val) {
      count[char.charCodeAt(0) - "a".charCodeAt(0)]++; // increment the count each time the character is found
    }
    const key = count.join("#"); // making a unique key by joining array

    if (!acc[key]) {
      acc[key] = [];
    }
    acc[key].push(val);
    return acc;
  }, {});
  return Object.values(ans);
}

Solution using map

var groupAnagrams = function (strs) {
  const hashMap = new Map();

  for (const str of strs) {
    let numericalRepresentationOfStrings = new Array(26).fill(0);

    for (const char of str) {
      let charValue = char.charCodeAt(0) - 97;

      numericalRepresentationOfStrings[charValue]++;
    }

    let key = numericalRepresentationOfStrings.join(",");

    let value = hashMap.get(key) || [];
    value.push(str);

    hashMap.set(key, value);
  }

  let result = [];
  for (const value of hashMap.values()) {
    result.push(value);
  }

  return result;
};