Solving Leet code 49. Group Anagrams Day150

Solving Leet code 49. Group Anagrams Day150

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2 min read

Today #Day150 of #365DaysOfCode, I am Solving Leet code 49. Group Anagrams.

Question

Given an array of strings strs, group the anagrams together. You can return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

Input: strs = ["eat","tea","tan","ate","nat","bat"]
Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

Example 2:

Input: strs = [""]
Output: [[""]]

Example 3:

Input: strs = ["a"]
Output: [["a"]]

Constraints:

  • 1 <= strs.length <= 10<sup>4</sup>

  • 0 <= strs[i].length <= 100

  • strs[i] consists of lowercase English letters.

Answer Leetcode:

var groupAnagrams = (words, map = new Map()) => {
    if (!words.length) return [];

    groupWords(words, map);    /* Time O(N * (K * log(K)) | Space O(N * K) */

    return [ ...map.values() ];/* Time O(N)               | Space O(N * K) */
};

var groupWords = (words, map) => {
    for (const original of words) {/* Time O(N) */
        const sorted = reorder(original);/* Time O(K * log(K)) | Space O(K) */
        const values = map.get(sorted) || [];

        values.push(original);           /*                    | Space O(N) */
        map.set(sorted, values);         /*                    | Space O(N * K) */
    }
}

const reorder = (str) => str
    .split('')                         /* Time O(K)          | Space O(K) */
    .sort((a, b) => a.localeCompare(b))/* Time O(K * log(K)) | Space O(1 || log(K)) */
    .join('');                         /* Time O(K)          | Space O(K) */

The groupAnagrams function takes in an array of words and creates a new Map object to store the grouped anagrams. If the input array is empty, it returns an empty array.

The groupWords function iterates over each word in the input array and calls the reorder function on it to get a sorted version of the word. It then checks if this sorted version already exists as a key in the Map object. If it does, it adds the original word to its value (which is an array). If it doesn’t exist yet, it creates a new key-value pair with the sorted version as the key and an array containing only the original word as its value.

Finally, groupAnagrams returns all values from map which are arrays containing grouped anagrams.

The reorder function takes in a string and returns a new string with its characters sorted alphabetically.

Conclusion

I solved the Leet code 49. Group Anagrams.

Source: 49. Group Anagrams[Link]

Author: Dheeraj.y

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