Mastering Anagram Grouping in JavaScript
Efficiently Grouping Anagrams in JavaScript
Anagrams are words or phrases formed by rearranging the letters of another word or phrase. This seemingly simple concept presents a fun algorithmic challenge: given an array of strings, how can we efficiently group together all the anagrams? This post explores two approaches to solve this problem in JavaScript, comparing their time complexities and providing detailed explanations.
Method 1: Sorting for Anagram Identification
Our first approach leverages the fundamental property of anagrams: they have the same letters, just in a different order. If we sort the letters of each word, anagrams will have identical sorted forms. This insight allows us to use a Map to group anagrams efficiently.
/**
* @param {string[]} strs
* @return {string[][]}
*/
var groupAnagrams = function(strs) {
let res = new Map();
for (let s of strs) {
const key = s.split('').sort().join('');
if (!res.has(key)) {
res.set(key, []);
}
res.get(key).push(s);
}
return Array.from(res.values());
};
This method has a time complexity of O(n * m log m), where n is the number of strings and m is the maximum length of a string. The dominant factor is the sorting step (m log m) for each string.
Method 2: Character Counting for Optimized Anagram Detection
While sorting is straightforward, we can achieve better performance using character counting. Instead of sorting, we create a frequency array representing the count of each character (a-z) in a string. Anagrams will share the same character frequency array. This leads to a more optimized solution.
/**
* @param {string[]} strs
* @return {string[][]}
*/
var groupAnagrams = function(strs) {
let res = new Map();
for (let s of strs) {
const count = new Array(26).fill(0);
for (c of s) {
count[c.charCodeAt(0) - "a".charCodeAt(0)] += 1;
}
const key = count.join(',');
if (!res.has(key)) {
res.set(key, []);
}
res.get(key).push(s);
}
return Array.from(res.values());
};
This character counting approach boasts a time complexity of O(n * m), significantly faster than the sorting method for larger input sets, as it avoids the logarithmic overhead of sorting.
Understanding Key JavaScript Concepts
charCodeAt(index)
This method is crucial in the character counting method. It returns the Unicode value of the character at a given index. For example, "a".charCodeAt(0) returns 97, the Unicode value of 'a'. This allows us to efficiently index into our frequency array.
res.get(key).push(s);
This line retrieves the array associated with the key in the Map and efficiently pushes the current string s onto it. Because arrays in JavaScript are reference types, the Map is updated in place; there's no need for a separate set operation.
Array.from(res.values())
The res.values() method returns an iterator of the values (arrays of anagrams) in the Map. Array.from() converts this iterator into a standard array, fulfilling the function's requirement of returning a string[][].
Conclusion
We've explored two distinct methods for grouping anagrams in JavaScript. The character counting method offers superior performance due to its linear time complexity, making it the preferred choice for larger datasets. Understanding the underlying principles of character encoding and JavaScript's data structures is key to crafting efficient and elegant solutions to this classic algorithmic problem. Try implementing these methods and experimenting with various input sizes to observe the performance differences firsthand!
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