Given an array of strings, 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, using all the original letters exactly once.
Example:
Input: strings = ["eat","tea","tan","ate","nat","bat"] Output: [["bat"],["nat","tan"],["ate","eat","tea"]]
Your algorithm should run in O(n * L * log L) time and use O(n * L) extra space.
The core challenge of this problem is to identify and group words that are anagrams of each other. An anagram is a rearrangement of the letters of a word to form another word. For example, "eat" and "tea" are anagrams because they contain the same letters.
This problem has practical applications in text analysis, cryptography, and data clustering.
Potential pitfalls include not correctly identifying anagrams or inefficiently grouping them, leading to suboptimal performance.
To solve this problem, we need to group words that are anagrams. A naive solution might involve comparing each word with every other word, but this would be inefficient with a time complexity of O(n^2 * L). Instead, we can use a more optimized approach:
1. **Sort and Hash**: The idea is to sort each word and use the sorted word as a key in a hash map. All anagrams will have the same sorted key.
2. **Hash Map**: Use a hash map to group words by their sorted version.
Here is a step-by-step breakdown of the algorithm:
// Function to group anagrams
function groupAnagrams(strs) {
// Initialize an empty hash map
const anagramGroups = {};
// Iterate over each word in the input array
for (let str of strs) {
// Sort the characters of the word to form the key
const sortedStr = str.split('').sort().join('');
// If the key is not in the hash map, add it with an empty list as its value
if (!anagramGroups[sortedStr]) {
anagramGroups[sortedStr] = [];
}
// Append the original word to the list corresponding to the sorted key
anagramGroups[sortedStr].push(str);
}
// Return the values of the hash map
return Object.values(anagramGroups);
}
// Example usage
const strings = ["eat", "tea", "tan", "ate", "nat", "bat"];
console.log(groupAnagrams(strings)); // Output: [["bat"],["nat","tan"],["ate","eat","tea"]]
The time complexity of this approach is O(n * L * log L), where n is the number of words and L is the maximum length of a word. This is because sorting each word takes O(L * log L) time, and we do this for each of the n words.
The space complexity is O(n * L) because we store each word in the hash map.
Potential edge cases include:
Examples:
Input: [] Output: [] Input: ["abc"] Output: [["abc"]] Input: ["abc", "bca", "cab"] Output: [["abc", "bca", "cab"]]
To test the solution comprehensively, consider the following test cases:
Use a testing framework like Jest or Mocha for automated testing.
When approaching such problems:
Practice solving similar problems to improve your problem-solving skills.
Grouping anagrams is a common problem that can be efficiently solved using sorting and hashing. Understanding the problem, considering different approaches, and analyzing their complexities are crucial steps in developing an optimal solution.
Keep practicing and exploring different problems to enhance your algorithmic thinking and coding skills.
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