Link

Video

https://youtu.be/7KHKFqLPMbs

Problem and Constraints

Given a set of strings. Return anagrams in group.

All Approaches and Explanations in English

O(n) time complexity. O(n) space complexity

O(N*26) time complexity.

compare the count of letters in each word and put group them in a map

Code, if any

Works but okay, okay

class Solution {
    
    private String getLetterCount(String str){
        int[] count = new int[26];
        
        for(char c: str.toCharArray()){
            count[c-'a']++;
        }
        return Arrays.toString(count);
    }
    
    public List<List<String>> groupAnagrams(String[] strs) {
        Map<String, List<String>> groups = new HashMap<>();
        for(String str: strs){
            String letterCount = getLetterCount(str);
            List<String> list = groups.getOrDefault(letterCount, new ArrayList<>());
            list.add(str);
            groups.put(letterCount, list);
        }
        
        return new ArrayList<>(groups.values());   
    }
}

Optimizations: rather than: Arrays.toString(count); could have been: new String(count)

rather than:

            List<String> list = groups.getOrDefault(letterCount, new ArrayList<>());
            list.add(str);
            groups.put(letterCount, list);

could have been:

letterCount could be renamed as hash. #DDD

groups.computeIfAbsent(letterCount, k-> new ArrayList<>())
groups.get(letterCount).add(str);