Character Array Cohesion — Problem Statement & Solution Guide
Problem Description
Given an array of lowercase alphabetic strings, define two strings as 'cohesive' if they are anagrams of each other. Your task is to group all strings in the input array into distinct clusters such that every string within a cluster is an anagram of every other string in that same cluster. Strings that do not share an anagram relationship with any other string in the input must each form their own singleton group. Return the collection of these groups. The order of the groups in the output array does not matter, nor does the order of strings within each group.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Character Array Cohesion"
WHY DOES IT MATTER?
Anagram grouping exemplifies the "canonical representation" pattern, which is essential for reducing complex equivalence relations to simple hashable keys. Mastering this pattern enables engineers to solve a wide range of clustering, deduplication, and similarity‑search problems efficiently.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that character frequency vectors provide a constant‑size, order‑independent signature, turning a potentially quadratic comparison problem into linear time by leveraging hash tables.
REAL-WORLD CONNECTION
In distributed caching systems, objects with identical content are often stored under a hash of their canonical form to avoid duplication—similar to how CDN edge nodes deduplicate files using content hashes. Grouping anagrams mirrors this deduplication by collapsing many permutations into a single identifier.
During an interview, compute the key first (sorted string or frequency array) and immediately insert into a map; avoid building auxiliary structures like nested loops. Keep the implementation concise: one pass to generate keys, one pass to collect groups.
COMPLEXITY AT A GLANCE
O(N·K)O(N·K)Core Theory — Why This Approach?
Grouping strings by anagram relationship is a classic example of using a canonical representation to achieve O(N·K) time, where N is the number of strings and K is the average length of each string. The naive method of pairwise comparison incurs O(N^2·K) time because each string would need to be compared against every other string, which quickly becomes infeasible for large datasets. By sorting the characters of each string or by counting character frequencies, we can transform each word into a deterministic key that is identical for all its anagrams. This key can then be used as a hash map entry, allowing constant‑time insertion and lookup, which collapses the problem to linear time relative to the input size.
The optimal paradigm leverages the concept of "bucket sort" in the abstract sense: each bucket corresponds to a unique anagram signature. When the signature is generated via a 26‑element frequency array (for lowercase English letters), we avoid the O(K log K) cost of sorting each string, achieving O(K) per string. The overall algorithm thus runs in O(N·K) time and O(N·K) space for storing the groups, which is optimal because any solution must at least read the entire input. This approach also scales well with modern hardware caches because the frequency array is a fixed small size, leading to high constant‑factor performance.
Understanding why the canonical key works is crucial: two strings are anagrams if and only if they have identical character counts. This bijective mapping between the multiset of characters and the frequency vector guarantees that no two non‑anagram strings share the same key, eliminating false positives. Consequently, the hash map groups strings perfectly without additional verification steps, making the solution both elegant and efficient for production‑grade workloads.
Interview Questions on This Problem
Q1How would you modify the solution if the input strings could contain uppercase letters and digits in addition to lowercase alphabets?
Extend the frequency vector to cover all possible characters (e.g., 62 slots for a‑z, A‑Z, 0‑9) or use a sorted string as the key. The rest of the algorithm remains unchanged; you still map each string to its canonical representation and group via a hash map.
Q2What are the trade‑offs between using a sorted‑string key versus a character‑count key for grouping anagrams?
Sorting each string costs O(K log K) time, while counting characters costs O(K) time with a fixed‑size array, making the count method faster for longer strings. However, the sorted‑string key is simpler to implement and works for Unicode characters without pre‑defining a fixed alphabet size, at the expense of higher time complexity.
Q3Can you design a streaming version of this algorithm that processes strings one by one and updates groups without storing the entire input array?
Yes. Maintain a hash map from canonical keys to lists of strings. For each incoming string, compute its key on the fly, insert it into the appropriate bucket, and optionally emit the bucket when a certain size or condition is met. This uses O(G·K) space where G is the number of distinct groups, not the total number of strings.
Examples
Input
words = ["abc", "bca", "cab", "xyz"]
Output
[["abc", "bca", "cab"], ["xyz"]]
Explanation: The strings "abc", "bca", and "cab" all contain the same character frequencies: {a:1, b:1, c:1}. Thus, they form one cohesive group. The string "xyz" has a unique character frequency profile {x:1, y:1, z:1} that does not match any other string, so it forms a singleton group.
Input
words = ["listen", "silent", "enlist", "hello"]
Output
[["listen", "silent", "enlist"], ["hello"]]
Explanation: "listen", "silent", and "enlist" are anagrams of each other, sharing the character counts {e:1, i:1, l:2, n:1, s:1, t:1}. They are grouped together. "hello" has a distinct character count {e:1, h:1, l:2, o:1} and forms its own group.
Input
words = ["a", "b", "a", "c"]
Output
[["a", "a"], ["b"], ["c"]]
Explanation: The two instances of "a" are anagrams of each other (identical strings), so they form a group. "b" and "c" are unique in their character composition and form separate singleton groups.
Constraints
- 1 <= words.length <= 10^4
- 1 <= words[i].length <= 100
- words[i] consists of lowercase English letters only.
Optimal Approach & Strategy
Generate a canonical key for each string (sorted characters or frequency vector) and store strings in a hash map keyed by that signature, achieving O(N·K) time.
Brute Force Approach
Compare every pair of strings to check if they are anagrams, which requires O(N^2·K) time and repeated sorting or counting for each comparison.
Verified Code Solutions
function groupAnagrams(strs) {
const anagrams = {};
for (const str of strs) {
const sortedStr = str.split('').sort().join('');
if (!anagrams[sortedStr]) anagrams[sortedStr] = [];
anagrams[sortedStr].push(str);
}
return Object.values(anagrams);
}class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
unordered_map<string, vector<string>> anagrams;
for (const string& str : strs) {
string sortedStr = str;
sort(sortedStr.begin(), sortedStr.end());
if (anagrams.find(sortedStr) == anagrams.end()) anagrams[sortedStr] = {};
anagrams[sortedStr].push_back(str);
}
vector<vector<string>> result;
for (const auto& pair : anagrams) result.push_back(pair.second);
return result;
}
}class Solution {
public List<List<String>> groupAnagrams(String[] strs) {
Map<String, List<String>> anagrams = new HashMap<>();
for (String str : strs) {
char[] chars = str.toCharArray();
Arrays.sort(chars);
String sortedStr = new String(chars);
if (!anagrams.containsKey(sortedStr)) anagrams.put(sortedStr, new ArrayList<>());
anagrams.get(sortedStr).add(str);
}
return new ArrayList<>(anagrams.values());
}
}def group_anagrams(strs):
anagrams = {}
for str in strs:
sorted_str = ''.join(sorted(str))
if sorted_str not in anagrams:
anagrams[sorted_str] = []
anagrams[sorted_str].append(str)
return list(anagrams.values())function groupAnagrams(strs) {
const anagrams = {};
for (const str of strs) {
const sortedStr = str.split('').sort().join('');
if (!anagrams[sortedStr]) anagrams[sortedStr] = [];
anagrams[sortedStr].push(str);
}
return Object.values(anagrams);
}Solve in Interative Editor
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