Group Anagrams

📋

Group Anagrams

Edge cases:

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IDE

def groupAnagrams(strs: list[str]) -> list[list[str]]:public List<List<String>> groupAnagrams(String[] strs)vector<vector<string>> groupAnagrams(vector<string>& strs)function groupAnagrams(strs)

def groupAnagrams(strs):
    # Write your solution here
    pass

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        // Write your solution here
        return new ArrayList<>();
    }
}

#include <vector>
#include <string>
using namespace std;

vector<vector<string>> groupAnagrams(vector<string>& strs) {
    // Write your solution here
    return {};
}

function groupAnagrams(strs) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: [['eat', 'tea'], ['tan', 'nat'], ['ate'], ['bat']]Grouping strings incorrectly by comparing only first characters or partial keys, causing 'ate' to be separated.✅ Use full sorted string as key for grouping all anagrams together.

Wrong: [['aabb', 'baba', 'abab'], ['bbaa'], ['aaab', 'abaa']]Failing to group all anagrams with repeated characters together due to partial key or set usage.✅ Use sorted strings as keys to capture exact character counts and group all anagrams correctly.

Wrong: [['abc', 'cab', 'bac', 'abcabc', 'bcaabc', 'aabbcc']]Merging groups with different character counts by ignoring multiplicity (0/1 vs unbounded confusion).✅ Use sorted strings or frequency count tuples as keys to distinguish groups by exact character counts.

Wrong: Timeout or no output on large inputUsing brute force pairwise comparison (O(n^2)) instead of hashing.✅ Use hash map with sorted string keys to achieve O(n*k log k) time complexity.

Wrong: [['a'], ['b'], ['c'], ['a'], ['b'], ['c']]Failing to group identical strings together; possibly using reference equality or not grouping at all.✅ Group strings by their sorted keys and append all matching strings to the same group.

✓

Test Cases

Try testing your solution with empty and minimal inputs to ensure base cases are handled.

Consider tricky inputs where strings have similar characters but different counts to catch grouping errors.

Focus on algorithmic complexity; avoid nested loops and use hashing to handle large inputs efficiently.

t1_01basic

Input["eat","tea","tan","ate","nat","bat"]

Expected[["eat","tea","ate"],["tan","nat"],["bat"]]

⏱ Performance - must finish in 2000ms

The words 'eat', 'tea', and 'ate' are anagrams of each other, as are 'tan' and 'nat'. 'bat' has no anagram in the list.

💡 Group words by a key that represents their letter composition.

💡 Use sorted strings as keys in a hash map to group anagrams.

💡 Sort each string and use it as a dictionary key to collect anagrams.

Why it failed: Failed to group anagrams correctly; likely did not use sorted string as key or mishandled grouping. Fix by sorting each string before using as key.

✓ Correctly groups anagrams using sorted string keys.

t1_02basic

Input["abc","bca","cab","xyz","zyx","zxy"]

Expected[["abc","bca","cab"],["xyz","zyx","zxy"]]

⏱ Performance - must finish in 2000ms

Two groups of anagrams: 'abc', 'bca', 'cab' and 'xyz', 'zyx', 'zxy'.

💡 Check if your grouping handles multiple groups correctly.

💡 Ensure each group key is consistent for all anagrams.

💡 Sort strings and group by sorted keys to form correct clusters.

Why it failed: Incorrect grouping; possibly used a single group or failed to distinguish keys. Fix by using sorted strings as keys for all input strings.

✓ Correctly groups multiple anagram sets.

t2_01edge

Input[]

Expected[]

⏱ Performance - must finish in 2000ms

Empty input array returns empty list.

💡 Consider what happens when input is empty.

💡 Check if your code handles zero-length input gracefully.

💡 Return an empty list immediately if input is empty.

Why it failed: Code crashes or returns non-empty output on empty input. Fix by adding a check for empty input and returning [].

✓ Handles empty input correctly by returning [].

t2_02edge

Input[""]

Expected[[""]]

⏱ Performance - must finish in 2000ms

Single empty string forms one group containing that string.

💡 Check how empty strings are handled as keys.

💡 Sorting an empty string results in an empty string key.

💡 Group all empty strings together using their sorted key ''.

Why it failed: Fails to group empty strings correctly; likely key generation fails on empty strings. Fix by allowing empty strings as valid keys.

✓ Correctly groups single empty string.

t2_03edge

Input["a","b","c"]

Expected[["a"],["b"],["c"]]

⏱ Performance - must finish in 2000ms

No anagrams present; each string forms its own group.

💡 Verify that distinct single-character strings form separate groups.

💡 Each string's sorted key is itself, so groups should be distinct.

💡 Ensure no merging of groups with different keys.

Why it failed: Incorrect grouping merges distinct strings; fix by using sorted string as key without merging different keys.

✓ Correctly forms separate groups for distinct strings.

t2_04edge

Input["","",""]

Expected[["","",""]]

⏱ Performance - must finish in 2000ms

All strings are empty; groups all empty strings together.

💡 Empty strings have the same sorted key ''.

💡 All empty strings should be grouped in one list.

💡 Use sorted string keys to group identical empty strings.

Why it failed: Fails to group multiple empty strings together; fix by using sorted string keys and appending all matching strings.

✓ Correctly groups multiple empty strings.

t3_01corner

Input["abc","bca","abcd","bcda","abdc","xyz"]

Expected[["abc","bca"],["abcd","bcda","abdc"],["xyz"]]

⏱ Performance - must finish in 2000ms

Groups with different lengths and overlapping characters; tests correct key grouping by sorted strings.

💡 Check that keys differentiate strings by length and character count.

💡 Sorting strings of different lengths produces different keys.

💡 Use sorted strings as keys to separate 'abc' and 'abcd' groups.

Why it failed: Grouping merges strings of different lengths incorrectly; fix by using full sorted string as key, not partial or length-only keys.

✓ Correctly separates groups by sorted string keys including length differences.

t3_02corner

Input["aabb","baba","abab","bbaa","aaab","abaa"]

Expected[["aabb","baba","abab","bbaa"],["aaab","abaa"]]

⏱ Performance - must finish in 2000ms

Tests grouping with repeated characters and similar but different counts.

💡 Sorting strings with repeated characters produces identical keys for anagrams.

💡 Ensure keys reflect character counts exactly, not just presence.

💡 Use sorted strings as keys to distinguish 'aabb' group from 'aaab' group.

Why it failed: Fails to distinguish groups with different character counts; fix by using sorted strings as keys instead of sets or frequency counts without order.

✓ Correctly groups anagrams with repeated characters.

t3_03corner

Input["abc","cab","bac","abcabc","bcaabc","aabbcc"]

Expected[["abc","cab","bac"],["abcabc","bcaabc","aabbcc"]]

⏱ Performance - must finish in 2000ms

Tests confusion between 0/1 and unbounded counts; 'abc' and 'abcabc' groups differ by character counts.

💡 Check that keys reflect exact character counts, not just presence.

💡 Avoid treating multiple occurrences as single occurrence.

💡 Use sorted strings or frequency tuples as keys to separate groups.

Why it failed: Merges groups with different character counts due to ignoring multiplicity; fix by using sorted strings or frequency counts as keys.

✓ Correctly separates groups with different character counts.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this"}

Expectednull

⏱ Performance - must finish in 2000ms

Input size n=100000 with strings length up to 100; solution must run in O(n*k log k) or better within 2 seconds.

💡 Sorting each string is O(k log k), do it once per string.

💡 Use hash map to group by sorted keys efficiently.

💡 Avoid nested loops or pairwise comparisons to prevent O(n^2) time.

Why it failed: TLE due to O(n^2) or inefficient grouping; fix by using hash map with sorted string keys instead of pairwise comparisons.

✓ Solution runs within time limit using efficient hashing and sorting.

Practice

1. You are given an array of integers and need to determine if any value appears at least twice. Which approach guarantees the most efficient detection of duplicates in terms of time complexity?

easy

A. Sort the array and then scan once to check if any adjacent elements are equal.

B. Use a hash set to track seen elements and check membership while iterating once.

C. Use nested loops to compare every pair of elements for equality.

D. Use dynamic programming to build a table of duplicate occurrences.

Solution

1. Step 1: Understand the problem requirements

   The problem requires detecting duplicates efficiently in an array.

2. Step 2: Compare approaches based on time complexity

   Sorting then scanning is O(n log n), nested loops is O(n²), and dynamic programming is unnecessary here. Using a hash set allows O(n) time with O(n) space by checking membership during one pass.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Hash set membership check is O(1) average, enabling linear time detection [OK]

Hint: Hash set membership check enables linear time detection [OK]

Common Mistakes:

• Confusing sorting approach as optimal
• Thinking nested loops are efficient enough
• Misapplying DP to a membership problem

2. You need to design a data structure that supports inserting elements, deleting elements, and retrieving a random element, all in average constant time. Which approach best guarantees this performance?

easy

A. Using a balanced binary search tree to maintain elements and their indices

B. Combining a hash map for element indices with an array to store elements, using a swap-delete trick on removal

C. Using a linked list for elements and a hash set for membership checks

D. Maintaining a sorted array and performing binary search for insertions and deletions

Solution

1. Step 1: Understand operation requirements

   Insert, delete, and getRandom must all be O(1) average time.

2. Step 2: Evaluate data structures

   Balanced BST or sorted array have O(log n) or worse for insert/delete. Linked list lacks O(1) random access. Hash map + array with swap-delete achieves O(1) for all operations.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Hash map + array with swap-delete is the classic O(1) solution [OK]

Hint: Hash map + array with swap-delete enables O(1) ops [OK]

Common Mistakes:

• Thinking balanced BST gives O(1) insert/delete
• Using linked list for O(1) random access
• Assuming sorted array can do O(1) insert/delete

3. Consider the following Python function to check if two strings are isomorphic:

def isIsomorphic(s: str, t: str) -> bool:
    if len(s) != len(t):
        return False
    s_to_t = {}
    t_to_s = {}
    for c1, c2 in zip(s, t):
        if c1 in s_to_t:
            if s_to_t[c1] != c2:
                return False
        else:
            s_to_t[c1] = c2
        if c2 in t_to_s:
            if t_to_s[c2] != c1:
                return False
        else:
            t_to_s[c2] = c1
    return True

print(isIsomorphic("egg", "add"))

What is the output of this code?

easy

A. True

B. None

C. Raises an exception

D. False

Solution

1. Step 1: Trace first iteration with c1='e', c2='a'

   Both maps are empty, so s_to_t['e'] = 'a' and t_to_s['a'] = 'e'.

2. Step 2: Trace second and third iterations

   Second iteration: c1='g', c2='d', maps updated similarly. Third iteration: c1='g' already maps to 'd', and c2='d' maps back to 'g', so no conflict.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Mapping is consistent for all characters, so returns True [OK]

Hint: Check mapping consistency at each iteration [OK]

Common Mistakes:

• Confusing output with 'foo' and 'bar'
• Assuming length mismatch causes exception

4. Consider the following buggy get method for the time-based key-value store. Which line contains the subtle bug that causes incorrect results when querying timestamps less than the smallest stored timestamp?

medium

A. Line 6: Returning empty string if i == 0

B. Line 2: Checking if key not in store

C. Line 5: Using bisect_left instead of bisect_right

D. Line 7: Returning values[i-1]

Solution

1. Step 1: Understand difference between bisect_left and bisect_right

   bisect_left returns insertion index to the left, which can cause off-by-one errors when timestamp equals stored timestamps.

2. Step 2: Identify impact on retrieval

   Using bisect_left causes incorrect index i for timestamps equal to stored ones, leading to wrong value or empty string returned.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   bisect_right is required to get correct insertion point for retrieval [OK]

Hint: bisect_right needed to get correct insertion index [OK]

Common Mistakes:

• Using bisect_left instead of bisect_right
• Incorrect boundary check for i==0
• Returning wrong index value

5. Suppose the time-based key-value store is extended to allow multiple values to be set at the same timestamp for a key, and get should return the most recently set value at or before the timestamp. Which modification to the data structure and retrieval algorithm is correct?

hard

A. Store multiple values per timestamp in a list and modify binary search to find the last value in that list.

B. Store timestamps with duplicates in a list and use bisect_right to find insertion index, then return the last inserted value at that timestamp.

C. Store all (timestamp, value) pairs in insertion order and perform linear search backward on get.

D. Use a hash map from (key, timestamp) to a stack of values and pop on get.

Solution

1. Step 1: Understand requirement for multiple values per timestamp

   Multiple values at same timestamp require storing duplicates and retrieving the most recent one.

2. Step 2: Analyze retrieval with duplicates

   Using bisect_right on timestamps list with duplicates returns insertion index after all duplicates; returning value at i-1 gives the last inserted value at that timestamp.

3. Step 3: Confirm data structure modification

   Storing timestamps with duplicates in order and values aligned allows correct retrieval without linear search or complex structures.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   bisect_right with duplicates returns correct index for last inserted value [OK]

Hint: bisect_right handles duplicates by returning insertion after last duplicate [OK]

Common Mistakes:

• Using linear search causing inefficiency
• Using stacks requiring pop on get
• Storing multiple values in nested lists complicates retrieval