Group Anagrams Using Hash Map in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Grouping Anagrams by Sorted Key

What is the output of the following C code that groups anagrams using a hash map keyed by sorted strings?

DSA C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

// Simple hash map node for demonstration
typedef struct Node {
    char *key;
    char **words;
    int count;
    int capacity;
    struct Node *next;
} Node;

// Function to sort characters in a string
void sortString(char *str) {
    int len = strlen(str);
    for (int i = 0; i < len - 1; i++) {
        for (int j = i + 1; j < len; j++) {
            if (str[i] > str[j]) {
                char temp = str[i];
                str[i] = str[j];
                str[j] = temp;
            }
        }
    }
}

// Insert word into hash map node
void insertWord(Node *node, char *word) {
    if (node->count == node->capacity) {
        node->capacity *= 2;
        node->words = realloc(node->words, node->capacity * sizeof(char*));
    }
    node->words[node->count++] = word;
}

// Simple hash function
unsigned int hash(char *str) {
    unsigned int h = 0;
    while (*str) {
        h = h * 31 + *str++;
    }
    return h % 10;
}

// Find or create node in hash map
Node* findOrCreate(Node **map, char *key) {
    unsigned int h = hash(key);
    Node *curr = map[h];
    while (curr) {
        if (strcmp(curr->key, key) == 0) return curr;
        curr = curr->next;
    }
    Node *newNode = malloc(sizeof(Node));
    newNode->key = strdup(key);
    newNode->capacity = 2;
    newNode->count = 0;
    newNode->words = malloc(newNode->capacity * sizeof(char*));
    newNode->next = map[h];
    map[h] = newNode;
    return newNode;
}

int main() {
    char *words[] = {"eat", "tea", "tan", "ate", "nat", "bat"};
    int n = 6;
    Node *map[10] = {0};

    for (int i = 0; i < n; i++) {
        char *copy = strdup(words[i]);
        sortString(copy);
        Node *node = findOrCreate(map, copy);
        insertWord(node, words[i]);
        free(copy);
    }

    // Print groups
    for (int i = 0; i < 10; i++) {
        Node *curr = map[i];
        while (curr) {
            printf("Group: ");
            for (int j = 0; j < curr->count; j++) {
                printf("%s ", curr->words[j]);
            }
            printf("\n");
            curr = curr->next;
        }
    }
    return 0;
}

A

Group: bat 
Group: tan 
Group: nat 
Group: eat 
Group: tea 
Group: ate 

B

Group: eat tea ate 
Group: bat 
Group: tan nat 

C

Group: bat 
Group: tan nat 
Group: eat tea ate 

D

Group: bat tan nat eat tea ate 

💡 Hint

Think about how sorting the characters in each word groups anagrams together.

The code sorts each word to create a key. Words with the same sorted key are grouped together. The hash map stores these groups. The output prints groups of anagrams.

🧠 Conceptual

intermediate

Why Use Sorted Strings as Keys in Grouping Anagrams?

Why do we use the sorted version of a word as the key in a hash map when grouping anagrams?

A Because sorting words makes them shorter and easier to store.

B Because sorted strings are unique identifiers for anagram groups.

C Because sorted strings are always the same length as original words.

D Because sorting words removes duplicate letters.

💡 Hint

Think about what makes anagrams similar.

Anagrams have the same letters in different orders. Sorting letters puts them in a standard order, so all anagrams share the same sorted string key.

🔧 Debug

advanced

Identify the Error in Hash Map Grouping Code

What error will this code produce when grouping anagrams using a hash map keyed by sorted strings?

DSA C

char *words[] = {"abc", "bca", "cab"};
Node *map[5] = {0};

for (int i = 0; i < 3; i++) {
    char *copy = strdup(words[i]);
    sortString(copy);
    Node *node = findOrCreate(map, copy);
    insertWord(node, copy);
    free(copy);
}

A No error; code runs correctly.

B Segmentation fault because map size is too small.

C Memory leak because 'copy' is never freed.

D Use after free error due to freeing 'copy' after inserting it.

💡 Hint

Check what happens to 'copy' after inserting it into the node.

The code inserts 'copy' pointer into the node, then frees 'copy'. Later access to that pointer causes use after free error.

❓ Predict Output

advanced

Output of Group Anagrams with Different Hash Function

What is the output of this code that groups anagrams using a hash map with a hash function returning constant 0?

DSA C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct Node {
    char *key;
    char **words;
    int count;
    int capacity;
    struct Node *next;
} Node;

void sortString(char *str) {
    int len = strlen(str);
    for (int i = 0; i < len - 1; i++) {
        for (int j = i + 1; j < len; j++) {
            if (str[i] > str[j]) {
                char temp = str[i];
                str[i] = str[j];
                str[j] = temp;
            }
        }
    }
}

void insertWord(Node *node, char *word) {
    if (node->count == node->capacity) {
        node->capacity *= 2;
        node->words = realloc(node->words, node->capacity * sizeof(char*));
    }
    node->words[node->count++] = word;
}

unsigned int hash(char *str) {
    return 0; // constant hash to force collisions
}

Node* findOrCreate(Node **map, char *key) {
    unsigned int h = hash(key);
    Node *curr = map[h];
    while (curr) {
        if (strcmp(curr->key, key) == 0) return curr;
        curr = curr->next;
    }
    Node *newNode = malloc(sizeof(Node));
    newNode->key = strdup(key);
    newNode->capacity = 2;
    newNode->count = 0;
    newNode->words = malloc(newNode->capacity * sizeof(char*));
    newNode->next = map[h];
    map[h] = newNode;
    return newNode;
}

int main() {
    char *words[] = {"eat", "tea", "tan", "ate", "nat", "bat"};
    int n = 6;
    Node *map[1] = {0};

    for (int i = 0; i < n; i++) {
        char *copy = strdup(words[i]);
        sortString(copy);
        Node *node = findOrCreate(map, copy);
        insertWord(node, words[i]);
        free(copy);
    }

    Node *curr = map[0];
    while (curr) {
        printf("Group: ");
        for (int j = 0; j < curr->count; j++) {
            printf("%s ", curr->words[j]);
        }
        printf("\n");
        curr = curr->next;
    }
    return 0;
}

A

Group: bat 
Group: tan nat 
Group: eat tea ate 

B

Group: eat tea ate 
Group: bat 
Group: tan nat 

C

Group: bat tan nat eat tea ate 

D

Group: eat tea ate bat tan nat 

💡 Hint

All keys hash to the same bucket, so groups are stored in a linked list.

The constant hash 0 puts all nodes in bucket 0. New nodes are prepended: 'aet' first, then 'ant' -> 'aet', then 'abt' -> 'ant' -> 'aet'. Printing traverses: 'bat', 'tan nat', 'eat tea ate'.

🚀 Application

expert

Number of Unique Anagram Groups

Given the list of words {"listen", "silent", "enlist", "google", "gogole", "abc", "cab"}, how many unique anagram groups will be formed using a hash map keyed by sorted strings?

A 3

B 6

C 5

D 4

💡 Hint

Group words by their sorted letters and count distinct groups.

The groups are: {"listen", "silent", "enlist"} (sorted 'eilnst'), {"google", "gogole"} (sorted 'eggloo'), {"abc", "cab"} (sorted 'abc'). 3 unique groups.

Practice

1. What is the main idea behind grouping anagrams using a hash map?

easy

A. Use the length of words as keys to group anagrams

B. Store words in a list without any sorting or grouping

C. Group words by their first character only

D. Use sorted strings as keys to group words with the same letters

Solution

1. Step 1: Understand anagrams and their properties

   Anagrams are words made of the same letters in any order, so sorting letters gives a unique key for each group.

2. Step 2: Use sorted strings as hash map keys

   By sorting each word and using it as a key, all anagrams map to the same key and can be grouped together.

3. Final Answer:

   Use sorted strings as keys to group words with the same letters -> Option D

4. Quick Check:

   Sorted word keys group anagrams [OK]

Hint: Sort each word and use as key to group anagrams [OK]

Common Mistakes:

• Grouping by word length only
• Using first character as key
• Not sorting words before grouping

2. Which of the following is the correct way to declare a hash map (dictionary) in C for grouping anagrams?

easy

A. map anagramGroups;

B. unordered_map anagramGroups;

C. unordered_map anagramGroups;

D. vector<string> anagramGroups;

Solution

1. Step 1: Identify the data types needed for grouping anagrams

   We need a map from a string key (sorted word) to a list of strings (anagrams).

2. Step 2: Choose the correct C++ STL container

   unordered_map fits this need perfectly.

3. Final Answer:

   unordered_map anagramGroups; -> Option B

4. Quick Check:

   Map from string to vector<string> groups anagrams [OK]

Hint: Use unordered_map for grouping anagrams [OK]

Common Mistakes:

• Using map which mismatches key and value types
• Using vector<string> alone without mapping keys
• Using unordered_map which is unrelated

3. What is the output of the following C++ code snippet? vector<string> words = {"eat", "tea", "tan", "ate", "nat", "bat"}; unordered_map groups; for (auto &word : words) { string sortedWord = word; sort(sortedWord.begin(), sortedWord.end()); groups[sortedWord].push_back(word); } for (auto &pair : groups) { sort(pair.second.begin(), pair.second.end()); for (auto &w : pair.second) cout << w << " "; cout << endl; }

medium

A. bat
nat tan
eat tea ate

B. eat tea ate
bat
tan nat

C. tea ate eat
bat
tan nat

D. ate eat tea
bat
nat tan

Solution

1. Step 1: Group words by sorted keys

   Words "eat", "tea", "ate" all sort to "aet" and group together; "tan", "nat" to "ant"; "bat" to "abt".

2. Step 2: Sort each group and print

   Each group is sorted alphabetically before printing: "ate eat tea", "bat", "nat tan".

3. Final Answer:

   bat nat tan eat tea ate -> Option A

4. Quick Check:

   Sorted keys group anagrams; groups sorted before print [OK]

Hint: Sort words and groups before printing for consistent output [OK]

Common Mistakes:

• Not sorting groups before printing
• Printing groups in wrong order
• Confusing sorted keys with original words

4. Identify the bug in this C++ code for grouping anagrams: vector<string> words = {"abc", "bca", "cab"}; unordered_map groups; for (auto word : words) { string key = word; sort(word.begin(), word.end()); groups[key].push_back(word); }

medium

A. Sorting is done on 'word' but key uses unsorted string

B. Using unordered_map instead of map causes error

C. Loop variable should be reference to avoid copies

D. Missing initialization of groups map

Solution

1. Step 1: Check how key and sorted string are assigned

   Key is assigned before sorting, so key is unsorted original word.

2. Step 2: Sorting is done on 'word' but key remains unsorted

   Groups use unsorted key, so anagrams won't group correctly.

3. Final Answer:

   Sorting is done on 'word' but key uses unsorted string -> Option A

4. Quick Check:

   Key must be sorted string for correct grouping [OK]

Hint: Sort string before assigning key for hash map [OK]

Common Mistakes:

• Assigning key before sorting
• Sorting wrong variable
• Not using reference in loop (less critical here)

5. Given a list of words, how can you modify the hash map grouping approach to handle case-insensitive anagrams (e.g., "Eat" and "tea" grouped together)?

hard

A. Group words by their first letter ignoring case

B. Use word length as key instead of sorted string

C. Convert all words to lowercase before sorting and grouping

D. Sort words without changing case and group normally

Solution

1. Step 1: Understand case-insensitive grouping

   "Eat" and "tea" differ in case but are anagrams ignoring case.

2. Step 2: Normalize words by converting to lowercase before sorting

   This ensures keys match regardless of original case.

3. Final Answer:

   Convert all words to lowercase before sorting and grouping -> Option C

4. Quick Check:

   Lowercase normalization groups case-insensitive anagrams [OK]

Hint: Lowercase words before sorting to group case-insensitive anagrams [OK]

Common Mistakes:

• Grouping by length only
• Ignoring case normalization
• Sorting without case conversion