Frequency Counter Pattern Using Hash Map in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a hash map for counting frequencies.

DSA C

int [1][256] = {0};

Drag options to blanks, or click blank then click option'

Amap

Bcount

Cfreq

Dfrequency

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is too long or unclear.

Not initializing the array to zero.

The variable freq is commonly used to store frequency counts in an array representing ASCII characters.

2fill in blank

medium

Complete the code to increase the frequency count of a character in the string.

DSA C

freq[(unsigned char)str[i]][1];

Drag options to blanks, or click blank then click option'

A++

B--

C+=

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using -- which decreases the count.

Using = which overwrites the count.

The ++ operator increases the count by 1 for the character at position i.

3fill in blank

hard

Fix the error in the condition to check if a character appears more than once.

DSA C

if (freq[(unsigned char)str[i]] [1] 1) {

Drag options to blanks, or click blank then click option'

A<=

B>

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using == which only checks for exactly one occurrence.

Using <= or < which check for less or equal.

The condition freq[...] > 1 checks if the character appears more than once.

4fill in blank

hard

Fill both blanks to complete the loop that counts character frequencies in a string.

DSA C

for (int [1] = 0; [2] < length; [2]++) {
    freq[(unsigned char)str[i]]++;
}

Drag options to blanks, or click blank then click option'

Ai

Bj

Clength

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables in the loop condition and increment.

Using a variable not declared in the loop.

The variable i is used as the loop counter to access each character in the string.

5fill in blank

hard

Fill all three blanks to create a frequency counter that prints characters appearing more than once.

DSA C

for (int [1] = 0; [1] < 256; [1]++) {
    if (freq[[1]] [2] 1) {
        printf("%c appears %d times\n", (char)[1], freq[[1]]);
    }
}

Drag options to blanks, or click blank then click option'

Ai

B>

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables in the loop and inside the if condition.

Using the wrong comparison operator in the if condition.

The loop variable i iterates over all possible characters. The condition freq[i] > 1 checks for characters appearing more than once. The variable i is used to print the character.

Practice

1. What is the main purpose of using a hash map in the Frequency Counter Pattern?

easy

A. To reverse the order of elements in the collection

B. To sort the elements in ascending order

C. To remove duplicate elements from the collection

D. To count how many times each element appears in a collection

Solution

1. Step 1: Understand the role of a hash map

   A hash map stores key-value pairs, allowing quick lookup and update of counts.

2. Step 2: Apply to frequency counting

   Using a hash map, we can track how many times each element appears by incrementing counts.

3. Final Answer:

   To count how many times each element appears in a collection -> Option D

4. Quick Check:

   Frequency counting = Counting elements [OK]

Hint: Hash map keys track elements, values track counts [OK]

Common Mistakes:

• Thinking hash map sorts elements
• Confusing frequency counting with removing duplicates
• Using arrays instead of hash maps for counting

2. Which of the following is the correct syntax to declare a hash map in C for counting frequencies of characters?

easy

A. map freq;

B. int freq[256] = {0};

C. unordered_map freq;

D. HashMap freq = new HashMap();

Solution

1. Step 1: Recognize C language limitations

   C does not have built-in hash map types like C++ STL's map or unordered_map.

2. Step 2: Use array for frequency counting in C

   We use an integer array of size 256 (for ASCII chars) initialized to zero to count frequencies.

3. Final Answer:

   int freq[256] = {0}; -> Option B

4. Quick Check:

   C uses arrays for frequency counting [OK]

Hint: In C, use arrays for frequency counts, not map types [OK]

Common Mistakes:

• Using C++ STL syntax in C code
• Trying to declare map or unordered_map in C
• Not initializing frequency array to zero

3. What will be the output of the following C code snippet?

char str[] = "aabbcc";
int freq[256] = {0};
for (int i = 0; str[i] != '\0'; i++) {
    freq[(int)str[i]]++;
}
printf("%d", freq['b']);

medium

A. 0

B. 1

C. 2

D. 3

Solution

1. Step 1: Count occurrences of 'b' in the string

   The string "aabbcc" has two 'b' characters at positions 2 and 3.

2. Step 2: Check frequency array value for 'b'

   freq['b'] is incremented twice, so freq['b'] = 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Count of 'b' = 2 [OK]

Hint: Count characters manually, then check freq array [OK]

Common Mistakes:

• Confusing character ASCII codes
• Forgetting null terminator in loop condition
• Printing freq for wrong character

4. Identify the error in the following C code that tries to count character frequencies:

char str[] = "hello";
int freq[256];
for (int i = 0; str[i] != '\0'; i++) {
    freq[(int)str[i]]++;
}
printf("%d", freq['l']);

medium

A. freq array is not initialized to zero

B. Loop condition should be i < strlen(str)

C. Casting str[i] to int is incorrect

D. printf format specifier should be %c

Solution

1. Step 1: Check initialization of freq array

   The freq array is declared but not initialized, so it contains garbage values.

2. Step 2: Effect on frequency counting

   Incrementing uninitialized values leads to incorrect counts; freq should be zeroed before use.

3. Final Answer:

   freq array is not initialized to zero -> Option A

4. Quick Check:

   Initialize freq array to zero before counting [OK]

Hint: Always zero frequency array before counting [OK]

Common Mistakes:

• Assuming uninitialized arrays start at zero
• Using wrong printf format for integers
• Misunderstanding loop conditions with null terminator

5. Given two strings s1 = "listen" and s2 = "silent", which approach using frequency counters correctly checks if they are anagrams in C?

hard

A. Count frequencies of characters in both strings using arrays, then compare arrays element-wise

B. Sort both strings and compare them character by character

C. Check if lengths are equal, then check if first characters match

D. Use a hash map to store characters of s1 only and check if s2 contains them

Solution

1. Step 1: Use frequency arrays for both strings

   Create two arrays freq1 and freq2 of size 256, initialize to zero, count characters in s1 and s2 respectively.

2. Step 2: Compare frequency arrays element-wise

   If all counts match for every character, s1 and s2 are anagrams; otherwise, they are not.

3. Final Answer:

   Count frequencies of characters in both strings using arrays, then compare arrays element-wise -> Option A

4. Quick Check:

   Frequency arrays equal means anagrams [OK]

Hint: Compare frequency arrays to confirm anagrams [OK]

Common Mistakes:

• Only counting one string's characters
• Comparing only string lengths or first characters
• Assuming sorting is the only way to check anagrams