Frequency Counter Pattern Using Hash Map in DSA C - Execution Trace

Concept Flow - Frequency Counter Pattern Using Hash Map

Start with empty hash map

↓

Read next element from input

↓

Check if element in hash map?

No→Add element with count=1

↓

Increment count by 1

↓

Repeat for all elements

↓

Hash map now has frequency counts

↓

Use frequency data as needed

↓

End

Start with an empty hash map, read each element, update its count, and finally have frequency counts for all elements.

Execution Sample

DSA C

int arr[] = {2, 3, 2, 4, 3};
int n = 5;
HashMap freq = {};
for (int i = 0; i < n; i++) {
  freq[arr[i]] = (freq[arr[i]] ? freq[arr[i]] : 0) + 1;
}

Counts how many times each number appears in the array using a hash map.

Execution Table

Step	Operation	Element Processed	Hash Map State	Visual State
1	Start	None	{}	{}
2	Process element	2	{2:1}	{2:1}
3	Process element	3	{2:1, 3:1}	{2:1, 3:1}
4	Process element	2	{2:2, 3:1}	{2:2, 3:1}
5	Process element	4	{2:2, 3:1, 4:1}	{2:2, 3:1, 4:1}
6	Process element	3	{2:2, 3:2, 4:1}	{2:2, 3:2, 4:1}
7	End	None	{2:2, 3:2, 4:1}	{2:2, 3:2, 4:1}

💡 All elements processed, frequency counts complete.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
freq	{}	{2:1}	{2:1, 3:1}	{2:2, 3:1}	{2:2, 3:1, 4:1}	{2:2, 3:2, 4:1}	{2:2, 3:2, 4:1}
element	None	2	3	2	4	3	None

Key Moments - 3 Insights

Why do we add the element with count 1 if it is not in the hash map?

Because the first time we see an element, its count starts at 1. This is shown in execution_table row 2 where element 2 is added with count 1.

Why do we increment the count instead of replacing it?

Because we want to keep track of how many times the element appears. Incrementing updates the count correctly, as seen in execution_table rows 4 and 6 where counts increase.

What happens if the input array is empty?

The hash map remains empty, as no elements are processed. This is like execution_table row 1 where freq is empty at start and no further steps occur.

Visual Quiz - 3 Questions

Look at the execution_table, what is the hash map state after processing the third element?

A {2:2, 3:2}

B {2:2, 3:1}

C {2:1, 3:1}

D {2:1, 3:2}

💡 Hint

Check row 4 in execution_table where the third element (2) is processed.

At which step does the element 4 get added to the hash map?

A Step 5

B Step 3

C Step 4

D Step 6

💡 Hint

Look at execution_table row 5 where element 4 is processed and added.

If the input array had one more '3' at the end, what would be the final count of 3 in the hash map?

A 4

B 2

C 3

D 1

💡 Hint

Currently count of 3 is 2 at final step (row 7), adding one more increments it by 1.

Concept Snapshot

Frequency Counter Pattern Using Hash Map:
- Start with empty hash map
- For each element, check if in map
- If not, add with count 1
- If yes, increment count
- Result: hash map with element frequencies
- Useful for counting duplicates or comparing data

Full Transcript

The Frequency Counter Pattern uses a hash map to count how many times each element appears in a list. We start with an empty hash map. For each element in the input, we check if it is already in the hash map. If it is not, we add it with a count of 1. If it is already there, we increase its count by 1. After processing all elements, the hash map holds the frequency of each element. This pattern helps quickly count duplicates or compare data sets efficiently.

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