Frequency Counter Pattern Using Hash Map in DSA C - Time & Space Complexity

Time Complexity: Frequency Counter Pattern Using Hash Map

O(n)

Understanding Time Complexity

We want to understand how fast the frequency counter pattern runs when counting items.

How does the time needed grow as the input list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

void frequencyCounter(char arr[], int freq[], int size) {
    for (int i = 0; i < size; i++) {
        freq[(unsigned char)arr[i]]++;
    }
}

int main() {
    char arr[] = "aabbcc";
    int freq[256] = {0};
    int size = strlen(arr);
    frequencyCounter(arr, freq, size);
    for (int i = 0; i < 256; i++) {
        if (freq[i] > 0) {
            printf("%c: %d\n", i, freq[i]);
        }
    }
    return 0;
}
    

This code counts how many times each character appears in the input string using a frequency array as a hash map.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop over the input array to count frequencies.
• How many times: Exactly once for each character in the input (size times).

How Execution Grows With Input

Each character is counted once, so the work grows directly with input size.

Input Size (n)	Approx. Operations
10	About 10 increments
100	About 100 increments
1000	About 1000 increments

Pattern observation: The number of operations grows in a straight line as input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to count frequencies grows directly with the number of items in the input.

Common Mistake

[X] Wrong: "Using a hash map makes counting slower because of extra lookups."

[OK] Correct: Hash maps or arrays let us find and update counts instantly, so counting is still fast and linear.

Interview Connect

Knowing how frequency counting works and its time cost helps you solve many problems efficiently in interviews.

Self-Check

"What if we used a linked list instead of a hash map to count frequencies? How would the time complexity change?"

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