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

Concept Flow - Frequency Counter Pattern Using Hash Map

Start with empty hash map

↓

For each element in array

↓

Check if element in hash map?

No→Add element with count 1

↓

Increment count

↓

Move to next element

↓

End

↓

Hash map with frequencies ready

We start with an empty hash map and go through each element in the array. If the element is new, add it with count 1. If it exists, increase its count. At the end, the hash map shows frequency of each element.

Execution Sample

DSA C++

std::unordered_map<int,int> freq;
int arr[] = {2,3,2,4,3};
for(int x : arr) {
  freq[x]++;
}

This code 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	-	{}	{}
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	-	{2:2, 3:2, 4:1}	{2:2, 3:2, 4:1}

💡 All elements processed, frequency map complete

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	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	-	2	3	2	4	3	-

Key Moments - 3 Insights

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

When the element is new (not in the map), we start counting it from 1 because we just saw it once. See execution_table step 2 where element 2 is added with count 1.

Why do we increment the count instead of adding a new entry when the element is already in the map?

If the element exists, it means we saw it before, so we increase its count by 1 to track total occurrences. See execution_table step 4 where element 2's count goes from 1 to 2.

What happens if the array is empty?

If the array is empty, the loop never runs and the hash map stays empty. See execution_table step 1 and 7 where the map is empty at start and end.

Visual Quiz - 3 Questions

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

A {2:2, 3:1}

B {2:1, 3:1}

C {2:2, 3:2}

D {2:1, 3:2}

💡 Hint

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

At which step does the frequency of element 3 become 2?

A Step 4

B Step 6

C Step 3

D Step 5

💡 Hint

Look at execution_table row 6 where element 3 is processed again and count increases.

If the input array was empty, what would be the final hash map state?

A {0:1}

B null

C {}

D {-1:1}

💡 Hint

Refer to key_moments explanation about empty array and execution_table start/end states.

Concept Snapshot

Frequency Counter Pattern Using Hash Map:
- Use a hash map to count occurrences of elements.
- For each element, add with count 1 if new.
- If exists, increment count.
- Result: hash map with element frequencies.
- Useful for quick frequency checks.

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 map. For each element, if it is not in the map, we add it with count 1. If it is already there, we increase its count by 1. This way, after processing all elements, the map shows the frequency of each element. For example, processing the array {2,3,2,4,3} results in {2:2, 3:2, 4:1}. This pattern helps quickly find counts without nested loops.

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

B. To sort the elements in ascending order

C. To store elements without any duplicates

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

   We use the hash map keys as elements and values as their counts to track frequency.

3. Final Answer:

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

4. Quick Check:

   Frequency Counter = 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 hash map to reverse order

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

easy

A. std::vector<int> freq;

B. std::unordered_map freq;

C. std::unordered_set<char> freq;

D. std::map freq;

Solution

1. Step 1: Identify correct container for frequency counting

   We need a map from characters to integers to store counts.

2. Step 2: Check syntax correctness

   std::unordered_map freq; declares a hash map with char keys and int values.

3. Final Answer:

   std::unordered_map freq; -> Option B

4. Quick Check:

   unordered_map = frequency map [OK]

Hint: Use unordered_map for frequency counting [OK]

Common Mistakes:

• Using map with wrong key-value types
• Using unordered_set which stores only keys
• Using vector which is not a map

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

std::string s = "aabbcc";
std::unordered_map freq;
for (char c : s) {
    freq[c]++;
}
for (auto &p : freq) {
    std::cout << p.first << ":" << p.second << " ";
}

medium

A. a:2 b:2 c:2

B. a:1 b:1 c:1

C. a:3 b:3 c:3

D. a:2 b:3 c:1

Solution

1. Step 1: Count frequencies of each character in string

   Characters 'a', 'b', and 'c' each appear twice in "aabbcc".

2. Step 2: Print each character and its count

   The loop prints each character and its frequency, so output is "a:2 b:2 c:2 ".

3. Final Answer:

   a:2 b:2 c:2 -> Option A

4. Quick Check:

   Count 'a','b','c' = 2 each [OK]

Hint: Count each char's occurrences in string [OK]

Common Mistakes:

• Assuming counts are 1 instead of 2
• Mixing counts between characters
• Confusing output format

4. Identify the error in this C++ code snippet that tries to count frequencies of integers in a vector:

std::vector<int> nums = {1, 2, 2, 3};
std::unordered_map freq;
for (int i = 0; i < nums.size(); i++) {
    freq[nums[i]] = freq[nums[i]] + 1;
}
for (auto p : freq) {
    std::cout << p.first << ":" << p.second << " ";
}

medium

A. Loop should use range-based for instead of index

B. Using freq[nums[i]] = freq[nums[i]] + 1; causes runtime error

C. No error; code works correctly

D. freq should be declared as std::unordered_set<int>

Solution

1. Step 1: Check frequency counting logic

   freq[nums[i]] = freq[nums[i]] + 1; correctly increments count for each element.

2. Step 2: Verify container types and loops

   Using index loop and unordered_map is valid and works as intended.

3. Final Answer:

   No error; code works correctly -> Option C

4. Quick Check:

   Increment freq with freq[key]++ works [OK]

Hint: freq[key]++ increments count safely [OK]

Common Mistakes:

• Thinking freq[key] = freq[key] + 1 causes error
• Confusing unordered_map with unordered_set
• Believing range-based for is mandatory

5. Given two strings s1 and s2, which approach using frequency counters correctly checks if s2 is an anagram of s1?

std::unordered_map freq1, freq2;
for (char c : s1) freq1[c]++;
for (char c : s2) freq2[c]++;
// What next?

hard

A. Compare freq1 and freq2 maps for equality

B. Check if freq1 contains all keys of freq2 only

C. Check if freq2 has more keys than freq1

D. Sort s1 and s2 and compare strings

Solution

1. Step 1: Build frequency maps for both strings

   freq1 and freq2 store counts of each character in s1 and s2 respectively.

2. Step 2: Compare frequency maps for exact match

   If freq1 and freq2 are equal, s2 is an anagram of s1 because they have same chars with same counts.

3. Final Answer:

   Compare freq1 and freq2 maps for equality -> Option A

4. Quick Check:

   Anagram check = freq maps equal [OK]

Hint: Anagrams have identical frequency maps [OK]

Common Mistakes:

• Only checking keys without counts
• Assuming more keys means anagram
• Using sorting instead of frequency maps