Frequency Counter Pattern Using Hash Map in DSA Java - 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 list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public static Map<Integer, Integer> frequencyCounter(int[] arr) {
    Map<Integer, Integer> freqMap = new HashMap<>();
    for (int num : arr) {
        freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
    }
    return freqMap;
}
    

This code counts how many times each number appears in the array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each element in the array once.
• How many times: Exactly once for each item in the input array.

How Execution Grows With Input

As the array gets bigger, the code checks each number once and updates the count.

Input Size (n)	Approx. Operations
10	About 10 checks and updates
100	About 100 checks and updates
1000	About 1000 checks and updates

Pattern observation: The number of operations grows directly with the input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to count frequencies grows in a straight line with the number of items.

Common Mistake

[X] Wrong: "Using a hash map makes the code slower because of extra lookups inside the loop."

[OK] Correct: Hash map lookups and updates happen in constant time on average, so they do not slow down the loop significantly.

Interview Connect

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

Self-Check

"What if we used a nested loop to count frequencies instead of a hash map? 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 count how many times each element appears in a collection

B. To sort elements in ascending order

C. To store elements without duplicates

D. To reverse the order of elements

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

   By using elements as keys and counts as values, it tracks how many times each element appears.

3. Final Answer:

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

4. Quick Check:

   Frequency counting = count 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 way to declare a HashMap to count frequencies of characters in Java?

easy

A. HashMap freq = new HashMap<>();

B. HashMap freq = new HashMap<>();

C. HashMap<String> freq = new HashMap<>();

D. HashMap<Character> freq = new HashMap<>();

Solution

1. Step 1: Identify correct generic types for HashMap

   Keys are characters, so use Character; values are counts, so use Integer.

2. Step 2: Check syntax correctness

   HashMap freq = new HashMap<>(); is valid Java syntax.

3. Final Answer:

   HashMap freq = new HashMap<>(); -> Option B

4. Quick Check:

   Correct generics and syntax = HashMap freq = new HashMap<>(); [OK]

Hint: Use wrapper classes Character and Integer for HashMap keys and values [OK]

Common Mistakes:

• Using primitive types like int in generics
• Missing value type in HashMap declaration
• Incorrect generic syntax

3. What will be the output of the following Java code?

import java.util.HashMap;
public class Main {
  public static void main(String[] args) {
    String s = "banana";
    HashMap freq = new HashMap<>();
    for (char c : s.toCharArray()) {
      freq.put(c, freq.getOrDefault(c, 0) + 1);
    }
    System.out.println(freq.get('a'));
  }
}

medium

A. 2

B. null

C. 1

D. 3

Solution

1. Step 1: Count occurrences of 'a' in "banana"

   String "banana" has 'a' at positions 1, 3, and 5, total 3 times.

2. Step 2: Check code logic for frequency counting

   Code uses getOrDefault to count each character's frequency correctly.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Count of 'a' in "banana" = 3 [OK]

Hint: Count characters manually, then verify with getOrDefault logic [OK]

Common Mistakes:

• Counting 'a' incorrectly as 2 or 1
• Confusing getOrDefault behavior
• Expecting null for existing keys

4. Identify the error in the following Java code snippet that tries to count frequencies of integers:

import java.util.HashMap;
public class Main {
  public static void main(String[] args) {
    int[] nums = {1, 2, 2, 3};
    HashMap freq = new HashMap<>();
    for (int num : nums) {
      freq.put(num, freq.get(num) + 1);
    }
    System.out.println(freq);
  }
}

medium

A. Using freq.get(num) without checking if key exists causes NullPointerException

B. HashMap declaration is incorrect

C. The for loop syntax is invalid

D. Printing freq directly causes syntax error

Solution

1. Step 1: Analyze freq.get(num) usage inside loop

   freq.get(num) returns null if key not present, so null + 1 causes NullPointerException.

2. Step 2: Identify missing default value handling

   Code should use freq.getOrDefault(num, 0) to avoid null on first occurrence.

3. Final Answer:

   Using freq.get(num) without checking if key exists causes NullPointerException -> Option A

4. Quick Check:

   NullPointerException due to missing default value [OK]

Hint: Use getOrDefault to avoid null when key missing [OK]

Common Mistakes:

• Ignoring possibility of null from get()
• Thinking HashMap declaration is wrong
• Assuming printing freq causes error

5. Given two integer arrays arr1 and arr2, which Java code snippet correctly checks if arr2 contains the same frequency of elements as arr1 using the Frequency Counter Pattern?

hard

A. HashMap freq1 = new HashMap<>();
HashMap freq2 = new HashMap<>();
for (int n : arr1) freq1.put(n, freq1.getOrDefault(n, 0) + 1);
for (int n : arr2) freq2.put(n, freq2.getOrDefault(n, 0) + 1);
return freq1.size() == freq2.size();

B. HashMap freq = new HashMap<>();
for (int n : arr1) freq.put(n, freq.getOrDefault(n, 0) + 1);
for (int n : arr2) freq.put(n, freq.getOrDefault(n, 0) - 1);
return freq.size() == 0;

C. HashMap freq = new HashMap<>();
for (int n : arr1) freq.put(n, freq.getOrDefault(n, 0) + 1);
for (int n : arr2) freq.put(n, freq.getOrDefault(n, 0) - 1);
for (int count : freq.values()) if (count != 0) return false;
return true;

D. HashMap freq = new HashMap<>();
for (int n : arr1) freq.put(n, freq.getOrDefault(n, 0) + 1);
for (int n : arr2) freq.put(n, freq.getOrDefault(n, 0) + 1);
return freq.size() == arr1.length;

Solution

1. Step 1: Understand frequency counting for both arrays

   Count frequencies of arr1 elements positively, then subtract frequencies for arr2 elements.

2. Step 2: Verify all counts are zero

   If all counts are zero, frequencies match; otherwise, they differ.

3. Final Answer:

   Code that increments for arr1, decrements for arr2, then checks zero counts -> Option C

4. Quick Check:

   Increment then decrement counts, check zero [OK]

Hint: Add counts for arr1, subtract for arr2, verify all zero [OK]

Common Mistakes:

• Checking sizes of two separate maps instead of contents
• Adding counts for both arrays instead of subtracting
• Checking map size instead of counts