Character Frequency Counting in DSA Javascript - Time & Space Complexity

Time Complexity: Character Frequency Counting

O(n)

Understanding Time Complexity

Counting how often each character appears in a string is a common task. Understanding how the time needed grows with the string length helps us write efficient code.

We want to know: how does the work increase as the string gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

const countChars = (str) => {
  const freq = {};
  for (let i = 0; i < str.length; i++) {
    const char = str[i];
    freq[char] = (freq[char] || 0) + 1;
  }
  return freq;
};
    

This code counts how many times each character appears in the input string.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character in the string once.
• How many times: Exactly once for each character, so as many times as the string length.

How Execution Grows With Input

As the string gets longer, the code checks each character once, so the work grows directly with the string size.

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 in a straight line with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to count characters grows directly in proportion to the string length.

Common Mistake

[X] Wrong: "Since we update counts inside the loop, it might be slower than linear time."

[OK] Correct: Each update is a simple operation done once per character, so it does not add extra loops or nested work.

Interview Connect

Knowing how to analyze simple loops like this builds your confidence for more complex problems. It shows you understand how work grows with input size, a key skill in interviews.

Self-Check

"What if we counted characters using nested loops to compare each character with every other? How would the time complexity change?"

Practice

1. What does character frequency counting mean in simple terms?

easy

A. Sorting characters alphabetically

B. Counting how many times each character appears in a string

C. Replacing characters with numbers

D. Removing duplicate characters from a string

Solution

1. Step 1: Understand the meaning of frequency

   Frequency means how often something appears.

2. Step 2: Apply to characters in a string

   Counting frequency means counting how many times each character shows up in the string.

3. Final Answer:

   Counting how many times each character appears in a string -> Option B

4. Quick Check:

   Frequency counting = Counting characters [OK]

Hint: Frequency means count occurrences of each character [OK]

Common Mistakes:

• Confusing frequency with sorting
• Thinking frequency means removing duplicates
• Mixing frequency with character replacement

2. Which of the following JavaScript code snippets correctly initializes an empty object to store character frequencies?

easy

A. const freq = '';

B. const freq = [];

C. const freq = {};

D. const freq = null;

Solution

1. Step 1: Identify the data structure for frequency counting

   We use an object to store characters as keys and counts as values.

2. Step 2: Check correct syntax for empty object

   Empty object is created with curly braces {} in JavaScript.

3. Final Answer:

   const freq = {}; -> Option C

4. Quick Check:

   Empty object = {} [OK]

Hint: Use curly braces {} to create empty object [OK]

Common Mistakes:

• Using array [] instead of object {}
• Using empty string '' instead of object
• Assigning null which is not an object

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

const str = 'hello';
const freq = {};
for (const ch of str) {
  freq[ch] = (freq[ch] || 0) + 1;
}
console.log(freq);

medium

A. {"h":1,"e":1,"l":2,"o":1}

B. {"h":1,"e":1,"l":1,"o":1}

C. {"h":0,"e":0,"l":0,"o":0}

D. SyntaxError

Solution

1. Step 1: Loop through each character in 'hello'

   Characters are 'h', 'e', 'l', 'l', 'o'.

2. Step 2: Count frequency using freq[ch] = (freq[ch] || 0) + 1

   For each character, add 1 to its count or start at 1 if not present.

3. Final Answer:

   {"h":1,"e":1,"l":2,"o":1} -> Option A

4. Quick Check:

   Count 'l' twice, others once [OK]

Hint: Use freq[ch] = (freq[ch] || 0) + 1 to count [OK]

Common Mistakes:

• Counting 'l' only once
• Getting syntax error from code
• Confusing keys and values in output

4. Find the error in this code that counts character frequency:

const str = 'abc';
const freq = {};
for (let i = 0; i <= str.length; i++) {
  const ch = str[i];
  freq[ch] = (freq[ch] || 0) + 1;
}
console.log(freq);

medium

A. Loop condition should be i < str.length, not i <= str.length

B. freq should be initialized as an array, not object

C. Use let instead of const for freq

D. Missing semicolon after freq[ch] assignment

Solution

1. Step 1: Check loop range

   Loop runs from i = 0 to i <= str.length, which means i goes up to str.length (3 for 'abc').

2. Step 2: Identify out-of-bounds access

   str[3] is undefined, so freq[undefined] is updated causing wrong key.

3. Final Answer:

   Loop condition should be i < str.length, not i <= str.length -> Option A

4. Quick Check:

   Array index max is length - 1 [OK]

Hint: Use i < length to avoid out-of-bounds errors [OK]

Common Mistakes:

• Using <= instead of < in loop
• Initializing freq as array instead of object
• Ignoring undefined character access

5. Given the string "aabbccddeeffgghhii", which code snippet correctly counts only characters that appear more than once and stores them in freq object?

hard

A. const freq = {};
for (const ch of str) {
freq[ch] = (freq[ch] || 0) + 1;
}
const filtered = {};
for (const ch in freq) {
if (freq[ch] > 1) filtered[ch] = freq[ch];
}

B. const freq = {};
for (const ch of str) {
if (freq[ch]) freq[ch]++;
else freq[ch] = 1;
}
for (const ch in freq) {
if (freq[ch] < 2) freq[ch] = 0;
}

C. const freq = {};
for (const ch of str) {
freq[ch] = 1;
}
for (const ch in freq) {
if (freq[ch] > 1) delete freq[ch];
}

D. const freq = {};
for (const ch of str) {
freq[ch] = (freq[ch] || 0) + 1;
}
for (const ch in freq) {
if (freq[ch] <= 1) delete freq[ch];
}

Solution

1. Step 1: Count all character frequencies

   Use freq[ch] = (freq[ch] || 0) + 1 to count all characters.

2. Step 2: Delete characters with frequency <= 1 from freq

   Iterate and delete keys where freq[ch] <= 1 to leave only duplicates in freq.

3. Final Answer:

   Code snippet that counts all then deletes singles from freq -> Option D

4. Quick Check:

   Delete freq <= 1 from freq [OK]

Hint: Count all, then delete characters with count <= 1 [OK]

Common Mistakes:

• Deleting keys while iterating freq object
• Setting frequency to 0 instead of removing
• Overwriting counts incorrectly