First Non Repeating Character Using Hash in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the purpose of using a hash (dictionary) in finding the first non-repeating character?

A hash (dictionary) stores the count of each character in the string, allowing quick lookup to identify characters that appear only once.

beginner

How do you identify the first non-repeating character after counting frequencies?

By scanning the string from left to right and checking the count of each character in the hash, the first character with count 1 is the first non-repeating character.

intermediate

Why is the time complexity of this approach O(n)?

Because we traverse the string twice: once to count characters and once to find the first non-repeating character, each operation is O(n), so total is O(n).

beginner

What happens if all characters repeat in the string?

If all characters repeat, the function returns a special value (like -1 or None) indicating no non-repeating character exists.

beginner

Show the Python code snippet to count characters using a dictionary.

counts = {} for ch in s: counts[ch] = counts.get(ch, 0) + 1

What data structure is best to count character frequencies for this problem?

A Stack

B Queue

C Hash (dictionary)

D Linked List

A hash (dictionary) allows fast counting and lookup of character frequencies.

What is the first step in finding the first non-repeating character?

A Remove duplicates

B Sort the string

C Reverse the string

D Count frequency of each character

Counting frequencies helps identify which characters appear only once.

If the string is 'aabbcc', what should the function return?

A 'a'

B -1 or None

C 'b'

D 'c'

All characters repeat, so no non-repeating character exists.

What is the time complexity of this approach?

A O(n)

B O(n^2)

C O(log n)

D O(1)

Two passes over the string make the time complexity linear, O(n).

Which of these is NOT required to find the first non-repeating character?

A Sorting the string

B Traversing the string

C Counting character frequency

D Checking counts in hash

Sorting is not needed; it changes the order and is unnecessary.

Explain how to find the first non-repeating character in a string using a hash.

Think about counting first, then scanning.

Describe the time complexity and why this method is efficient.

Focus on number of traversals and hash operations.

Practice

1. What does the "First Non Repeating Character Using Hash" algorithm primarily do?

easy

A. Finds the first character in a string that appears only once

B. Counts the total number of characters in a string

C. Sorts characters of a string alphabetically

D. Reverses the string

Solution

1. Step 1: Understand the problem goal

   The algorithm aims to find the first character in the string that does not repeat anywhere else.

2. Step 2: Identify the main operation

   It uses a hash (dictionary) to count occurrences and then finds the first character with count 1.

3. Final Answer:

   Finds the first character in a string that appears only once -> Option A

4. Quick Check:

   First unique character = first non repeating character [OK]

Hint: Look for the first character with count one in the hash [OK]

Common Mistakes:

• Thinking it counts total characters instead of unique
• Confusing sorting with finding unique characters
• Assuming it reverses the string

2. Which of the following Python code snippets correctly initializes a hash (dictionary) to count characters in a string s?

easy

A. counts = []; for ch in s: counts[ch] += 1

B. counts = {}; for ch in s: counts[ch] = counts.get(ch, 0) + 1

C. counts = set(); for ch in s: counts.add(ch)

D. counts = 0; for ch in s: counts += 1

Solution

1. Step 1: Check dictionary initialization and update

   counts = {}; for ch in s: counts[ch] = counts.get(ch, 0) + 1 correctly creates an empty dictionary and updates counts using get with default 0.

2. Step 2: Analyze other options

   counts = []; for ch in s: counts[ch] += 1 uses a list which cannot be indexed by characters; counts = set(); for ch in s: counts.add(ch) uses a set which only stores unique items without counts; counts = 0; for ch in s: counts += 1 uses an integer which cannot store counts per character.

3. Final Answer:

   counts = {}; for ch in s: counts[ch] = counts.get(ch, 0) + 1 -> Option B

4. Quick Check:

   Dictionary with get method counts chars correctly [OK]

Hint: Use dict.get(key, 0) to count characters safely [OK]

Common Mistakes:

• Using list instead of dictionary for counts
• Using set which does not count duplicates
• Trying to increment integer without keys

3. What is the output of the following code?

def first_non_repeating(s):
    counts = {}
    for ch in s:
        counts[ch] = counts.get(ch, 0) + 1
    for ch in s:
        if counts[ch] == 1:
            return ch
    return '_'

print(first_non_repeating('swiss'))

medium

A. s

B. _

C. i

D. w

Solution

1. Step 1: Count character occurrences in 'swiss'

   Counts: s=3, w=1, i=1

2. Step 2: Find first character with count 1 in order

   Order: s (3), w (1) -> first non repeating is 'w'

3. Final Answer:

   w -> Option D

4. Quick Check:

   First unique char in 'swiss' is 'w' [OK]

Hint: Check counts then scan string for first count 1 char [OK]

Common Mistakes:

• Returning first unique char incorrectly as 's'
• Confusing 'i' as first unique instead of 'w'
• Returning '_' when unique char exists

4. The following code is intended to find the first non repeating character but has a bug. What is the bug?

def first_non_repeating_bug(s):
    counts = {}
    for ch in s:
        counts[ch] += 1
    for ch in s:
        if counts[ch] == 1:
            return ch
    return '_'

medium

A. The for loop should iterate over counts.keys() instead of s

B. The return '_' statement is unreachable

C. counts dictionary is not initialized with default values before increment

D. The function does not return anything

Solution

1. Step 1: Analyze counts dictionary usage

   The code tries to increment counts[ch] without initializing counts[ch], causing KeyError.

2. Step 2: Identify missing initialization

   Must initialize counts[ch] to 0 before increment or use counts.get(ch, 0) + 1.

3. Final Answer:

   counts dictionary is not initialized with default values before increment -> Option C

4. Quick Check:

   Missing initialization causes KeyError [OK]

Hint: Initialize counts[ch] before increment or use get() [OK]

Common Mistakes:

• Assuming return '_' is unreachable
• Thinking loop over s is wrong
• Believing function returns nothing

5. Given the string s = 'aabbccddeeffg', which character will the first non repeating character algorithm return?
Note: The algorithm uses a hash to count characters and returns the first character with count 1 or '_' if none.

medium

A. g

B. a

C. _

D. f

Solution

1. Step 1: Count occurrences of each character in 'aabbccddeeffg'

   a=2, b=2, c=2, d=2, e=2, f=2, g=1

2. Step 2: Find first character with count 1 in string order

   All characters except 'g' repeat twice; 'g' appears once at the end.

3. Final Answer:

   g -> Option A

4. Quick Check:

   Only 'g' is unique in the string [OK]

Hint: Check counts; last unique char is first non repeating here [OK]

Common Mistakes:

• Choosing 'a' or 'f' which repeat twice
• Choosing '_' when unique char exists
• Ignoring order of characters