First Non Repeating Character Using Hash in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a HashMap to store character counts.

DSA Java

Map<Character, Integer> countMap = new HashMap<>();
for (char ch : str.toCharArray()) {
    countMap.[1](ch, countMap.getOrDefault(ch, 0) + 1);
}

Drag options to blanks, or click blank then click option'

Aget

Bput

Cremove

DcontainsKey

Attempts:

3 left

💡 Hint

Common Mistakes

Using get() instead of put() to update the map.

Trying to remove keys instead of updating counts.

We use put() to add or update the count of each character in the HashMap.

2fill in blank

medium

Complete the code to check if a character count is 1 (non-repeating).

DSA Java

for (char ch : str.toCharArray()) {
    if (countMap.get(ch) [1] 1) {
        return ch;
    }
}

Drag options to blanks, or click blank then click option'

A<

B!=

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of == which checks for non-equality.

Using > or < which do not specifically check for count 1.

We check if the count is exactly 1 to find the first non-repeating character.

3fill in blank

hard

Fix the error in the code to return a default value when no non-repeating character is found.

DSA Java

return [1];

Drag options to blanks, or click blank then click option'

A'\0'

Bnull

C'#'

D' '

Attempts:

3 left

💡 Hint

Common Mistakes

Returning null which is invalid for char return type.

Returning space ' ' which is a valid character.

Returning '\0' (null character) is a common way to indicate no valid character found.

4fill in blank

hard

Fill both blanks to complete the method signature and initialize the HashMap.

DSA Java

public static char [1](String str) {
    Map<Character, Integer> [2] = new HashMap<>();

Drag options to blanks, or click blank then click option'

AfirstNonRepeatingChar

BcountMap

CfindChar

DcharCount

Attempts:

3 left

💡 Hint

Common Mistakes

Using generic or unclear method names.

Using variable names that do not indicate their purpose.

The method is named firstNonRepeatingChar and the map variable is countMap.

5fill in blank

hard

Fill all three blanks to complete the loop that updates counts and finds the first non-repeating character.

DSA Java

for (char ch : str.toCharArray()) {
    countMap.[1](ch, countMap.getOrDefault(ch, 0) [2] 1);
}
for (char ch : str.toCharArray()) {
    if (countMap.get(ch) [3] 1) {
        return ch;
    }
}

Drag options to blanks, or click blank then click option'

Aput

B+

C==

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using get() instead of put() to update counts.

Using subtraction or wrong comparison operators.

We use put() to update counts, add 1 to the current count, and check if count equals 1.

Practice

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

easy

A. Reverses the string using a hash map

B. Counts the total number of characters in a string

C. Sorts characters of a string in alphabetical order

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

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: Role of hash map

   A hash map stores character counts to quickly check which characters appear once.

3. Final Answer:

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

4. Quick Check:

   First non repeating character = Finds the first character in a string that appears only once [OK]

Hint: Look for the first unique character using counts [OK]

Common Mistakes:

• Thinking it counts total characters only
• Confusing sorting with finding unique character
• Assuming it reverses the string

2. Which Java collection is best suited to store character counts for this problem?

easy

A. ArrayList<Character>

B. HashMap

C. LinkedList<Character>

D. TreeSet<Character>

Solution

1. Step 1: Identify data structure needs

   We need to map each character to its count, so a key-value structure is required.

2. Step 2: Choose appropriate Java collection

   HashMap stores characters as keys and their counts as values efficiently.

3. Final Answer:

   HashMap -> Option B

4. Quick Check:

   Use HashMap for character counts [OK]

Hint: Use HashMap for key-value pairs of char and count [OK]

Common Mistakes:

• Using list which doesn't map keys to counts
• Using TreeSet which stores unique sorted elements only
• Using LinkedList which is not key-value based

3. What is the output of the following Java code snippet?

String s = "swiss";
HashMap map = new HashMap<>();
for(char c : s.toCharArray()) {
  map.put(c, map.getOrDefault(c, 0) + 1);
}
for(char c : s.toCharArray()) {
  if(map.get(c) == 1) {
    System.out.println(c);
    break;
  }
}

medium

A. w

B. s

C. i

D. No output

Solution

1. Step 1: Count character frequencies

   Characters and counts: s=3, w=1, i=1.

2. Step 2: Find first character with count 1

   Iterate string: 's'(3), 'w'(1) -> first non repeating is 'w'.

3. Final Answer:

   w -> Option A

4. Quick Check:

   First unique char in "swiss" = w [OK]

Hint: Check counts then first unique char in order [OK]

Common Mistakes:

• Choosing 's' which repeats
• Choosing 'i' which is unique but appears after 'w'
• Assuming no output if not careful

4. Identify the error in this code snippet for finding the first non repeating character:

String s = "hello";
HashMap map = new HashMap<>();
for(int i = 0; i < s.length(); i++) {
  map.put(s.charAt(i), 1);
}
for(int i = 0; i < s.length(); i++) {
  if(map.get(s.charAt(i)) == 1) {
    System.out.println(s.charAt(i));
    break;
  }
}

medium

A. The map.put always sets count to 1, ignoring repeats

B. The loop should start from 1, not 0

C. The map should use Character keys, not int

D. The break statement is missing

Solution

1. Step 1: Analyze map population logic

   The code sets count to 1 for every character without incrementing, so repeats are not counted.

2. Step 2: Effect on finding non repeating character

   Since all counts are 1, the code incorrectly treats all characters as unique.

3. Final Answer:

   The map.put always sets count to 1, ignoring repeats -> Option A

4. Quick Check:

   Count update missing causes wrong results [OK]

Hint: Increment count instead of setting 1 every time [OK]

Common Mistakes:

• Not incrementing counts in map
• Wrong loop indices
• Confusing key types in map

5. Given the string "aabbccddeeffg", which character will be the first non repeating character using the hash method? What if the string is "aabbccddeeffggh"?

hard

A. First: g, Second: g

B. First: a, Second: h

C. First: g, Second: h

D. First: h, Second: h

Solution

1. Step 1: Analyze first string "aabbccddeeffg"

   All characters except 'g' appear twice; 'g' appears once, so first non repeating is 'g'.

2. Step 2: Analyze second string "aabbccddeeffggh"

   Now 'g' appears twice, 'h' appears once, so first non repeating is 'h'.

3. Final Answer:

   First: g, Second: h -> Option C

4. Quick Check:

   Check counts carefully for each string [OK]

Hint: Check counts carefully; last unique changes with input [OK]

Common Mistakes:

• Assuming first character is always unique
• Ignoring updated counts in second string
• Confusing repeated characters as unique