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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the frequency array for all ASCII characters.

DSA C

int freq[[1]] = {0};

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

A256

B64

C128

D512

Attempts:

3 left

💡 Hint

Common Mistakes

Using 128 instead of 256 causes out-of-bound errors for extended ASCII characters.

The frequency array size should be 256 to cover all ASCII characters.

2fill in blank

medium

Complete the code to increment the frequency count of each character in the string.

DSA C

freq[(unsigned char)str[[1]]]++;

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

Aindex

Bj

Ck

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable not declared or not used in the loop.

The loop variable 'i' is used to access each character in the string.

3fill in blank

hard

Fix the error in the loop condition to find the first non-repeating character.

DSA C

for (int [1] = 0; str[i] != '\0'; i++) {

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

Ak

Bj

Ci

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables in initialization and condition causes errors.

The loop variable should be 'i' to match the condition and increment.

4fill in blank

hard

Fill both blanks to check if the character frequency is 1 and return it.

DSA C

if (freq[(unsigned char)str[[1]]] [2] 1) {
    return str[i];
}

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

Ai

B==

C!=

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causes wrong character to be returned.

We check if the frequency of the character at index 'i' is equal to 1.

5fill in blank

hard

Fill all three blanks to complete the function that returns the first non-repeating character or '_' if none found.

DSA C

char firstNonRepeatingChar(char *str) {
    int freq[[1]] = {0};
    for (int [2] = 0; str[[2]] != '\0'; [2]++) {
        freq[(unsigned char)str[[2]]]++;
    }
    for (int i = 0; str[i] != '\0'; i++) {
        if (freq[(unsigned char)str[i]] [3] 1) {
            return str[i];
        }
    }
    return '_';
}

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

A256

Bi

C==

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong array size, wrong loop variable, or wrong comparison operator.

The frequency array size is 256, the loop variable is 'i', and the frequency check uses '=='.

Practice

1. What is the main purpose of using a hash (like a dictionary or map) to find the first non-repeating character in a string?

easy

A. To reverse the string quickly

B. To sort the characters alphabetically

C. To remove duplicate characters from the string

D. To count the frequency of each character efficiently

Solution

1. Step 1: Understand the role of hash in counting

   A hash (dictionary) stores characters as keys and their counts as values, allowing quick frequency counting.

2. Step 2: Identify why frequency counting helps

   Knowing how many times each character appears helps find the first character with count 1, which is the first non-repeating character.

3. Final Answer:

   To count the frequency of each character efficiently -> Option D

4. Quick Check:

   Hash counts characters = frequency count [OK]

Hint: Use hash to count chars, then find first with count 1 [OK]

Common Mistakes:

• Thinking hash sorts characters
• Assuming hash reverses string
• Using hash to remove duplicates only

2. Which of the following C code snippets correctly initializes a hash map (using an array) to count characters for ASCII strings?

easy

A. int count[256] = {0};

B. int count[] = NULL;

C. int count[256]; // uninitialized

D. int count = 256;

Solution

1. Step 1: Check correct array initialization

   In C, to count ASCII chars, an int array of size 256 initialized to zero is needed: int count[256] = {0};

2. Step 2: Identify incorrect options

   int count[] = NULL; sets pointer to NULL, no array. int count[256]; // uninitialized is uninitialized, may contain garbage. int count = 256; is a single int, not array.

3. Final Answer:

   int count[256] = {0}; -> Option A

4. Quick Check:

   Array size 256 with zero init = correct hash map [OK]

Hint: Use int count[256] = {0} for ASCII char frequency [OK]

Common Mistakes:

• Not initializing array to zero
• Using pointer instead of array
• Declaring single int instead of array

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

char str[] = "swiss";
int count[256] = {0};
for (int i = 0; str[i] != '\0'; i++) {
    count[(int)str[i]]++;
}
for (int i = 0; str[i] != '\0'; i++) {
    if (count[(int)str[i]] == 1) {
        printf("%c", str[i]);
        break;
    }
}

medium

A. s

B. i

C. w

D. No output

Solution

1. Step 1: Count character frequencies in "swiss"

   Characters: s(3 times), w(1 time), i(1 time)

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

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

3. Final Answer:

   w -> Option C

4. Quick Check:

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

Hint: Check counts, print first char with count 1 [OK]

Common Mistakes:

• Printing first char without checking count
• Confusing 'i' and 'w' counts
• Ignoring character order

4. Identify the bug in this C code snippet that tries to find the first non-repeating character:

char str[] = "hello";
int count[256];
for (int i = 0; str[i] != '\0'; i++) {
    count[(int)str[i]]++;
}
for (int i = 0; str[i] != '\0'; i++) {
    if (count[(int)str[i]] == 1) {
        printf("%c", str[i]);
        break;
    }
}

medium

A. count array is not initialized to zero

B. Loop condition is incorrect

C. Casting char to int is wrong

D. printf format specifier is incorrect

Solution

1. Step 1: Check count array initialization

   count array is declared but not initialized, so it contains garbage values causing wrong counts.

2. Step 2: Explain impact of uninitialized array

   Incrementing garbage values leads to incorrect frequency counts, so first non-repeating char cannot be found correctly.

3. Final Answer:

   count array is not initialized to zero -> Option A

4. Quick Check:

   Uninitialized count array causes wrong frequency [OK]

Hint: Always initialize count array to zero before counting [OK]

Common Mistakes:

• Forgetting to initialize count array
• Assuming default zero initialization
• Misusing loop conditions

5. Given the string "aabbccddeffg", which character will be identified as the first non-repeating character using a hash map counting method? Consider the code below:

char str[] = "aabbccddeffg";
int count[256] = {0};
for (int i = 0; str[i] != '\0'; i++) {
    count[(int)str[i]]++;
}
for (int i = 0; str[i] != '\0'; i++) {
    if (count[(int)str[i]] == 1) {
        printf("%c", str[i]);
        break;
    }
}

hard

A. g

B. e

C. d

D. f

Solution

1. Step 1: Count frequencies of each character

   Characters and counts: a(2), b(2), c(2), d(2), e(1), f(2), g(1)

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

   Order: a,b,c,d,e,f,g; e and g have count 1, but e appears before g.

3. Step 3: Check carefully the string order

   String is "aabbccddeffg"; 'e' appears once at position 8, 'g' at position 11. So first non-repeating is 'e'.

4. Final Answer:

   e -> Option B

5. Quick Check:

   First unique char in string = e [OK]

Hint: Check counts and order carefully to find first unique char [OK]

Common Mistakes:

• Choosing last unique char instead of first
• Ignoring character order in string
• Miscounting character frequencies