First Non Repeating Character Using Hash in DSA C - Time & Space Complexity

Time Complexity: First Non Repeating Character Using Hash

O(n)

Understanding Time Complexity

We want to know how the time to find the first non-repeating character changes as the input string gets longer.

How does the number of steps grow when the string size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <stdio.h>
#include <string.h>

char firstNonRepeatingChar(char *str) {
    int count[256] = {0};
    int i;
    int len = strlen(str);

    for (i = 0; i < len; i++) {
        count[(unsigned char)str[i]]++;
    }

    for (i = 0; i < len; i++) {
        if (count[(unsigned char)str[i]] == 1) {
            return str[i];
        }
    }

    return '\0';
}

This code counts how many times each character appears, then finds the first character that appears only once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two loops over the input string.
• How many times: Each loop runs once over all characters, so twice over the string length.

How Execution Grows With Input

As the string gets longer, the number of steps grows roughly twice as fast because we scan the string two times.

Input Size (n)	Approx. Operations
10	About 20 steps
100	About 200 steps
1000	About 2000 steps

Pattern observation: The operations grow linearly with input size, doubling because of two passes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the first non-repeating character grows directly in proportion to the string length.

Common Mistake

[X] Wrong: "Because there are two loops, the time complexity is O(n²)."

[OK] Correct: The loops run one after another, not nested, so their times add up, not multiply.

Interview Connect

Understanding how to analyze simple loops like this helps you explain your code clearly and shows you know how to reason about efficiency.

Self-Check

"What if we used a nested loop to check each character against all others? How would the time complexity change?"

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