String Basics and Memory Representation in DSA C - Time & Space Complexity

Time Complexity: String Basics and Memory Representation

O(n)

Understanding Time Complexity

We want to understand how the time to work with strings grows as the string gets longer.

Specifically, how does reading or copying a string change when the string size changes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

void copyString(char *dest, const char *src) {
    int i = 0;
    while (src[i] != '\0') {
        dest[i] = src[i];
        i++;
    }
    dest[i] = '\0';
}
    

This code copies a string from source to destination character by character until it reaches the end.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Copying each character from source to destination.
• How many times: Once for every character in the string until the end character '\0'.

How Execution Grows With Input

As the string gets longer, the number of characters copied grows directly with its length.

Input Size (n)	Approx. Operations
10	About 10 character copies
100	About 100 character copies
1000	About 1000 character copies

Pattern observation: The work grows in a straight line with the string length.

Final Time Complexity

Time Complexity: O(n)

This means the time to copy the string grows directly with the number of characters in the string.

Common Mistake

[X] Wrong: "Copying a string takes the same time no matter how long it is."

[OK] Correct: Each character must be copied one by one, so longer strings take more time.

Interview Connect

Understanding how string operations scale helps you reason about performance in real programs and shows you know how memory and loops work together.

Self-Check

"What if we used a built-in function like strcpy instead of a loop? How would the time complexity change?"

Practice

1. Which of the following correctly describes how strings are stored in C?

easy

A. As an array of characters ending with a null character '\0'.

B. As a linked list of characters.

C. As a single integer representing the string length.

D. As a floating-point number encoding characters.

Solution

1. Step 1: Understand C string storage

   In C, strings are stored as arrays of characters with a special null character '\0' marking the end.

2. Step 2: Compare options with this knowledge

   Only As an array of characters ending with a null character '\0'. correctly states this; others describe different data structures or types.

3. Final Answer:

   As an array of characters ending with a null character '\0'. -> Option A

4. Quick Check:

   C strings end with '\0' = A [OK]

Hint: Remember C strings always end with '\0' to mark their end [OK]

Common Mistakes:

• Thinking strings are linked lists in C
• Assuming strings store length explicitly
• Confusing string storage with other data types

2. Which of the following is the correct way to declare and initialize a string literal in C?

easy

A. char str = "hello";

B. char str[5] = 'hello';

C. char str[] = "hello";

D. char str[] = {'h','e','l','l','o'};

Solution

1. Step 1: Check string declaration syntax

   In C, to store a string literal, use an array of chars with double quotes and square brackets.

2. Step 2: Evaluate each option

   A uses char (single char) with double quotes (wrong). B uses single quotes for multiple chars (wrong). D misses null terminator '\0'. C correctly declares and initializes with null terminator.

3. Final Answer:

   char str[] = "hello"; -> Option C

4. Quick Check:

   Use char array with double quotes = C [OK]

Hint: Use char array with double quotes for string literals [OK]

Common Mistakes:

• Using single quotes for strings
• Not including null terminator
• Declaring char variable instead of array

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

char str[] = "abc";
printf("%c", str[3]);

medium

A. c

B. \0 (null character, prints nothing visible)

C. a

D. Runtime error

Solution

1. Step 1: Understand string indexing and null terminator

   The string "abc" is stored as {'a','b','c','\0'}. Index 3 is the null character '\0'.

2. Step 2: Determine printf output for str[3]

   Printing '%c' with '\0' prints nothing visible (null character), no error occurs.

3. Final Answer:

   \0 (null character, prints nothing visible) -> Option B

4. Quick Check:

   Index 3 is '\0' = A [OK]

Hint: Remember string ends with '\0' at last index + 1 [OK]

Common Mistakes:

• Assuming str[3] is 'a' or 'c'
• Expecting runtime error for accessing null char
• Confusing null char with printable character

4. Identify the error in the following C code snippet:

char str[5] = "hello";
str[5] = '!';

medium

A. Array index out of bounds when assigning str[5].

B. Incorrect use of single quotes for string.

C. Missing null terminator in string initialization.

D. No error, code runs fine.

Solution

1. Step 1: Analyze array size and indexing

   Array str has size 5, valid indices are 0 to 4. Accessing str[5] is out of bounds.

2. Step 2: Check string initialization and assignment

   "hello" fits exactly 5 chars including '\0'? No, 'hello' needs 6 chars including '\0'. But main error is str[5] assignment which is invalid index.

3. Final Answer:

   Array index out of bounds when assigning str[5]. -> Option A

4. Quick Check:

   Index 5 invalid for size 5 array = D [OK]

Hint: Array indices go from 0 to size-1 only [OK]

Common Mistakes:

• Ignoring out of bounds access
• Confusing array size with string length
• Thinking str[5] is valid index

5. Given the code below, what will be the output?

char str1[] = "abc";
char *str2 = str1;
str2[1] = 'z';
printf("%s", str1);

hard

A. abc

B. Runtime error

C. a\0c

D. azc

Solution

1. Step 1: Understand pointer and array relation

   str2 points to str1 array. Changing str2[1] changes str1[1].

2. Step 2: Apply the character change and print

   str2[1] = 'z' changes 'b' to 'z'. Printing str1 shows "azc".

3. Final Answer:

   azc -> Option D

4. Quick Check:

   Pointer modifies original array = B [OK]

Hint: Pointer to array modifies original string content [OK]

Common Mistakes:

• Thinking str1 remains unchanged
• Expecting null character insertion
• Assuming runtime error due to pointer