String Basics and Memory Representation in DSA Go - 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.

How does the computer handle each character and how long does it take?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

import "fmt"

func printChars(s string) {
    for i := 0; i < len(s); i++ {
        fmt.Printf("%c ", s[i])
    }
    fmt.Println()
}

func main() {
    printChars("hello")
}

This code prints each character of a string one by one by accessing each character in order.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character of the string.
• How many times: Exactly once for each character in the string (length n).

How Execution Grows With Input

As the string gets longer, the number of steps grows directly with the number of characters.

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

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 process the string grows directly with its length.

Common Mistake

[X] Wrong: "Accessing a character in a string is instant and does not depend on string length."

[OK] Correct: While accessing a character by index is fast, looping through all characters means the total time depends on how many characters there are.

Interview Connect

Understanding how string operations scale helps you write efficient code and explain your reasoning clearly in interviews.

Self-Check

"What if we used recursion to print each character instead of a loop? How would the time complexity change?"

Practice

1. What does a string represent in Go language?
var s string = "hello"

easy

A. A sequence of bytes representing characters

B. A mutable array of characters

C. A pointer to an integer

D. A floating-point number

Solution

1. Step 1: Understand Go string type

   In Go, a string is a sequence of bytes that represent characters, usually UTF-8 encoded.

2. Step 2: Check mutability and type

   Strings in Go are immutable, so they cannot be changed after creation. They are not arrays or pointers to integers.

3. Final Answer:

   A sequence of bytes representing characters -> Option A

4. Quick Check:

   Go strings = immutable byte sequences [OK]

Hint: Remember: Go strings are immutable byte sequences [OK]

Common Mistakes:

• Thinking strings are mutable arrays
• Confusing strings with pointers
• Assuming strings store integers

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

easy

A. var s string = 'hello'

B. var s string = "hello"

C. var s = hello

D. string s = "hello"

Solution

1. Step 1: Check string literal syntax in Go

   Go uses double quotes "" for string literals, single quotes '' are for rune (character) literals.

2. Step 2: Validate declaration syntax

   Correct declaration uses var keyword, type string, and double quotes for value.

3. Final Answer:

   var s string = "hello" -> Option B

4. Quick Check:

   Use double quotes for strings in Go [OK]

Hint: Use double quotes "" for strings, single quotes '' for runes [OK]

Common Mistakes:

• Using single quotes for strings
• Omitting quotes around string value
• Using incorrect variable declaration syntax

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

package main
import "fmt"
func main() {
    s := "GoLang"
    fmt.Println(len(s))
}

medium

A. 6

B. 7

C. 5

D. Error

Solution

1. Step 1: Understand the string value

   The string "GoLang" has 6 characters: G, o, L, a, n, g.

2. Step 2: Use len function on string

   len(s) returns the number of bytes in the string. Since all characters are ASCII, length is 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   len("GoLang") = 6 [OK]

Hint: len() returns byte count for ASCII strings [OK]

Common Mistakes:

• Counting characters incorrectly
• Expecting len() to count runes instead of bytes
• Assuming error due to missing imports

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

package main
import "fmt"
func main() {
    s := "hello"
    s[0] = 'H'
    fmt.Println(s)
}

medium

A. No error; code runs fine

B. Missing import for fmt package

C. Incorrect string declaration syntax

D. Strings are immutable; cannot assign to s[0]

Solution

1. Step 1: Check string mutability in Go

   Strings in Go are immutable, so you cannot change individual bytes like s[0].

2. Step 2: Analyze the assignment statement

   Attempting s[0] = 'H' causes a compile-time error because strings cannot be modified.

3. Final Answer:

   Strings are immutable; cannot assign to s[0] -> Option D

4. Quick Check:

   Strings immutable = no s[0] assignment [OK]

Hint: Remember: Go strings cannot be changed after creation [OK]

Common Mistakes:

• Trying to modify string bytes directly
• Ignoring immutability errors
• Confusing strings with byte slices

5. Given the string s := "Go😊Lang", what will len(s) return and why?

hard

A. 7, because len counts characters

B. 8, because emoji counts as two bytes

C. 10, because emoji is 4 bytes plus other characters

D. 6, because emoji is ignored

Solution

1. Step 1: Analyze string content and encoding

   The string "Go😊Lang" has 7 characters: G, o, 😊, L, a, n, g. The emoji 😊 is a Unicode character encoded in UTF-8 using 4 bytes.

2. Step 2: Calculate total byte length

   Characters G, o, L, a, n, g are 1 byte each (6 bytes). The emoji is 4 bytes. Total bytes = 6 + 4 = 10 bytes.

3. Step 3: Verify len(s) output

   len(s) returns the number of bytes, so it returns 10, not the number of characters.

4. Final Answer:

   10, because emoji is 4 bytes plus other characters -> Option C

5. Quick Check:

   len counts bytes, emoji uses multiple bytes [OK]

Hint: len() counts bytes, not characters; emojis use multiple bytes [OK]

Common Mistakes:

• Assuming len counts characters, not bytes
• Ignoring multi-byte encoding of emojis
• Thinking emoji counts as one byte