String Traversal and Character Access in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of rune traversal in Go string

What is the output of the following Go code that traverses a string using range and prints each rune's integer value?

DSA Go

package main
import "fmt"
func main() {
    s := "Go😊"
    for _, r := range s {
        fmt.Print(int(r), " ")
    }
}

A 71 111 128522

B 71 111 55357 56842

C 71 111 63

D 71 111 128522 32

💡 Hint

Remember that range over a string in Go iterates over runes, not bytes.

The range loop over string s iterates over runes (Unicode code points). 'G' is 71, 'o' is 111, and the emoji 😊 is a single rune with code point 128522.

❓ Predict Output

intermediate

Indexing bytes vs runes in Go strings

What is the output of this Go code that prints bytes and runes of a string containing multibyte characters?

DSA Go

package main
import "fmt"
func main() {
    s := "Hi😊"
    fmt.Print("Bytes: ")
    for i := 0; i < len(s); i++ {
        fmt.Print(s[i], " ")
    }
    fmt.Print("\nRunes: ")
    for _, r := range s {
        fmt.Print(r, " ")
    }
}

A

Bytes: 72 105 63 
Runes: 72 105 128522 

B

Bytes: 72 105 128522 
Runes: 72 105 128522 

C

Bytes: 72 105 240 159 152 138 
Runes: 72 105 55357 56842 

D

Bytes: 72 105 240 159 152 138 
Runes: 72 105 128522 

💡 Hint

Bytes represent raw UTF-8 encoding, runes represent Unicode code points.

The emoji 😊 is encoded in UTF-8 as 4 bytes: 240, 159, 152, 138. The range loop reads it as one rune 128522.

🧠 Conceptual

advanced

Why direct indexing on Go strings can be misleading

Why can using direct indexing (s[i]) on a Go string with multibyte characters lead to incorrect character access?

A Because s[i] accesses bytes, not runes, so multibyte characters are split into multiple bytes.

B Because s[i] returns runes, but runes are not Unicode characters.

C Because Go strings are mutable and indexing changes the string.

D Because s[i] returns the character's position, not its value.

💡 Hint

Think about UTF-8 encoding and how Go stores strings.

Go strings are byte slices. Indexing returns a byte, not a full Unicode character (rune). Multibyte characters are multiple bytes, so indexing can split them.

❓ Predict Output

advanced

Output of string slicing with multibyte characters in Go

What is the output of this Go code that slices a string containing an emoji?

DSA Go

package main
import "fmt"
func main() {
    s := "Hi😊"
    fmt.Println(s[0:2])
    fmt.Println(s[2:])
}

A

Hi
😊

B

Hi
panic: slicing out of range

C

Hi
�

D

Hi
�😊

💡 Hint

Remember slicing works on bytes, not runes, so slicing inside a multibyte character can cause invalid UTF-8.

The first slice s[0:2] is bytes for 'H' and 'i'. The second slice s[2:] starts inside the emoji's 4-byte UTF-8 sequence, causing invalid UTF-8 and printing replacement character �.

🔧 Debug

expert

Identify the error in rune counting code

What error will this Go code produce when counting runes in a string, and why?

DSA Go

package main
import "fmt"
func main() {
    s := "Hello😊"
    count := 0
    for i := 0; i < len(s); i++ {
        if s[i] >= 128 {
            count++
        }
    }
    fmt.Println(count)
}

A It prints 1, correctly counting the emoji as one rune.

B It prints 4, counting bytes with value >= 128, not runes.

C It causes a runtime panic due to invalid indexing.

D It prints 0, because s[i] is a rune and none are >= 128.

💡 Hint

Consider what s[i] returns and how multibyte characters are encoded.

s[i] accesses bytes, so the emoji's 4 bytes with values >= 128 are counted individually, resulting in 4 instead of 1 rune.

Practice

1. What does the expression len(s) return when s is a string in Go?

easy

A. The first character of the string

B. The number of characters (runes) in the string

C. The last character of the string

D. The number of bytes in the string

Solution

1. Step 1: Understand what len does on strings in Go

   In Go, len(s) returns the number of bytes, not characters, because strings are byte slices.

2. Step 2: Differentiate bytes and characters

   Characters (runes) can be multiple bytes, so len counts bytes, not runes.

3. Final Answer:

   The number of bytes in the string -> Option D

4. Quick Check:

   len(s) = bytes count [OK]

Hint: Remember: len counts bytes, not characters in Go strings [OK]

Common Mistakes:

• Thinking len counts characters (runes)
• Confusing bytes with characters
• Assuming len returns a character

2. Which of the following is the correct way to access the third character of a string s in Go?

easy

A. s[3]

B. s[2]

C. s.charAt(2)

D. s.get(2)

Solution

1. Step 1: Recall Go string indexing

   Go strings can be indexed like byte slices, starting at 0, so the third character is at index 2.

2. Step 2: Check syntax options

   Only s[2] is valid Go syntax for accessing a byte at index 2.

3. Final Answer:

   s[2] -> Option B

4. Quick Check:

   Indexing starts at 0, third char = s[2] [OK]

Hint: Indexing starts at 0; third char is s[2] in Go [OK]

Common Mistakes:

• Using 1-based indexing like s[3]
• Using methods like charAt or get which don't exist in Go
• Confusing bytes with runes

3. What is the output of the following Go code?

package main
import "fmt"
func main() {
  s := "GoLang"
  fmt.Println(string(s[1]))
}

medium

A. o

B. G

C. 1

D. Syntax error

Solution

1. Step 1: Identify the character at index 1

   String "GoLang" indexed at 1 is byte for 'o'.

2. Step 2: Convert byte to string for printing

   Using string(s[1]) converts byte 'o' to string "o".

3. Final Answer:

   o -> Option A

4. Quick Check:

   s[1] = 'o' printed as string [OK]

Hint: s[index] gives byte; convert to string to print char [OK]

Common Mistakes:

• Printing byte as number instead of string
• Confusing index 1 with 0
• Expecting syntax error

4. What is wrong with this Go code snippet?

package main
import "fmt"
func main() {
  s := "hello"
  for i := 0; i <= len(s); i++ {
    fmt.Printf("%c", s[i])
  }
}

medium

A. Loop condition causes index out of range error

B. fmt.Printf format specifier is incorrect

C. String s is not initialized properly

D. for loop syntax is invalid

Solution

1. Step 1: Analyze loop range

   The loop runs from i=0 to i=len(s) inclusive, but valid indices are 0 to len(s)-1.

2. Step 2: Identify error cause

   At i=len(s), s[i] causes index out of range runtime error.

3. Final Answer:

   Loop condition causes index out of range error -> Option A

4. Quick Check:

   Index must be less than len(s) [OK]

Hint: Loop index must be < len(s), not <= len(s) [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Ignoring runtime index out of range error
• Misunderstanding string indexing

5. Given a string s containing Unicode characters, which Go code snippet correctly prints each character (rune) on a new line?

hard

A. for i := 0; i < len(s); i++ { fmt.Println(string(s[i])) }

B. for i := range s { fmt.Println(s[i]) }

C. for _, r := range s { fmt.Println(string(r)) }

D. for i := 0; i <= len(s); i++ { fmt.Println(s[i]) }

Solution

1. Step 1: Understand Unicode handling in Go strings

   Go strings are byte slices; indexing gives bytes, not runes. To get runes, use range loop.

2. Step 2: Check each option

   for _, r := range s { fmt.Println(string(r)) } uses for _, r := range s which iterates runes correctly and prints each rune as string.

3. Final Answer:

   for _, r := range s { fmt.Println(string(r)) } -> Option C

4. Quick Check:

   Use range to iterate runes in string [OK]

Hint: Use range loop to get runes, not byte indexing [OK]

Common Mistakes:

• Using byte indexing for Unicode characters
• Looping with <= len(s)
• Printing bytes directly without conversion