String Traversal and Character Access in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print each character of the string word using a for loop.

DSA Go

word := "hello"
for [1] := 0; [1] < len(word); [1]++ {
    fmt.Printf("%c ", word[[1]])
}

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

Ai

Bj

Ck

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in the for loop parts.

Starting the index from 1 instead of 0.

Using a variable not declared or unrelated.

The variable 'i' is commonly used as the index in loops. It starts at 0 and goes up to len(word)-1 to access each character.

2fill in blank

medium

Complete the code to print the ASCII value of each character in the string text.

DSA Go

text := "GoLang"
for i := 0; i < len(text); i++ {
    fmt.Println([1])
}

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

Atext

Bstring(text[i])

Ctext[i]

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the whole string instead of a character.

Converting the character to string which prints the character, not ASCII.

Printing the index variable instead of the character.

Accessing text[i] gives the byte value (ASCII) of the character at index i.

3fill in blank

hard

Fix the error in the code to correctly print each character of sentence as a rune (Unicode character).

DSA Go

sentence := "Hi!"
for i := 0; i < len(sentence); i++ {
    fmt.Printf("%c ", [1])
}

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

Arune(sentence[i])

Bsentence

Csentence[i]

Dstring(sentence[i])

Attempts:

3 left

💡 Hint

Common Mistakes

Printing sentence[i] directly which may not handle Unicode correctly.

Using string(sentence[i]) which converts to string, not rune.

Printing the whole sentence instead of characters.

Casting to rune ensures the byte is treated as a Unicode character for printing.

4fill in blank

hard

Fill both blanks to create a map of characters to their counts from the string input.

DSA Go

input := "banana"
counts := make(map[rune]int)
for _, [1] := range input {
    counts[[2]]++
}

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

Ach

Bi

Didx

Attempts:

3 left

💡 Hint

Common Mistakes

Using the index variable instead of the character as map key.

Using different variable names in the loop and map access.

Not using rune type for map keys.

The variable 'ch' holds each character (rune) from the string. We use it as the key to count occurrences.

5fill in blank

hard

Fill all three blanks to create a slice of runes from str and print each rune.

DSA Go

str := "code"
runes := []rune([1])
for [2], [3] := range runes {
    fmt.Printf("%c ", r)
}

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

Astr

Bi

Cr

Drunes

Attempts:

3 left

💡 Hint

Common Mistakes

Using the string directly in the loop without conversion.

Using wrong variable names in the loop.

Not converting string to runes causing incorrect character handling.

Convert string to rune slice with []rune(str). Use 'i' and 'r' as index and rune in the loop.

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