String Traversal and Character Access in DSA Go

Mental Model

A string is a sequence of characters you can look at one by one from start to end.

Analogy: Imagine reading a sentence letter by letter with your finger pointing to each character in order.

S -> t -> r -> i -> n -> g -> null
↑

Dry Run Walkthrough

Input: string: "Go!"

Goal: Visit each character in the string and print it separately

Step 1: Start at the first character

G -> o -> ! -> null
↑

Why: We begin reading from the first character

Step 2: Access character 'G' and print it

G -> o -> ! -> null
  ↑

Why: We want to see the first character

Step 3: Move to next character 'o' and print it

G -> o -> ! -> null
    ↑

Why: Continue reading the string in order

Step 4: Move to next character '!' and print it

G -> o -> ! -> null
      ↑

Why: Print the last character

Step 5: Reached end of string, stop

G -> o -> ! -> null
        ↑

Why: No more characters to read

Result:

G -> o -> ! -> null
Printed characters: G\no\n!\n

Annotated Code

DSA Go

package main

import "fmt"

func main() {
    s := "Go!"
    for i := 0; i < len(s); i++ {
        // Access character at index i
        c := s[i]
        fmt.Printf("%c\n", c) // Print character
    }
}

for i := 0; i < len(s); i++ {

loop through each index of the string from start to end

c := s[i]

access the character at current index i

fmt.Printf("%c\n", c)

print the character followed by a newline

Output

G

o

!

Complexity Analysis

Time: O(n) because we visit each character once where n is string length

Space: O(1) because we only store one character at a time during traversal

vs Alternative: Compared to converting string to a slice of runes, this uses less memory but may not handle multi-byte characters correctly

Edge Cases

empty string ""

loop does not run, no output printed

DSA Go

for i := 0; i < len(s); i++ {

string with one character "A"

prints single character 'A' then stops

DSA Go

for i := 0; i < len(s); i++ {

When to Use This Pattern

When you need to process or examine each character in a string one by one, use string traversal with indexing.

Common Mistakes

Mistake: Using indexing without considering multi-byte characters in Go
Fix: Use range loop over string to handle Unicode characters correctly or convert to rune slice

Summary

It visits each character in a string one by one using index positions.

Use it when you want to read or process every character in order.

Remember a string is like a line of characters you can point to and read one at a time.

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