String Traversal and Character Access in DSA Typescript

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 from left to right to understand or find something.

"H" -> "e" -> "l" -> "l" -> "o" -> null
 ↑
 current position at first character

Dry Run Walkthrough

Input: string: "Hello", traverse and print each character

Goal: Visit each character in the string one by one and print it

Step 1: Start at first character 'H'

"H" -> "e" -> "l" -> "l" -> "o" -> null
 ↑

Why: We begin reading from the first character

Step 2: Print 'H' and move to next character 'e'

"H" -> "e" -> "l" -> "l" -> "o" -> null
    ↑

Why: After reading one character, move forward to continue

Step 3: Print 'e' and move to next character 'l'

"H" -> "e" -> "l" -> "l" -> "o" -> null
       ↑

Why: Keep moving forward to read all characters

Step 4: Print 'l' and move to next character 'l'

"H" -> "e" -> "l" -> "l" -> "o" -> null
          ↑

Why: Continue until the end of the string

Step 5: Print 'l' and move to next character 'o'

"H" -> "e" -> "l" -> "l" -> "o" -> null
             ↑

Why: Almost at the last character

Step 6: Print 'o' and reach end of string

"H" -> "e" -> "l" -> "l" -> "o" -> null
                ↑

Why: Finished reading all characters

Result:

"H" -> "e" -> "l" -> "l" -> "o" -> null
Printed characters: H e l l o

Annotated Code

DSA Typescript

class StringTraversal {
  static traverseAndPrint(str: string): void {
    for (let i = 0; i < str.length; i++) {
      console.log(str[i]); // print current character
    }
  }
}

// Driver code
const inputString = "Hello";
StringTraversal.traverseAndPrint(inputString);

for (let i = 0; i < str.length; i++) {

loop through each character index from start to end

console.log(str[i]); // print current character

access and print character at current index

Output

H

e

l

l

o

Complexity Analysis

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

Space: O(1) because we only use a few variables regardless of string size

vs Alternative: Compared to recursive traversal, this iterative approach uses less memory and is simpler

Edge Cases

empty string ""

loop does not run, nothing is printed

DSA Typescript

for (let i = 0; i < str.length; i++) {

string with one character "A"

prints single character 'A' once

DSA Typescript

for (let i = 0; i < str.length; 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 <= in loop condition causing out-of-bounds access
Fix: Use < operator to stop before string length

Mistake: Trying to modify string characters directly (strings are immutable)
Fix: Use a new string or array to build modified results

Summary

It visits each character in a string one by one to read or process it.

Use it when you want to check or use every character in a string.

Remember strings are like arrays of letters you can access by position.

Practice

1. What does the expression str[2] return when str = "hello"?

easy

A. The character 'e'

B. The character 'l'

C. The substring 'll'

D. An error because strings are not indexable

Solution

1. Step 1: Understand string indexing

   Strings in TypeScript can be accessed by index starting at 0, so index 2 refers to the third character.

2. Step 2: Identify the character at index 2 in "hello"

   The characters are: h(0), e(1), l(2), l(3), o(4). So index 2 is 'l'.

3. Final Answer:

   The character 'l' -> Option B

4. Quick Check:

   str[2] = 'l' [OK]

Hint: Remember string index starts at 0 [OK]

Common Mistakes:

• Thinking index starts at 1
• Confusing character with substring
• Assuming strings are not indexable

2. Which of the following is the correct way to loop through each character of a string text in TypeScript?

easy

A. for (let char of text) { console.log(char); }

B. for (let char in text) { console.log(char); }

C. for (let i = 0; i <= text.length; i++) { console.log(text[i]); }

D. for (let i = 1; i <= text.length; i++) { console.log(text[i]); }

Solution

1. Step 1: Understand looping over strings

   Using for...of loops over characters directly, which is the cleanest way.

2. Step 2: Evaluate the options

   The index-based loop using <= text.length causes an out-of-bounds access. for...in iterates over string indices ('0', '1', etc.) rather than characters. for...of correctly iterates over characters. Starting the index at 1 skips the first character and risks out-of-bounds.

3. Final Answer:

   for (let char of text) { console.log(char); } -> Option A

4. Quick Check:

   Use for...of to access characters directly [OK]

Hint: Use for...of to loop characters directly [OK]

Common Mistakes:

• Using for...in which loops keys, not values
• Starting index at 1 instead of 0
• Confusing index with character in loops

3. What is the output of the following code?

const s = "abcde";
let result = "";
for (let i = 0; i < s.length; i += 2) {
  result += s[i];
}
console.log(result);

medium

A. "bdf"

B. ""

C. "abcde"

D. "ace"

Solution

1. Step 1: Analyze the loop increments and indexing

   The loop starts at 0 and increments by 2 each time, so indexes are 0, 2, 4.

2. Step 2: Collect characters at indexes 0, 2, 4

   Characters are s[0] = 'a', s[2] = 'c', s[4] = 'e'. Concatenated result is "ace".

3. Final Answer:

   "ace" -> Option D

4. Quick Check:

   Indexes 0,2,4 characters combined = "ace" [OK]

Hint: Check loop step size and indexes accessed [OK]

Common Mistakes:

• Assuming loop increments by 1
• Confusing indexes with characters
• Expecting all characters concatenated

4. Identify the error in the following code snippet that tries to print each character of word:

const word = "test";
for (let i = 1; i <= word.length; i++) {
  console.log(word[i]);
}

medium

A. Missing declaration of variable i

B. Using <= instead of < causes infinite loop

C. Indexing starts at 1, skipping first character and causing undefined at end

D. Strings cannot be accessed by index in TypeScript

Solution

1. Step 1: Check loop index range

   The loop starts at 1 and goes up to word.length (4), so indexes accessed are 1,2,3,4.

2. Step 2: Identify indexing issues

   Index 0 is skipped (first character 't'), and index 4 is out of bounds (undefined). So first character is missed and last access is invalid.

3. Final Answer:

   Indexing starts at 1, skipping first character and causing undefined at end -> Option C

4. Quick Check:

   String indexes start at 0, not 1 [OK]

Hint: Loop indexes for strings start at 0 and go to length-1 [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using <= instead of < in loop condition
• Assuming strings are not indexable

5. Given a string input, which code snippet correctly reverses the string by traversing it from end to start?

hard

A. let rev = '';
for (let i = input.length - 1; i >= 0; i--) {
rev += input[i];
}
console.log(rev);

B. let rev = '';
for (let i = 0; i < input.length; i++) {
rev += input[input.length - i];
}
console.log(rev);

C. let rev = '';
for (let i = input.length; i > 0; i--) {
rev += input[i];
}
console.log(rev);

D. let rev = '';
for (let i = input.length; i >= 0; i--) {
rev += input[i];
}
console.log(rev);

Solution

1. Step 1: Understand correct index range for reverse traversal

   Indexes go from input.length - 1 down to 0 inclusive to cover all characters in reverse.

2. Step 2: Verify the index ranges

   Looping from input.length - 1 down to 0 inclusive correctly captures all characters in reverse. A forward loop using input.length - i first accesses an out-of-bounds index. Starting at input.length immediately accesses invalid index. Decrementing from input.length to 0 also starts with out-of-bounds.

3. Final Answer:

   let rev = ''; for (let i = input.length - 1; i >= 0; i--) { rev += input[i]; } console.log(rev); -> Option A

4. Quick Check:

   Loop from length-1 down to 0 for reverse traversal [OK]

Hint: Start loop at length-1 and go down to 0 [OK]

Common Mistakes:

• Starting loop at length instead of length-1
• Using wrong index inside loop causing undefined
• Loop condition off by one causing missing characters