String Traversal and Character Access in DSA Typescript - Execution Trace

Concept Flow - String Traversal and Character Access

Start at index = 0

↓

Check: index < string length?

No→END

Yes↓

Access character at index

↓

Process character (e.g., print)

↓

Increment index by 1

↩Back to Check

This flow shows how we start from the first character of a string and move through each character one by one until we reach the end.

Execution Sample

DSA Typescript

const str = "hello";
for(let i = 0; i < str.length; i++) {
  console.log(str[i]);
}

This code prints each character of the string "hello" one by one.

Execution Table

Step	Index i	Condition i < str.length	Character Accessed	Action
1	0	0 < 5 = true	"h"	Print 'h'
2	1	1 < 5 = true	"e"	Print 'e'
3	2	2 < 5 = true	"l"	Print 'l'
4	3	3 < 5 = true	"l"	Print 'l'
5	4	4 < 5 = true	"o"	Print 'o'
6	5	5 < 5 = false	N/A	Exit loop

💡 Index i reaches 5, condition 5 < 5 is false, loop ends.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
i	0	1	2	3	4	5	5
Character Accessed	"h"	"e"	"l"	"l"	"o"	N/A	N/A

Key Moments - 3 Insights

Why does the loop stop when i equals the string length?

Because the condition i < str.length becomes false at i = 5 (string length), so the loop exits as shown in step 6 of the execution_table.

How do we access each character in the string?

We use str[i] where i is the current index, as shown in the Character Accessed column in the execution_table.

What happens if we try to access str[i] when i equals the string length?

It is out of bounds and returns undefined or causes an error, but the loop prevents this by stopping before i reaches str.length, as shown in the exit condition.

Visual Quiz - 3 Questions

Look at the execution_table, what character is accessed at step 3?

A "l"

B "e"

C "h"

D "o"

💡 Hint

Check the 'Character Accessed' column at step 3 in the execution_table.

At which step does the loop condition become false?

A Step 5

B Step 6

C Step 4

D Step 3

💡 Hint

Look at the 'Condition i < str.length' column to find when it becomes false.

If the string was "hi" instead of "hello", how many times would the loop run?

A 2 times

B 5 times

C 3 times

D 1 time

💡 Hint

Check the length of the string and compare with the loop condition in the execution_table.

Concept Snapshot

String Traversal and Character Access:
- Use a loop from index 0 to string length - 1
- Access characters with str[i]
- Stop when index reaches string length
- Each iteration processes one character
- Prevent out-of-bounds by checking condition

Full Transcript

This lesson shows how to go through each character in a string one by one. We start at index zero and check if the index is less than the string's length. If yes, we access the character at that index and do something with it, like print it. Then we move to the next index. We repeat this until the index reaches the string length, then we stop. This way, we safely access all characters without errors.

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