String Traversal and Character Access in DSA C++ - Execution Trace

Concept Flow - String Traversal and Character Access

Start at index 0

↓

Check: index < string length?

No→Stop traversal

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 C++

std::string s = "hello";
for (int i = 0; i < s.length(); i++) {
    std::cout << s[i] << " ";
}

This code prints each character of the string 'hello' separated by spaces.

Execution Table

Step	i (index)	Condition i < length	Character Accessed	Output
1	0	0 < 5 (True)	'h'	h
2	1	1 < 5 (True)	'e'	e
3	2	2 < 5 (True)	'l'	l
4	3	3 < 5 (True)	'l'	l
5	4	4 < 5 (True)	'o'	o
6	5	5 < 5 (False)	N/A	Stop loop

💡 Loop stops when index i reaches 5, which is not less than string length 5.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
i	0	1	2	3	4	5	5
Output	""	"h "	"h e "	"h e l "	"h e l l "	"h e l l o "	"h e l l o "

Key Moments - 3 Insights

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

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

What does s[i] mean in the code?

s[i] accesses the character at position i in the string, for example s[0] is 'h' as shown in execution_table row 1.

Why do we start indexing from 0 instead of 1?

String indexing starts at 0 in C++, so the first character is at index 0, matching execution_table step 1.

Visual Quiz - 3 Questions

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

A 'e'

B 'l'

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 < length' column where it shows false in the execution_table.

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

A 2 steps

B 3 steps

C 5 steps

D 1 step

💡 Hint

Refer to the exit condition in the execution_table and relate it to the string length.

Concept Snapshot

String Traversal and Character Access:
- Use a loop from 0 to string length - 1
- Access characters with s[i]
- Loop stops when i equals string length
- Indexing starts at 0
- Commonly used to process or print each character

Full Transcript

This concept shows how to go through each character in a string one by one using a loop. We start at index 0, check if the index is less than the string length, then access the character at that index. We do something with that character, like print it, then move to the next index. When the index reaches the string length, the loop stops. This is how we read or process strings character by character in C++.

Practice

1. What does the expression str[i] do when str is a string in C++?

easy

A. Returns the length of the string

B. Accesses the character at index i in the string

C. Deletes the character at index i

D. Converts the string to uppercase

Solution

1. Step 1: Understand string indexing in C++

   In C++, strings are like arrays of characters. Using str[i] accesses the character at position i.

2. Step 2: Clarify what str[i] returns

   The expression returns the character stored at index i without modifying the string.

3. Final Answer:

   Accesses the character at index i in the string -> Option B

4. Quick Check:

   str[i] = character at i [OK]

Hint: Remember strings are arrays of chars; indexing gets a character [OK]

Common Mistakes:

• Thinking it returns string length
• Assuming it deletes characters
• Confusing with string case conversion

2. Which of the following is the correct way to loop through each character of a string str in C++?

easy

A. for (int i = 0; i <= str.length(); i++)

B. for (int i = 1; i < str.length(); i++)

C. for (int i = 0; i < str.length(); i++)

D. for (int i = 1; i <= str.length(); i++)

Solution

1. Step 1: Understand string length and indexing

   String indices start at 0 and go up to str.length() - 1. So loop must run while i < str.length().

2. Step 2: Check loop conditions

   for (int i = 0; i < str.length(); i++) uses i < str.length() starting from 0, which correctly covers all characters without going out of bounds.

3. Final Answer:

   for (int i = 0; i < str.length(); i++) -> Option C

4. Quick Check:

   Loop from 0 to length-1 [OK]

Hint: Loop from 0 to less than length to avoid out-of-bounds [OK]

Common Mistakes:

• Using <= causes out-of-bounds error
• Starting loop at 1 misses first character
• Confusing length with last index

3. What is the output of the following C++ code?

std::string s = "hello";
for (int i = 0; i < s.length(); i++) {
    if (s[i] == 'l') {
        std::cout << i << " ";
    }
}

medium

A. 0 1

B. 1 2

C. 3 4

D. 2 3

Solution

1. Step 1: Identify characters and their indices

   The string "hello" has characters: h(0), e(1), l(2), l(3), o(4).

2. Step 2: Check which indices have 'l'

   Characters at indices 2 and 3 are 'l', so those indices are printed.

3. Final Answer:

   2 3 -> Option D

4. Quick Check:

   Indices of 'l' = 2 3 [OK]

Hint: Check string indices carefully for matching characters [OK]

Common Mistakes:

• Confusing character positions
• Printing characters instead of indices
• Off-by-one errors in indexing

4. Identify the error in this code snippet that tries to print all characters of a string str:

for (int i = 0; i <= str.length(); i++) {
    std::cout << str[i];
}

medium

A. Using <= causes out-of-bounds access

B. Missing semicolon after for loop

C. str[i] cannot be used to access characters

D. Loop variable should be unsigned

Solution

1. Step 1: Analyze loop condition

   The loop runs while i <= str.length(), which means it tries to access str[str.length()].

2. Step 2: Understand string indexing limits

   Valid indices are from 0 to str.length() - 1. Accessing str[str.length()] is out-of-bounds and causes undefined behavior.

3. Final Answer:

   Using <= causes out-of-bounds access -> Option A

4. Quick Check:

   Loop must use <, not <= [OK]

Hint: Use < not <= to avoid accessing past string end [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Assuming string length is last index
• Ignoring out-of-bounds errors

5. Given a string str, how can you create a new string containing only the vowels from str using traversal and character access in C++?

hard

A. Loop through str, check if each character is a vowel, and append it to a new string

B. Use str.substr() to extract vowels directly

C. Replace all consonants with spaces in str

D. Sort the string and pick vowels from the start

Solution

1. Step 1: Traverse the string character by character

   Use a loop to access each character in str using indexing.

2. Step 2: Check if character is a vowel and build new string

   If the character is one of 'a', 'e', 'i', 'o', 'u' (case-insensitive), append it to a new string variable.

3. Step 3: Return or print the new string

   After the loop, the new string contains only vowels from the original string.

4. Final Answer:

   Loop through str, check if each character is a vowel, and append it to a new string -> Option A

5. Quick Check:

   Filter vowels by checking each character [OK]

Hint: Check each char; add vowels to new string [OK]

Common Mistakes:

• Trying to extract vowels without looping
• Using substr incorrectly
• Sorting string changes order