String Traversal and Character Access in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character access in a loop

What is the output of the following C++ code snippet?

DSA C++

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

A h-e-l-l-o-

B hello-

C h e l l o

D h-e-l-l-o

💡 Hint

Remember that s[i] accesses one character at a time and the loop prints a dash after each character.

The loop prints each character followed by a dash. Since the loop runs from 0 to s.size()-1, it prints all characters with a dash after each, including the last one.

❓ Predict Output

intermediate

Accessing characters with at() method

What will be the output of this C++ code?

DSA C++

std::string s = "world";
for (size_t i = 0; i < s.length(); ++i) {
    std::cout << s.at(i);
}

A w-o-r-l-d

B w o r l d

C dlrow

D world

💡 Hint

The at() method returns the character at the given index without extra characters.

The loop prints each character without spaces or separators, so the output is the string itself.

❓ Predict Output

advanced

Output of nested loops with string indexing

What is the output of this C++ code snippet?

DSA C++

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

A a ab abc

B a a b abc

C cba ba a

D a b c

💡 Hint

The inner loop prints characters from start up to the current outer loop index.

For i=0, prints 'a'; i=1 prints 'ab'; i=2 prints 'abc', each followed by a space.

❓ Predict Output

advanced

Result of modifying characters via indexing

What is the output after running this C++ code?

DSA C++

std::string s = "test";
s[1] = 'a';
s[3] = 'p';
std::cout << s;

A tesp

B tasp

C tast

D teap

💡 Hint

Characters at index 1 and 3 are replaced.

Original string is 'test'. s[1] = 'a' changes 'e' to 'a', s[3] = 'p' changes 't' to 'p'. Result is 'tasp'.

🧠 Conceptual

expert

Understanding string traversal with iterators

Given the following C++ code, what is the output?

DSA C++

std::string s = "data";
for (auto it = s.begin(); it != s.end(); ++it) {
    if (*it == 'a') {
        std::cout << '*';
    } else {
        std::cout << *it;
    }
}

A data

B d*ta

C d*t*

D d*at

💡 Hint

The code replaces every 'a' character with '*'.

The loop prints each character. When the character is 'a', it prints '*', else prints the character itself. So 'data' becomes 'd*t*'.

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