String Traversal and Character Access in DSA C++

Mental Model

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

Analogy: Like reading a sentence letter by letter to understand or find a specific letter.

S -> t -> r -> i -> n -> g -> \0
↑

Dry Run Walkthrough

Input: string: "cat"; goal: print each character on a new line

Goal: To visit each character in the string and print it separately

Step 1: Start at first character 'c'

c -> a -> t -> \0
↑

Why: We begin reading from the first character

Step 2: Print 'c' and move to next character 'a'

c -> a -> t -> \0
  ↑

Why: After processing 'c', move forward to continue traversal

Step 3: Print 'a' and move to next character 't'

c -> a -> t -> \0
    ↑

Why: Continue visiting each character in order

Step 4: Print 't' and move to null terminator '\0'

c -> a -> t -> \0
      ↑

Why: Reached last character, next is string end

Step 5: Stop traversal at '\0' (end of string)

c -> a -> t -> \0
        ↑

Why: Null terminator marks end, stop to avoid invalid access

Result:

c\na\nt\n

Annotated Code

DSA C++

#include <iostream>
#include <string>

void printChars(const std::string& str) {
    for (size_t i = 0; i < str.size(); ++i) {
        std::cout << str[i] << '\n'; // print current character
    }
}

int main() {
    std::string input = "cat";
    printChars(input);
    return 0;
}

for (size_t i = 0; i < str.size(); ++i) {

iterate over each character index from start to end

std::cout << str[i] << '\n';

access and print the character at current index

Output

c

a

t

Complexity Analysis

Time: O(n) because we visit each of the n characters once

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

vs Alternative: Direct indexing is simpler and faster than using iterators or copying characters

Edge Cases

empty string ""

The loop does not run and nothing is printed

DSA C++

for (size_t i = 0; i < str.size(); ++i) {

string with one character "x"

Prints the single character and stops

DSA C++

for (size_t i = 0; i < str.size(); ++i) {

When to Use This Pattern

When you need to process or examine each letter in a word or sentence, use string traversal to access characters one by one.

Common Mistakes

Mistake: Accessing characters beyond string length causing out-of-bounds error
Fix: Always check that index is less than string size before accessing

Summary

It visits each character in a string one by one.

Use it when you want to read or process every letter in a word or sentence.

Remember strings are sequences you can access by position from start to end.

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