String Traversal and Character Access in DSA Javascript - Time & Space Complexity

Time Complexity: String Traversal and Character Access

O(n)

Understanding Time Complexity

We want to understand how long it takes to look at each letter in a word or sentence.

How does the time grow when the word gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code goes through each letter in the string and prints it out one by one.

Identify Repeating Operations

• Primary operation: Accessing each character in the string one by one.
• How many times: Exactly once for each character in the string.

How Execution Grows With Input

As the string gets longer, the number of steps grows in the same way.

Input Size (n)	Approx. Operations
10	10 character accesses
100	100 character accesses
1000	1000 character accesses

Pattern observation: The work grows directly with the length of the string.

Final Time Complexity

Time Complexity: O(n)

This means the time to finish grows in a straight line as the string gets longer.

Common Mistake

[X] Wrong: "Accessing characters inside a string is instant and does not depend on string length."

[OK] Correct: Even though each access is quick, you still have to do it for every character, so total time grows with string length.

Interview Connect

Understanding how time grows when reading strings helps you explain and improve code that works with text data.

Self-Check

"What if we only accessed every other character in the string? How would the time complexity change?"

Practice

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

easy

A. The first character of the string, which is 'h'

B. The third character of the string, which is 'l'

C. The length of the string, which is 5

D. An error because strings cannot be accessed by index

Solution

1. Step 1: Understand zero-based indexing in strings

   Strings in JavaScript start counting characters from index 0, so index 2 is 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 third character of the string, which is 'l' -> Option B

4. Quick Check:

   str[2] = 'l' [OK]

Hint: Remember string indexes start at 0, so 2 means third character [OK]

Common Mistakes:

• Thinking index 2 is the second character
• Confusing string length with character access
• Believing strings can't be indexed like arrays

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

easy

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

B. for (let char of text) { console.log(text[char]); }

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

D. for (let i = 0; i < text.length; i++) { console.log(text.charAt(i+1)); }

Solution

1. Step 1: Check the loop range and indexing

   The loop should start at 0 and go up to less than text.length to cover all characters.

2. Step 2: Verify character access inside the loop

   Accessing characters by text[i] is correct. for (let i = 0; i < text.length; i++) { console.log(text[i]); } uses this correctly.

3. Final Answer:

   for (let i = 0; i < text.length; i++) { console.log(text[i]); } -> Option A

4. Quick Check:

   Loop from 0 to length-1 and access text[i] [OK]

Hint: Loop from 0 to length-1 and use text[i] to access chars [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using text[char] when char is a character, not index
• Accessing index i+1 causing off-by-one errors

3. What will be the output of the following code?

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

medium

A. "wrd"

B. "wolrd"

C. "orld"

D. "wrld"

Solution

1. Step 1: Understand the loop and condition

   The loop goes from 0 to 4 (length 5). It adds characters at even indexes (0,2,4) to result.

2. Step 2: Identify characters at even indexes

   Index 0: 'w', 2: 'r', 4: 'd'. So result becomes "wrd".

3. Final Answer:

   "wrd" -> Option A

4. Quick Check:

   Characters at even indexes = "wrd" [OK]

Hint: Pick characters where index % 2 is zero [OK]

Common Mistakes:

• Including characters at odd indexes
• Confusing index positions with characters
• Appending characters incorrectly

4. The following code is intended to print each character of word on a new line, but it prints an extra undefined. What is the problem?

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

medium

A. Strings cannot be accessed by index in JavaScript

B. console.log cannot print string characters

C. The variable i is not declared

D. The loop condition should be i < word.length instead of <=

Solution

1. Step 1: Analyze the loop condition

   The loop runs from i = 0 to i = word.length (which is 4). Index 4 is out of bounds for string indices 0-3.

2. Step 2: Identify the fix

   Change the condition to i < word.length to avoid accessing undefined index.

3. Final Answer:

   The loop condition should be i < word.length instead of <= -> Option D

4. Quick Check:

   Loop must stop before length to avoid undefined index [OK]

Hint: Use i < length not <= to avoid out-of-bounds [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Thinking strings can't be indexed
• Ignoring off-by-one errors

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

hard

A. let reversed = '';
for (let i = inputStr.length; i > 0; i--) {
reversed += inputStr[i];
}

B. let reversed = '';
for (let i = 0; i < inputStr.length; i++) {
reversed += inputStr[inputStr.length - i];
}

C. let reversed = '';
for (let i = inputStr.length - 1; i >= 0; i--) {
reversed += inputStr[i];
}

D. let reversed = '';
for (let i = inputStr.length - 1; i > 0; i--) {
reversed += inputStr[i];
}

Solution

1. Step 1: Understand string indexing and loop bounds

   String indices go from 0 to length-1. To reverse, start at length-1 down to 0 inclusive.

2. Step 2: Check each option's loop and indexing

   let reversed = ''; for (let i = inputStr.length - 1; i >= 0; i--) { reversed += inputStr[i]; } loops from length-1 to 0 inclusive and appends characters correctly. Others have off-by-one errors or wrong indices.

3. Final Answer:

   let reversed = ''; for (let i = inputStr.length - 1; i >= 0; i--) { reversed += inputStr[i]; } -> Option C

4. Quick Check:

   Loop backward from last index to 0 and append chars [OK]

Hint: Loop backward from length-1 to 0 and append characters [OK]

Common Mistakes:

• Using wrong loop bounds causing missing first/last char
• Accessing index equal to length (out of bounds)
• Stopping loop before reaching index 0