String algorithms are among the most common topics in JavaScript interviews.

At first glance, they seem straightforward. Reverse a string, check whether it’s a palindrome, or remove a character. None of these problems look particularly difficult.

However, these exercises teach much more than simple string manipulation. They reveal how well you understand JavaScript fundamentals, immutable data structures, algorithmic thinking, and time and space complexity.

In this article, we’ll explore three classic interview questions:

  • Reversing a string
  • Checking whether a string is a palindrome
  • Determining whether removing at most one character can produce a palindrome

Along the way, you’ll also learn why the two-pointer technique is one of the most valuable patterns in algorithm design.

Reversing a String

The classic problem is simple.

Given:

js
"abc"

Return:

js
"cba"

Unlike arrays, JavaScript strings don’t have a built-in reverse() method.

This doesn’t work:

js
const str = "abc";

// TypeError
str.reverse();

Instead, the standard solution converts the string into an array.

js
const str = "abc";

const reversed = str
  .split("")
  .reverse()
  .join("");

console.log(reversed); // "cba"

The process consists of three steps:

  1. split("") converts the string into an array.
  2. reverse() reverses the array.
  3. join("") combines everything back into a string.

This solution is clean, easy to read, and runs in O(n) time.

Its space complexity is also O(n) because a new array is created.

Why Does a String Have .length?

JavaScript strings are primitive values.

js
typeof "hello";
// "string"

Yet they still behave like objects.

js
const str = "hello";

console.log(str.length);
console.log(str[0]);

How is that possible?

Whenever you access a property or method on a primitive string, JavaScript temporarily wraps it in a String object behind the scenes.

Conceptually, it’s similar to this:

js
const temp = new String("hello");

temp.length;

The wrapper object exists only for the duration of the operation before being discarded.

This behavior lets primitive values expose useful methods while remaining lightweight.

Understanding call() and this

Another interview favorite involves Function.prototype.call().

Consider this example.

js
const person = {
  name: "Alice",

  say() {
    console.log(this.name);
  },
};

const anotherPerson = {
  name: "Bob",
};

person.say();

person.say.call(anotherPerson);

The first call prints:

text
Alice

The second prints:

text
Bob

call() doesn’t copy the function.

It simply changes what this refers to during execution.

This same mechanism powers one of JavaScript’s most reliable type-checking techniques.

js
Object.prototype.toString.call([]);

Result:

text
[object Array]

Because toString() executes with a different this, it can inspect many different object types.

Checking for a Palindrome

A palindrome reads the same in both directions.

Examples include:

text
level
racecar
abba

The simplest solution reverses the string and compares it with the original.

js
function isPalindrome(str) {
  return str ===
    str
      .split("")
      .reverse()
      .join("");
}

Although easy to understand, this approach creates another string and requires extra memory.

The Better Solution: Two Pointers

A palindrome is symmetrical.

Instead of reversing the entire string, compare characters from both ends.

js
function isPalindrome(str) {
  let left = 0;
  let right = str.length - 1;

  while (left < right) {
    if (str[left] !== str[right]) {
      return false;
    }

    left++;
    right--;
  }

  return true;
}

This solution still runs in O(n) time.

However, it only uses O(1) extra memory because no additional arrays or strings are created.

The two-pointer technique appears in countless interview questions involving:

  • symmetric data
  • sorted arrays
  • substrings
  • intervals
  • partitioning

Learning it early pays off repeatedly.

Remove One Character to Form a Palindrome

Now let’s make the problem slightly harder.

Given a string, determine whether removing at most one character makes it a palindrome.

Examples:

text
"aba"  → true

"abca" → true

"abc"  → false

The two-pointer approach still works.

Whenever the left and right characters match, continue toward the center.

If they don’t match, only two possibilities remain.

Skip the left character.

Or skip the right one.

js
function validPalindrome(str) {

  function isPalindrome(left, right) {

    while (left < right) {

      if (str[left] !== str[right]) {
        return false;
      }

      left++;
      right--;
    }

    return true;
  }

  let left = 0;
  let right = str.length - 1;

  while (left < right) {

    if (str[left] !== str[right]) {

      return (
        isPalindrome(left + 1, right) ||
        isPalindrome(left, right - 1)
      );
    }

    left++;
    right--;
  }

  return true;
}

The helper function lets us verify both possibilities without duplicating code.

Why Only Two Cases?

Suppose these two characters don’t match.

text
... x ........ y ...
    ↑          ↑
   left      right

If removing one character can solve the problem, then one of these mismatched characters must disappear.

Removing any other character won’t fix the current mismatch.

Therefore, we only need to test:

js
isPalindrome(left + 1, right);

or

js
isPalindrome(left, right - 1);

Nothing else needs to be checked.

That’s why the algorithm remains linear instead of becoming quadratic.

Complexity Analysis

Reverse String

  • Time: O(n)
  • Space: O(n)

Palindrome (Two Pointers)

  • Time: O(n)
  • Space: O(1)

Remove One Character

Although an additional palindrome check may occur after finding a mismatch, every character is still visited only a constant number of times.

The complexity remains:

  • Time: O(n)
  • Space: O(1)

Good Habits When Solving String Problems

As you practice more interview questions, several patterns emerge.

Remember that strings are immutable.

Most operations create new strings rather than modifying existing ones.

Prefer two pointers whenever possible.

They’re often faster and require less memory than repeatedly creating arrays or substrings.

Extract reusable logic.

Small helper functions usually make algorithms easier to understand and easier to test.

Always think about edge cases.

Test your solution with:

  • empty strings
  • single-character strings
  • already-valid inputs
  • repeated characters
  • very long strings

These scenarios often reveal hidden bugs.

Final Thoughts

String algorithms are much more than interview exercises.

They teach fundamental JavaScript concepts like immutable strings, wrapper objects, built-in APIs, function binding, and algorithm optimization.

More importantly, they encourage a problem-solving mindset.

Rather than reaching for the first solution that works, learn to identify patterns, simplify the problem, and choose algorithms that minimize both time and memory usage.

Master these classic questions, and you’ll be well prepared not only for coding interviews, but also for writing cleaner, more efficient JavaScript in everyday projects.