Functional & Parameterized Recursion

Observe how recursion checks if a string is a palindrome by comparing characters from outside inward.

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1function isPalindrome(i, str) {

2 if (i >= str.length / 2) return true;

3 if (str[i] !== str[str.length - 1 - i]) return false;

4 return isPalindrome(i + 1, str);

Recursion on Strings: Palindrome Check

A string is a palindrome if it reads the same backward as forward (e.g., "radar", "level", "madam").

To check this recursively, we compare characters from the outside inwards:

Index Mapping: At index $i$, the corresponding symmetrically opposite character is located at len - 1 - i.
Base Case: i >= len / 2. When pointer $i$ reaches or crosses the middle of the string, it means all outer pairs matched. We return true.
Mismatch check: If str[i] !== str[len - 1 - i], the string cannot be a palindrome. We return false immediately.
Recursive Step: If the characters match, call isPalindrome(i + 1, str).

Complexity Analysis

Time Complexity: O(N)
In the worst case (palindrome string), the function compares $N/2$ pairs of characters, executing in $O(N)$ linear time.

Space Complexity: O(N)
The call stack size reaches a maximum height of $N/2$ stack frames. Thus, the auxiliary space complexity is linear, $O(N)$.

algorithm.txt

123456789101112131415// String Palindrome Check in JavaScript
algorithm isPalindrome(i, str)
  // Base case: crossed middle index
  if i >= str.length / 2 then
    return true
  // Mismatch case
  if str[i]  NOT  =  str[str.length - 1 - i] then
    return false
  // Recursive case
  return isPalindrome(i + 1, str)

// Example call:
word ← "radar"
result ← isPalindrome(0, word); // Returns true
console.log(result)

Palindrome Check Quiz

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