Palindrome Check — Two Pointers

What does it solve?

Decide whether a string is a palindrome: it reads the same left-to-right and right-to-left. Place one pointer at the start and one at the end, compare, then step both inward.

Key insight: a palindrome mirrors around its center, so s[left] must equal s[right] at every step. The moment they differ, it can't be a palindrome, stop early. If the pointers cross without a clash, it is one. Each character is touched once → O(N), no extra memory.

The core idea

// converge from both ends
left = 0, right = n - 1
while left < right:
  if s[left] != s[right]:
    return false
  left++
  right--
return true

Complexity

Time complexity

O(N)

The two pointers together sweep the string once. At most N/2 comparisons before they meet in the middle.

Space complexity

O(1)

Only two index variables. No reversed copy, no extra buffer, the check happens in place.

Examples & edge cases

Live animation — two pointers converging from both ends

left pointer right pointer matched pair (settled) mismatch → not a palindrome center (pointers met)

String — comparing s[left] with s[right] —

Current comparison

left = — right = — compared = — verdict = —

Step — of —

left pointer

—

head

right pointer

—

tail

pairs checked

—

matched so far

Palindrome?

—

checking…

Select an example and press Play to start.

✓ Answer: —

Speed Medium

Step-by-step table

Step	Phase	left	right	s[left]	s[right]	Equal?	Action	Verdict

C++ implementation

valid_palindrome.cpp

#include <bits/stdc++.h>
using namespace std;

// True if s reads the same forwards and backwards
bool isPalindrome(string s) {
    int left = 0, right = (int)s.size() - 1;

    while (left < right) {
        if (s[left] != s[right])
            return false;   // mirror broken — stop early
        left++;
        right--;
    }
    return true;          // pointers crossed with no mismatch
}

// Variant: ignore case & non-alphanumerics (LeetCode 125)
bool isPalindromeClean(string s) {
    int l = 0, r = (int)s.size() - 1;
    while (l < r) {
        while (l < r && !isalnum(s[l])) l++;
        while (l < r && !isalnum(s[r])) r--;
        if (tolower(s[l]) != tolower(s[r])) return false;
        l++; r--;
    }
    return true;
}

int main() {
    cout << isPalindrome("racecar") << "\n"; // 1 (true)
    cout << isPalindrome("hello")   << "\n"; // 0 (false)
    return 0;
}

My Notes

📝 Notes ✅

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