Coding

Palindrome Check

Asked at

Adobe

1Give yourself 5 minutes

2Answer out loud, not in your head

3Then compare with the answer below

The problem

You are given a string s. Return true if the string is palindrome, otherwise false.

A string is called palindrome if it reads the same forward and backward.

Input : s = "hannah" Output : true Explanation : The given string when read backward is -> "hannah", which is same as when read forward. Hence answer is true.

Input : s = "aabbaaa" Output : false Explanation : The given string when read backward is -> "aaabbaa", which is not same as when read forward. Hence answer is false.

Input : s = "aabbccbbaa"

1 <= s.length <= 105
s consist of only uppercase and lowercase English characters.

cpp

class Solution{	
	public:		
		bool palindromeCheck(string& s){
			//your code goes here
		}
};

java

class Solution {   
    public boolean palindromeCheck(String s) {
        //your code goes here
    }
}

python

class Solution:    
    def palindromeCheck(self, s):
        #your code goes here

javascript

class Solution {
    palindromeCheck(s) {
        //your code goes here
    }
}

csharp

class Solution {
    public bool palindromeCheck(string s) {
        //your code goes here
    }
}

func palindromeCheck(s string) bool {
	//your code goes here
}

Stuck? Show a way to structure it+

01Set pointers at both ends of the string.
02Compare the characters at each pair.
03Return false on the first mismatch.
04Move inward and return true when pointers cross.

Reference answer

Use two pointers, one at the start and one at the end. While left < right, compare s[left] and s[right]; a mismatch proves the string is not a palindrome. Otherwise move inward. If all mirrored pairs match, return true. This is O(n) time and O(1) extra space. The loop invariant is that every pair outside the current pointers has already matched, so a first mismatch is sufficient evidence to exit early. The stated input includes uppercase and lowercase letters, so comparison is case-sensitive unless a different normalization rule is explicitly added. Empty and one-character strings are palindromes under the usual definition, though this record's constraints start at length one. Avoid reversing into a second string when a direct comparison is clearer and uses no extra allocation.

Then expect these follow-ups

How would you ignore punctuation and case?
Tests: input normalization
How would you find the longest palindromic substring?
Tests: string algorithms

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