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#125 Valid Palindrome

Posted on January 20, 2022January 20, 2022 By thecodepathshala 9 Comments on #125 Valid Palindrome

A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.

Given a string s, return true if it is a palindrome, or false otherwise.

Example 1:

Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.

Example 2:

Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.

Example 3:

Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.

Solution:

string removeNonAlphabetsAndConvertToUpperCase(string s)
    {
        for(int i=0; i<s.size(); i++)
        {
            if((s[i] < 'A' || s[i] > 'Z') && (s[i] < 'a' || s[i] > 'z') && (s[i] < '0' || s[i] > '9'))
            {
                s.erase(i,1);
                i--;
            }
        }
        transform(s.begin(), s.end(), s.begin(), ::toupper);
        return s;
    }

    string getReverse(string s)
    {
        string s1;
        for(int i = s.length()-1; i>=0; i--)
            s1.push_back(s.at(i));
        return s1;
    }
    bool isPalindrome(string s) {
        string s2 = removeNonAlphabetsAndConvertToUpperCase(s);
        string s1 = getReverse(s2);

        if(s1.compare(s2) == 0)
            return true;
        return false;
    }

Complete Code:

// function to check if it is palindrome or not
#include <bits/stdc++.h>
using namespace std;

string removeNonAlphabetsAndConvertToUpperCase(string s)
{
	for(int i=0; i<s.size(); i++)
	{
		if((s[i] < 'A' || s[i] > 'Z') && (s[i] < 'a' || s[i] > 'z') && (s[i] < '0' || s[i] > '9'))
		{
			s.erase(i,1);
			i--;
		}
	}
	transform(s.begin(), s.end(), s.begin(), ::toupper);
	return s;
}

string getReverse(string s)
{
	string s1;
	for(int i = s.length()-1; i>=0; i--)
		s1.push_back(s.at(i));
	return s1;
}
bool isPelendrome(string s) {
	string s2 = removeNonAlphabetsAndConvertToUpperCase(s);
	string s1 = getReverse(s2);
	
	cout << s1 << endl << s2 << endl;
	if(s1.compare(s2) == 0)
		return true;
	return false;
}

int main()
{
	string s = "0p";
	cout << isPelendrome(s) << endl;
	return 0;
}
Algo, Competitive Programming, DS & Algo, GFG, Leetcode Problems Tags:coding interview, easy, leetcode, string

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