C++ Basic Programs | Check Palindrome

Introduction:

In this tutorial, we are going to solve a basic problem on string. and this is really an important problem to understand few concepts related to problem solving.basically today we are going to solve a problem in which we will try to find whether a given string is palindrome string or not?

Problem Statement:

We have given a string s. and then we have to check whether it is palindrome string or not?

Input:string s = "sana"

Output: "No"

Input: string s = "abba"

Output: "yes"

Solution:

let's first of all understand, what palindrome means?

Anything is palindrome if we get the same result from left view and right view.

eg: "abba" is palindrome, cause you see it from right or left, it will be same.

eg: "Abba" is not a palindrome, cause from left it starts with "A", while from right, it starts with "a".

we can solve this problem by two methods:

1.Using two pointer(easy one)

2.Using Stack(it will take extra space).

so let's see the first method, which is more efficient. or you can see whenever there is palindrome question you can follow above mentioned two methods.

Full code:

#include<bits/stdc++.h>

using namespace std;

int main(){

string s;

cin>>s;

int n=s.size();

int flag=0;

for(int i=0; i<n;i++){

       if(s[i]!=s[n-1-i]){

           flag=1;

           break;

       }

}

if(flag==0){

cout<<"yes"<<endl;

}

else{

cout<<"No"<<endl;

}

}

so the idea is, in first iteration we will check first and last character of string, if it matches, then in second iteration we will check second character from the beginning and second character from last and so on.

and this is done by s[i]!=s[n-1-i] this piece of code. if there is a mismatch then it means the given string is not palindrome and we will stop iteration here. and set the flag value to 1.flag variable will keep track that we have found a mismatch or not. if we found a mismatch then we will set flag to 1.

Conclusion:

Start checking from both end at the same moment if both matches, then we will proceed further, otherwise we will stop the execution.

i hope you got my points.