Skip to main content

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.

Comments

Popular posts from this blog

Symmetric Tree | Interview Problem

Introduction: In this tutorial we are going to solve a good question which will clear your doubts on how BFS or level order traversal is useful to solve binary tree problems. this is basically a question from Leetcode, and it is really very good problem to practice BFS. Problem Statement: Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: But the following [1,2,2,null,3,null,3] is not: Link : Leetcode Link please make binary tree on paper by the given array above. and you will get a clear picture of the problem. what we are going to do in this problem. Solution: As we know we are going to apply BFS to solve this problem. actually you have to just get one thing to solve this problem. we will perform BFS, but with little change or modification. So first we will store the root->val in a vector or array of the left subtree. and during this we will first insert left child and then ...

Construct Binary Tree from preorder and inorder | Data Structure

Introduction: In this tutorial we are going to see how we can construct the binary tree from given preorder and inorder. Prerequisites: you should know about binary tree traversal and on paper you can draw binary tree from given preorder and inorder traversal. Inorder:left->root->right; Preorder:root->left->right; Problem Statement: we have given two arrays. one for preorder and another for inorder. by using these two array we have to built a binary tree. eg: preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] solution: Solution: We will follow recursive approach to solve this question.let's discuss how we can solve it. Trick: In the given preorder the very first element will be the root of the tree. then we will find root element in inorder also. and we know in inorder traversal we have left part then root and then right part of the tree. by using preorder we can get the root of the main tree and by using inorder and root we can get the left part and right part of the...

Linked List Data Structure | Creation and Traversal

Introduction: In this tutorial we will create our linked list. Before writing code let's understand key words related to linked list: 1. Head:    first node of the linked list is called the head of the list. and this node is most important. 2.Tail:       last node is called the tail of the list which must points to  null.            step1: For creating the list first of all let's declare the structure of Nodes of the list: struct Node{ int val; struct Node *next; } step2: write main function: int main(){     int n;     cout<<"enter the number of Nodes:";     cin>>n;     struct Node* head = NULL;     struct Node* temp;     int m;     while(n--){         cout<<"enter the data into nodes:";         cin>>m;    ...