Palindromic Numbers : HackerEarth Problem Solution

Palindromic Numbers : HackerEarth Problem Solution


C Implementation:

#include <stdio.h> int checkpalin(int N) { int number=N,rem; int out=0; while(N!=0) { out=out*10; rem=N%10; out=out+rem; N=N/10; } if(out==number) return 1; else return 0; } int main() { int T,L,R,i,status,count; scanf("%d",&T); while(T--) { scanf("%d %d",&L,&R); count=0; for(i=L;i<=R;i++) { status=checkpalin(i); if(status==1) count++; } printf("%d\n",count); } return 0; }

C++ Implementation:

#include <iostream> using namespace std; int checkpalin(int N) { int number=N,rem; int out=0; while(N!=0) { out=out*10; rem=N%10; out=out+rem; N=N/10; } if(out==number) return 1; else return 0; } int main() { int T,L,R,i,status,count; cin>>T; while(T--) { cin>>L>>R; count=0; for(i=L;i<=R;i++) { status=checkpalin(i); if(status==1) count++; } cout<<count<<"\n"; } return 0; }

Java Implementation:

import java.util.Scanner; class TestClass { public static void main(String args[] ) throws Exception { TestClass obj = new TestClass(); Scanner in = new Scanner(System.in); int T=in.nextInt(); while((T--)>0) { int L=in.nextInt(); int R=in.nextInt(); int count=0; for(int i=L;i<=R;i++) { int status=obj.checkpalin(i); if(status==1) count++; } System.out.println(count); } } int checkpalin(int N) { int number=N; int rem; int out=0; while(N!=0) { rem=N%10; out=(out*10)+rem; N=N/10; } if(number==out) return 1; else return 0; } }