Last Updated on March 28, 2022 by Ria Pathak
CONCEPTS USED:
Recursion
DIFFICULTY LEVEL:
Easy
PROBLEM STATEMENT(SIMPLIFIED):
Given a number N, check whether the number is palindrome or not using recursion. The palindrome number remains the same when its digits are reversed.
See original problem statement here
For Example :
Input : N = "12321"
Output : YESInput : N = "123"
Output : NO
Explanation : 123 is not equal to 321SOLVING APPROACH:
-
Find
reverseof a number –
1.1 Initialiseresultwith 0.
1.2 Whenever value of N becomes 0, returnresult.
1.3 If N is not 0, updateresultwithresult* 10 + N%10.
1.4 And recursively solve for N/10. -
Compare
reversewith N if both are equal, print YES else NO.
ALGORITHM:
result = 0
solve(N) :
if(N becomes 0)
return result
else
res = res * 10 + N %10
solve(N/10)SOLUTIONS:
#include <stdio.h>
//function for finding reverse of a number
int isPalin(int n,int res){
if(n == 0)
return res;
res = res*10 + n%10;
return isPalin(n/10,res);
}
int main()
{
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
int rev = isPalin(n,0);
if(n == rev)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
//function for finding reverse of a number
int isPalin(int n,int res){
if(n == 0)
return res;
res = res*10 + n%10;
return isPalin(n/10,res);
}
int main()
{
int t;cin>>t;
while(t--){
int n;cin>>n;
int rev = isPalin(n,0);
if(n == rev)
cout<<"YES\n";
else
cout<<"NO\n";
}
return 0;
}
import java.util.*;
import java.io.*;
public class Main {
//function for finding reverse of a number
static int isPalin(int n,int res){
if(n == 0)
return res;
res = res*10 + n%10;
return isPalin(n/10,res);
}
public static void main(String args[]) throws IOException {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t != 0){
int n = sc.nextInt();
int res = 0;
int rev = isPalin(n,res);
if(n == rev)
System.out.println("YES");
else
System.out.println("NO");
t--;
}
}
}
def isPalin( n, res):
if(n == 0):
return res
res = res * 10 + n % 10
return isPalin(n // 10, res)
for _ in range(int(input())):
n = int(input())
rev = isPalin(n, 0)
if n == rev:
print("YES")
else:
print("NO")
[forminator_quiz id="996"]
This article tried to discuss the Recursion. Hope this blog helps you understand the concept. To practice more problems on Recursion you can check out MYCODE | Competitive Programming