HackerEarth Palindromic Numbers problem solution

In this Leetcode Palindromic Numbers problem solution we have given A and B, count the numbers N such that A ≤ N ≤ B, and N is a palindrome.


HackerEarth Palindromic Numbers problem solution


HackerEarth Palindromic Numbers problem solution.

def check(s):
q=s[::-1]
if s==q: return True
return False
t=input()
for i in range(t):
ans=0
x=raw_input()
a=int(x.split()[0])
b=int(x.split()[1])
for j in range(a,b+1):
if check(str(j)):
ans += 1
print ans


Second solution

#include <iostream>
using namespace std;

int ispallin(int n){
int arr[10],count = 0;
while(n!=0){
arr[count++] = n%10;
n/=10;
}
int i=0,j=count-1;
while(i<j){
if(arr[i]!=arr[j]) return 0;
i++; j--;
}
return 1;
}

int main()
{
int t;
cin >> t;
while(t--){
int a,b;
cin >> a >> b;
int ans = 0,count=0;;
for(int i=a;i<=b;i++) if(ispallin(i)) count++;
cout << count << endl;
}
return 0;
}

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