In this HackerEarth Even Length Palindromic Number problem solution You have to design a new model which maps an even length palindromic number to some digit between 0 to 9.
The number is mapped to a digit  on the basis of the following criteria:
  1. x should appear a maximum number of times in the palindromic number, that is, among all digits in the number, x should appear a maximum number of times.
  2. If more than one digit appears a maximum number of times,  should be the smallest digit among them.

Given an integer N, you have to find the digit x for the Nth even length palindromic number.


HackerEarth Even Length Palindromic Number problem solution


HackerEarth Even Length Palindromic Number problem solution.

#include<bits/stdc++.h>
#define ll unsigned long long
#define ld long double
#define mp make_pair
#define pb push_back
#define si(x) scanf("%d",&x)
#define pi(x) printf("%d\n",x)
#define s(x) scanf("%lld",&x)
#define p(x) printf("%lld\n",x)
#define sc(x) scanf("%s",x)
#define pc(x) printf("%s",x)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define F first
#define S second
#define inf 4e18
#define prec(x) fixed<<setprecision(15)<<x
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define mem(x,y) memset(x,y,sizeof(x))
#define PQG priority_queue< int,std::vector<int>,std::greater<int> >
#define PQL priority_queue< int,std::vector<int>,std::less<int> >
#define PQPL priority_queue<pii ,vectosr< pii >, less< pii > >
#define PQPG priority_queue<pii ,vector< pii >, greater< pii > >
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)

using namespace std;


int main() {
#ifndef ONLINE_JUDGE
freopen("in09.txt","r",stdin);
freopen("out09.txt","w",stdout);
#endif
int q; cin>>q;
while(q--) {
string s; cin>>s;
int cnt[10];
mem(cnt,0);
for(char c:s) {
cnt[c-'0']++;
}
int mx=-1,dig;
for(int i=0;i<10;i++) {
if(cnt[i]>mx) {
mx=cnt[i]; dig=i;
}
}
cout<<dig<<endl;
}

return 0;
}


Second solution

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
ll n;
cin>>n;
int f[10]={0};
while(n)
{
f[n%10]++;
n/=10;
}
int index,ans=0;
for(int i=0;i<=9;i++)
{
if(ans<f[i])
{
ans=f[i];
index=i;
}
}
cout<<index<<"\n";
}
return 0;
}