HackerEarth Even Length Palindromic Number problem solution

YASH PAL, 31 July 202411 February 2026

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:

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.
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("%dn",x)
#define s(x) scanf("%lld",&x)
#define p(x) printf("%lldn",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;
}

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