HackerEarth A special date problem solution

In this HackerEarth, A special date problem solution A special birth date is D. Your task is to find the last palindrome date that comes before a date D. 

You are given a date D in the string format. The date format is as follows:

A valid date D is of length 8 and in the DDMMYYYY format.

A valid date ranges between 01 to 30.

A valid month ranges between 01 to 12.

A valid year ranges between 0001 to 9999.

HackerEarth A special date problem solution.

#include<bits/stdc++.h>
using namespace std;

#define     F           first
#define     S           second
#define     pb          push_back
#define     lb          lower_bound
#define     ub          upper_bound
#define     vi          vector<int>
#define     all(x)      x.begin(),x.end()
#define     fix         fixed<<setprecision(10)
#define     rep(i,a,b)  for(int i=int(a);i<=int(b);i++)
#define     repb(i,b,a) for(int i=int(b);i>=int(a);i--)
#define     FastIO      ios_base::sync_with_stdio(0),cin.tie(0)

typedef double db;
typedef long long ll;

const int N=2e5+5;
const int mod=1e9+7;

bool ok(int ty,int tm,int td,int y,int m,int d){
    if(ty>y) return 0;
    if(ty<y) return 1;
    if(tm>m) return 0;
    if(tm<m) return 1;
    return td<d;
}

void solve(){
    string s;
    cin>>s;
    int d=(s[0]-'0')*10+(s[1]-'0');
    int m=(s[2]-'0')*10+(s[3]-'0');
    int y=(s[4]-'0')*1000+(s[5]-'0')*100+(s[6]-'0')*10+(s[7]-'0');
    set<vi>S;
    rep(i,1,30) rep(j,1,12){
        int td=i,tm=j,ty=0;
        ty=tm%10*1000+tm/10*100+td%10*10+td/10;
        if(ok(ty,tm,td,y,m,d)){
            S.insert({ty,tm,td});
        }
    }
    if(!S.empty()){
        vi u=*S.rbegin();
        string ans;
        ans+=char('0'+u[2]/10);
        ans+=char('0'+u[2]%10);
        ans+=char('0'+u[1]/10);
        ans+=char('0'+u[1]%10);
        ans+=char('0'+u[0]/1000);
        ans+=char('0'+u[0]/100%10);
        ans+=char('0'+u[0]/10%10);
        ans+=char('0'+u[0]%10);
        cout<<ans<<'\n';
    }
    else cout<<"-1\n";
}

signed main(){
    FastIO;
    int t;
    cin>>t;
    while(t--) solve();
    return 0;
}

Second solution

t = int(input())


def is_valid(s):
    d = int(s[:2])
    m = int(s[2:4])
    y = int(s[4:])
    return 1 <= d <= 30 and 1 <= m <= 12 and 1 <= y <= 9999


def cmp(s):
    return s[4:] + s[2:4] + s[:2]


while t > 0:
    t -= 1
    c = d = input()
    d = d[7:3:-1] + d[4:8]
    while True:
        if is_valid(d) and cmp(d) <= cmp(c):
            print(d)
            break
        y = int(d[4:])
        if y <= 1:
            print(-1)
            break
        y -= 1
        y = f"{y:04}"
        d = y[::-1] + y