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HackerEarth Unique sorting problem solution

In this HackerEarth Unique sorting problem solution, You are given a string s consisting of numeric digits (0-9) and Latin alphabetic characters ('a'-'z') only.

Your task is to sort the string but it must be sorted in a unique manner because here even characters or digits have higher priority than odd characters or digits. Although, as usual, other characters or digits have the same priority level. Also, digits have higher priority than characters.

HackerEarth Unique sorting problem solution

HackerEarth Unique sorting problem solution.

using namespace std;

using namespace __gnu_pbds;
template<class T> using oset=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

#define F first
#define S second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define pii pair<int,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;

void solve(){
string s;
int cf[26]{0},df[10]{0};
for(char c:s){
if(isdigit(c)) df[c-'0']++;
else cf[c-'a']++;
string ans;
for(int i=1;i<26;i+=2){
rep(j,1,cf[i]) ans+=char('a'+i);
for(int i=0;i<26;i+=2){
rep(j,1,cf[i]) ans+=char('a'+i);
for(int i=1;i<10;i+=2){
rep(j,1,df[i]) ans+=char('0'+i);
for(int i=0;i<10;i+=2){
rep(j,1,df[i]) ans+=char('0'+i);
signed main(){
int t;
while(t--) solve();
return 0;

Second solution

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 14;
int main(){
ios::sync_with_stdio(0), cin.tie(0);
int t = 1;
cin >> t;
string s;
cin >> s;
sort(s.begin(), s.end(), [](char c, char d) -> bool{
if(isalpha(c) != isalpha(d))
return isalpha(c);
if(isalpha(c) ? (c - 'a') % 2 != (d - 'a') % 2 : (c - '0') % 2 != (d - '0') % 2)
return isalpha(c) ? (c - 'a') % 2 > (d - 'a') % 2 : (c - '0') % 2 > (d - '0') % 2;
return c < d;
cout << s << '\n';

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