HackerEarth Shil and Palindrome problem solution YASH PAL, 31 July 202413 February 2026 In this HackerEarth Shil and Palindrome problem solution Shil is your new boss and he likes palindromes very much. A palindrome is a string that can be read the same way in either direction, from the left to the right and from the right to the left. (ex. madam, aabaa, racecar) Given a string S, beautiful Palindrome is a lexicographical minimum palindrome that can be formed by rearranging all the characters of string S. In order to please your boss, find a beautiful Palindrome that can be formed with help of string S. String x is lexicographically less than string y, if either x is a prefix of y (and x≠y), or there exists such i (1≤i≤min(|x|,|y|)), that xi<yi, and for any j (1≤j<i) xj=yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages. HackerEarth Shil and Palindrome problem solution.#include<bits/stdc++.h>using namespace std;#define ll long long int#define INF 10000000#define f first#define pi pair<ll,ll>#define pb emplace_back#define mod 1000000007#define s second#define mp make_pair#define fr freopen("input-15.txt","r",stdin)#define fo freopen("output-15.txt","w",stdout)int main(){ string s; cin>>s; ll n=s.length(); ll f[26]={0}; string ans=string(n,'0'); for(int i=0;i<s.length();i++) f[s[i]-'a']++; if(n%2==0){ for(int i=0;i<26;i++){ if(f[i]%2!=0){ cout<<-1; return 0; } } } else{ ll cnt=0; char c; for(int i=0;i<26;i++){ if(f[i]%2!=0) cnt++,c=char(i+'a'); } if(cnt!=1){ cout<<-1; return 0; } ans[n/2]=c; f[c-'a']--; } ll j=0; for(int i=0;i<26;i++){ while(f[i]>0){ f[i]-=2; ans[j]=char(i+'a'); ans[n-j-1]=char(i+'a'); j++; } } cout<<ans;} coding problems solutions HackerEarth HackerEarth