problem no 31: string problem:Anagram palindrome (VIP).

 QUESTION:

Given a string S, Check if characters of the given string can be rearranged to form a palindrome. 
For example characters of “geeksogeeks” can be rearranged to form a palindrome “geeksoskeeg”, but characters of “geeksforgeeks” cannot be rearranged to form a palindrome.

Input:
First line consists of integer T  denoting the number of test cases. T testcases follow. For each testcase there are one line of input containing string S.

Output:
For each testcase, in a new line, print "Yes" if is possible to make it a palindrome, else "No".

Constraints:
1 <= T <= 100
1 <= |S| <= 1000

Example:
Input:
2
geeksogeeks
geeksforgeeks
Output:
Yes
No

 

ANSWER:

 

#include <iostream>
#include<cstring>
using namespace std;

int main() {
    int t;
    cin>>t;
    while(t--)
    {
       
        string s;
        cin>>s;
        int arr[128];
        for(int i=0;i<128;i++)
        {
            arr[i]=0;
        }
       if(s.size()%2==0)
       {
           for(int i=0;i<s.size();i++)
           {
               arr[(int)s[i]]=  arr[(int)s[i]]+1;
           }
           int countofoddnumber=0;
           for(int i=0;i<128;i++)
           {
               if((arr[i]%2)!=0)
               {
                   countofoddnumber+=1;
               }
           }
   
           if(countofoddnumber>0)
           {
               cout<<"No\n";
           }
           else
           {
               cout<<"Yes\n";
           }
       }
       else
       {
           for(int i=0;i<s.size();i++)
           {
               arr[(int)s[i]]=  arr[(int)s[i]]+1;
           }
           int countofoddnumber=0;
           for(int i=0;i<128;i++)
           {
               if((arr[i]%2)!=0)
               {
                   countofoddnumber+=1;
               }
           }
           if(countofoddnumber==1)
           {
               cout<<"Yes\n";
           }
           else
           {
               cout<<"No\n";
           }
       }
    }
    return 0;
}