Palindromic string

#1

question–>
https://practice.geeksforgeeks.org/problems/anagram-palindrome/0

code–>

#2

Hey @sheikhhaji18

#include <iostream>
#include<string>
#include<map>
using namespace std;

int main() {
	//code
	int t;
// 	map<char,int> m;not here but inside loop
	cin>>t;
	while(t--){
	    map<char,int> m;//declared here
	    string s;
	    cin>>s;
	    for(int i=0;i<s.length();i++){
	        if(m.count(s[i])==0){
	            m[s[i]]=1;
	        }
	        else{
	            m[s[i]]++;
	        }
	    }
	    bool flag=true;
        bool once=true;//added
	    for(auto x:m){
	        if(x.second%2!=0){ //updated this
	            // cout<<"No"<<endl; removed
                if(once)once=false;//added
                else{ //added
                    flag=false;
                    break;
                }
	        }
	    }
	    if(flag)
	    cout<<"Yes"<<endl;
        else cout<<"No"<<endl;//added here
	}
	return 0;
}

There can be atmost one char with odd freq for it to be anagram palindromic

#3

thx @Kartikkhariwal1

1 Like
#4

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