Codeforces round 665(C. Simple Strings)

Just check the neighbour of every element's left & right neighbour excluding first & last because they has only one neighbour.

So for first check the right neighbour & last check the left neighbour.

///Bismillahir Rahmanir Rahim

///Author: Tanvir Ahmmad

///CSE,Islamic University,Bangladesh

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<string>

#include<cstring>

#include<sstream>

#include<cmath>

#include<cstring>

#include<vector>

#include<queue>

#include<map>

#include<set>

#include<stack>

#include<vector>

#include<iterator>

#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array

/*every external angle sum=360 degree

  angle find using polygon hand(n) ((n-2)*180)/n*/

///Floor[Log(b)  N] + 1 = the number of digits when any number is represented in base b

using namespace std;

typedef long long ll;

int main()

{

    string str,chk="abc";

    cin>>str;

    for(ll i=1;i<str.size();i++)

    {

        if(str[i-1]==str[i])

        {

            if(str[i-1]!='a' && str[i+1]!='a') str[i]=chk[0];

            else if(str[i-1]!='b' && str[i+1]!='b') str[i]=chk[1];

            else str[i]=chk[2];

        }

    }

    cout<<str<<endl;

    return 0;

}

///Alhamdulillah

 

Problem Link:https:https://codeforces.com/problemset/problem/665/C