Codeforces #211 (Div. 2) C.Fixing Typos

problem link: http://codeforces.com/problemset/problem/363/CA simple greedy approach will give you the solution.

A simple greedy approach will give you the solution.

Special care has to be taken for testcases like aabbccddeeff, aaaabbc

Code is self explanatory.

	#include <vector>
	#include <list>
	#include <map>
	#include <set>
	#include <deque>
	#include <queue>
	#include <stack>
	#include <bitset>
	#include <algorithm>
	#include <functional>
	#include <numeric>
	#include <utility>
	#include <sstream>
	#include <iostream>
	#include <iomanip>
	#include <cstdio>
	#include <cmath>
	#include <cstdlib>
	#include <ctime>
	#include <cstring>
	#include <climits>
	#include <stdlib.h>
	#include <stdio.h>
	using namespace std;
	#define REP(i,n) for(int i=0; i<n; i++)
	#define FOR(i,st,end) for(int i=st;i<end;i++)
	#define db(x) cout << (#x) << " = " << x << endl;
	#define mp make_pair
	#define pb push_back
	#define MAX 10000005
	typedef long long int ll;
	//count in seperate vector and character in seperate vector
	//count1[i] denotes the count of character[i]
	vector<int>count1;
	vector<char>character;
	int main(){
	string str;
	cin>>str;
	REP(i,str.size()){
	char candidate=str[i];
	int c=1;
	for(int j=i+1;j<str.size();j++){
	if(candidate==str[j])
	c++;
	else
	break;
	}
	i+=(c-1);
	count1.pb(c);
	character.pb(candidate);
	}
	REP(i,count1.size()){
	if(count1[i]>=3)
	count1[i]=2;
	}
	REP(i,count1.size()-1){
	if(count1[i]==2&&count1[i+1]==2){
	count1[i+1]=1;
	}
	}
	REP(i,count1.size()){
	REP(j,count1[i]){
	cout<<character[i];
	}
	}
	}

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