SLPCYCLE - Editorial

PROBLEM LINK:

Practice
Contest: Division 3
Contest: Division 2
Contest: Division 1

Author: Daanish Mahajan
Tester: Istvan Nagy
Editorialist: Vichitr Gandas

DIFFICULTY:

SIMPLE

PREREQUISITES:

Greedy

PROBLEM:

Given a body needs H units of continuous sleep. If you sleep x ( < H) units then next time, we need 2*(H-x) units of continuous sleep.
Given a string of length L describing the day where S_i=0 means that unit of time is free and S_i=1 means that unit of time is occupied. Find if you would be able to achieve the sleep requirement.

EXPLANATION

As we know that if we sleep x(<H) units then we need to sleep 2*(H-x) units more next time.

So when we have some x(<H) units of free time available then should us sleep or not?
We should sleep only if, it reduces our remaining sleep units that is we should sleep x units only if 2*(H-x) < H.
=> 2*H - 2*x < H
=> 2*H - H < 2*x
=> H < 2*x
=> x > H/2

So iterate over the given string S, find count of continuous 0s, check if count > H/2, if yes then sleep for this time and update the H as following:
H = 2*(H-count)

If at any point we have H <= 0 then we are able to achieve the requirement otherwise no.

TIME COMPLEXITY:

O(L) per test case

SOLUTIONS:

Setter's Solution
#include<bits/stdc++.h>
using namespace std;
# define ll long long int  

const int maxl = 1e5, maxh = 1e5, maxt = 10;

int main()
{   
    int t; cin >> t;
    while(t--){
        int l, h; cin >> l >> h;
        string s; cin >> s;
        int req = h; int cnt = 0;
        for(int i = 0; i < l && req > 0; i++){
            if(s[i] == '0')cnt++;
            else{
                if(cnt >= req / 2 + 1){
                    req -= cnt; req *= 2;
                }
                cnt = 0;
            }
        }
        if(cnt >= req / 2 + 1 && req > 0){
            req -= cnt; req *= 2;
        }
        string ans = req <= 0 ? "YeS" : "No";
        cout << ans << endl;
    }
} 
Tester's Solution
 #include <iostream>
#include <cassert>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <random>

#ifdef HOME
	#include <windows.h>
#endif

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

template<class T> bool umin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool umax(T &a, T b) { return a < b ? (a = b, true) : false; }

using namespace std;

long long readInt(long long l, long long r, char endd) {
	long long x = 0;
	int cnt = 0;
	int fi = -1;
	bool is_neg = false;
	while (true) {
		char g = getchar();
		if (g == '-') {
			assert(fi == -1);
			is_neg = true;
			continue;
		}
		if ('0' <= g && g <= '9') {
			x *= 10;
			x += g - '0';
			if (cnt == 0) {
				fi = g - '0';
			}
			cnt++;
			assert(fi != 0 || cnt == 1);
			assert(fi != 0 || is_neg == false);

			assert(!(cnt > 19 || (cnt == 19 && fi > 1)));
		}
		else if (g == endd) {
			assert(cnt > 0);
			if (is_neg) {
				x = -x;
			}
			assert(l <= x && x <= r);
			return x;
		}
		else {
			assert(false);
		}
	}
}

string readString(int l, int r, char endd) {
	string ret = "";
	int cnt = 0;
	while (true) {
		char g = getchar();
		assert(g != -1);
		if (g == endd) {
			break;
		}
		cnt++;
		ret += g;
	}
	assert(l <= cnt && cnt <= r);
	return ret;
}
long long readIntSp(long long l, long long r) {
	return readInt(l, r, ' ');
}
long long readIntLn(long long l, long long r) {
	return readInt(l, r, '\n');
}
string readStringLn(int l, int r) {
	return readString(l, r, '\n');
}
string readStringSp(int l, int r) {
	return readString(l, r, ' ');
}


int main(int argc, char** argv) 
{
#ifdef HOME
	if(IsDebuggerPresent())
	{
		freopen("../in.txt", "rb", stdin);
		freopen("../out.txt", "wb", stdout);
	}
#endif
	int T = readIntLn(1, 10);
	forn(tc, T)
	{
		int L = readIntSp(1, 100'000);
		int H = readIntLn(1, 100'000);
		string S = readStringLn(L, L);
		int last = H;
		int rem = H;
		bool ok = false;
		char pr = '1';
		for (auto c : S)
		{
			assert(c == '0' || c == '1');
			if (c == '1')
			{
				if (pr == '0')
				{
					rem <<= 1;
					if (rem < last)
					{
						last = rem;
					}
				}
				pr = c;
				continue;
			}
			if (pr == '1')
			{
				rem = last;
			}
			pr = c;
			--rem;
			if (rem == 0)
				ok = true;
		}
		if (ok)
			printf("YES\n");
		else
			printf("NO\n");
	}
	assert(getchar() == -1);
	
	return 0;
}

Editorialist's Solution
/*
 * @author: vichitr
 * @date: 26th June 2021
 */

#include <bits/stdc++.h>
using namespace std;

void solve() {
	int H, L;
	cin >> L >> H;
	string S; cin >> S;
	int cnt = 0;
	for (char c : S) {
		if (c == '0')
			cnt++;
		else {
			if (cnt > H / 2) {
				H = 2 * (H - cnt);
			}
			cnt = 0;
			if(H < 0) break;
		}
	}
	if (cnt > H / 2)
		H = 2 * (H - cnt);
	if (H <= 0) cout << "YES\n";
	else cout << "NO\n";
}


int main() {

#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif

	int t = 1;
	cin >> t;
	while (t--)
		solve();
	return 0;
}

VIDEO EDITORIAL:

If you have other approaches or solutions, let’s discuss in comments.

3 Likes

Now, please tell me which test case is failing for my code.

Solution: 48302652 | CodeChef.

Will be really appreciated.
Thanks.

1
7 5
0100000
The output of this testcase is “YES”.
But your code output is “NO”.

3 Likes

guys please say where i have mistaken
t=int(input())

for _ in range(t):

h,l=[int(a) for a in input().split()]

re=int(l)

s=input()

z=0

for i in s:

    if i=="0":

        z=z+1

    else:

        if z!=0:

            if (re-z)!=0:

                re=2*(re-z)

            else:

                re=re-z

            z=0

            if re==0:

                break

if z!=0:

    if (re-z)!=0:

        re=2*(re-z)

    else:

        re=re-z

    z=0

if re==0:

    print("YES")

else:

    print("NO")

How will the output be “YES”. He 1st sleeps 1 unit so he has to sleep another 2*(5-1)=8 units but the remaining zeros are 5.

Can anybody please help me find which case is missing. Please! :pray:

#include <bits/stdc++.h>
using namespace std;

int main()
{
int t;
cin>>t;
while(t–)
{
int l,h;
cin>>l>>h;
string s;
cin>>s;

          map<int,int> mp;
          
          //storing what all contiguous unit of time is available
          for(int i=0;i<l;i++)
          {
                 int c=0;
                 while(i<l && s[i]=='0')
                 {
                        c++; i++;
                 }
                 if(c!=0) 
                 mp[c]++;
          }
          
        
          bool ans=false;
          while(mp.empty()==false)
          {
             
                 auto it=mp.lower_bound(h);
                 if(it==mp.end())
                 it--;
                        
                 h=2*(h-((*it).first));
                 ((*it).second)--;
                        
                 if(((*it).second)==0)
                 mp.erase((*it).first);
                 
                 if(h<=0)
                 {
                        ans=true;
                        break;
                 }
   
          }
         
         if(ans==true) cout<<"Yes\n";
          else cout<<"No\n";
   }
   return 0;

}

check with edge cases
https://www.codechef.com/viewsolution/48288067

didn’t consider the fact that he may have the option of not sleeping during free time. UGHH. But it wasn’t mentioned clearly ;-;

9 Likes

But what happens when he does not sleeps 1 unit and work for 1 unit and sleep afterward?
Think You will get your answer.

1 Like

You are computing H = 2 ( H - X ) every time x is less than H. That will give you a wrong answer. What you should do is :
if 2(H-X) < H
H = H-X

1 Like

what is wrong with my code?suggest some corner cases.

 #include<bits/stdc++.h>
    using namespace std;
    #define ll long long int
    #define mod 1000000007


    void solve()
    {
        ll L,H;
        cin>>L>>H;
        string s;
        cin>>s;
        int flag=0;
        ll count=0;
        
        ll i=0;
        for( i=0;i<L;i++)
        {
            if(s[i]=='1')
            {
                if(flag==1)
                {
                    H=2*(H-count);
                   // if(H==0)
                   // break;
                   flag=0;
                }
                count=0;
            }
            if(s[i]=='0' )
            {
                count++;
                flag=1;
            }
        
        }
        if(i==L && s[L-1]=='0')
        {
            H=H-count;
        }
       
        if(H>0)
        cout<<"NO"<<endl;
        else if(H<=0)
        cout<<"YES"<<endl;
    }





    int main()
    {
        ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

        ll T;
        cin>>T;
        while(T--)
        {
            solve();
        }
    }

I think quite differently -

1.push all the length of continuous zeroes in a vector (denotes that how much my user can sleep continuously)
2.Now if there are no zero answers is NO
3. else, let prev = h denotes we need this hrs to sleep, now iterate my vector, if my current value (which is basically the length of continuous zeroes ) is less than prev, then I will change my previous to prev = 2 * (prev-vc[i]); (as question said)

Also, one thing is to check sometimes while doing this (prev = 2 * (prev-vc[i]); ) , my prev is greater than h , so now we will set prev to h because from now we will check from that time frame no need to take previously.

otherwise(if vc[i]>prev) will set flag=1 means we get enough continuous hrs to sleep.

in last what I add is to check max continuous zeroes, we r checking here from starting and doing all the operations above may increase my prev, so I directly check whether the max continuous zeroes are greater than equal to h or not, if yes then I will simply sleep in that time frame only.

    lli n,h;
    cin>>n>>h;
    string s;
    cin>>s;
    lli cnt=0;
    vi vc;
    loop(i,n)
    {
        if(s[i]=='0')
            cnt++;
        else
        {
            if(cnt)
                vc.pb(cnt);
            cnt=0;
        }
    }
    if(cnt)
      vc.pb(cnt);
    if(sz(vc)==0)
    {
        print("NO");
    }
    else
    {
        lli prev = h;
        lli flag=0;
        loop(i,sz(vc))
        {
            if(vc[i]<prev)
            {
                prev = 2*(prev-vc[i]);
                if(prev>h)
                    prev = h;
            }
            else
                flag=1;
        }
        if(flag==1 or h<=(*max_element(all(vc))))
            print("YES");
        else
            print("NO");
    }
    
}
return 0;
}

solution : Solution: 48282764 | CodeChef

1 Like

Same with me too… :no_mouth:

Can anyone point out error in my code. I’m unable to understand where is my code failing?

#include
#define ll long long
using namespace std;

int main() {
// your code goes here
ll t,h,l;
cin>>t;
while(t–){
cin>>l>>h;
string s;
cin>>s;
ll count=0,flag=0;
if(h==0){
cout<<“YES”<<endl;
continue;}
for(ll i=0;i<l;i++){
if(s[i]==‘0’)
count++;
else{
if(h<=count){
flag=1;
break;}
if(count){
h=2*(h-count);}
count=0;
}}
if(h<=count || flag==1)
cout<<“YES”<<endl;
else
cout<<“NO”<<endl;
}
return 0;
}

can someone give a test case which my code might fail to pass because I am still trying but cannot find a test case on which I can improve my code? Any help will be recommended thanks:

My WA code link : Solution: 48288922 | CodeChef

PS: Ignore bad variable names, sorry for that

how will it be YES?

#include<bits/stdc++.h>
#define ll long long
using namespace std;
// ios_base :: sync_with_stdio(false);
// cin.tie(NULL);

//////////////////////// Sleep Cycle ///////////////////
void solve() {
   int l , h;
   string s;
   cin >> l >> h;
   cin >> s;
   int req = h;
   int count = 0;
   for(ll i = 0 ; i < s.length() ;) {
       if(s[i] == '0') {
           while((s[i] == '0') && (i < s.length())) {
               count ++;
               i++;
           }
           if(count >= req) {
               cout << "YES" << endl;
               return;
           } else {
               req = 2 * (req-count);
               count = 0;
           }
       } else {
           i++;
       }
     
   }
   cout << "NO" << endl;
}

int main() {
    ios_base :: sync_with_stdio(false);
    cin.tie(NULL);

    int tc;
    cin >> tc;
    while(tc--) {
        solve();
    }
}

Can anyone tell me what is wrong here ? I passed the sample cases but got WA after submitting.

include

using namespace std;
int n,h;
int t;
int a=0;

int main()
{
cin>>t;
while(t–){
cin>>n>>h;
char str[n] ;
cin>>str;
int count=0;
for(int i=0;i<n;i++){
// if(str[i]==‘0’){
// int j=i;
count=0;
while(str[i]==‘0’ && i<n){
count++;
if(count==h){
a=1;
break;
}
i++;

       }
       if(count!=0){
       h=2*(h-count);

   }

}
if(a==0){
cout<<“NO”<<endl;
}
else{
cout<<“YES”<<endl;
}

}

return 0;

}
where am I wrong??

https://www.codechef.com/viewsolution/48269643

please can you tell what’s wrong with my code?

Now that I realized my mistake, or rather lack of observation: 2(H-x) < H, I feel stupid.

1 Like