 Practice

Contest

Easy

# Pre-requisites:

Dynamic Programming

# Problem:

A W-String is defined as a string consisting of lower-case alphabets, and exactly 3 "#"s that divide the string into 4 non-empty contiguous parts, where each part consists of the same alphabet. Given a string S, find the length of the longest W-String that is a subsequence of S, with the added constraint that the three chosen #-positions in W, are consecutive #-positions in S.

# Explanation:

Let us suppose that we choose our three #-positions as P1, P2 and P3. After fixing P1, P2, P3, we greedily choose the most frequent character from [0, P1-1], [P1+1, P2-1], [P2+1, P3-1], [P3+1, N-1]. We further should disregard cases when any one of the above yields an “empty” string.

Let cnt[i][k] = Number of times character k occurs in S[0…i]. Then, given P1, P2, P3, we just need to calculate

max(cnt[P1][k] : 1<=k<=26) (greedy choice of character in [0, P1-1]

• 1 (# character)

• max(cnt[P2][k] - cnt[P1][k] : 1<=k<=26) (greedy choice of character in [P1+1, P2-1]

• 1 (# character)

• max(cnt[P3][k] - cnt[P2][k] : 1<=k<=26) (greedy choice of character in [P2+1, P3-1]

• 1 (# character)

• max(cnt[N-1][k] - cnt[P1][k] : 1<=k<=26) (greedy choice of character in [P3+1, N-1],

for all potential W-Strings (that is, they have all “greedy choice of character” values > 0).

Precomputing values of cnt[i][k] will take time O(N * 26). Further, we can iterate over all consecutive P1,P2,P3 in O(N) time, and each will lead to a calculation of 26 time. Thus overall complexity will be O(N * 26 * T). This can be sped up by a constant factor by precomputing MaxLeft(i) and MaxRight(i) which will find the maximum frequency of a single letter to the left or to the right of position i respectively.

# Setter’s Solution:

Can be found here

# Tester’s Solution:

Can be found here

4 Likes

plz tell me where i can reduce steps to reduce my time complexity,…

http://www.codechef.com/viewsolution/2275695

http://www.codechef.com/viewsolution/2276497

Plz explain me why a#b#c#d#e is not a w-string…
it is clear that a#b#c#d is a sub-sequence of a#b#c#d#e and a#b#c#d is a w-string.
This shows that a#b#c#d#e is a w-string…
Plz tell the violated condition…

@satya8081 >> This was the most common misconception during the contest. a#b#c#d#e is NOT a W string, but a#b#c#d is a W string. The string given by Ryuk `need not` be a W string, your task was to find out the length of the maximum subsequence of the given string S, such that it is a W string. Please don’t get confused. Read the conditions for a string to be a W string, once again, carefully.

For people who are asking why there code was wrong, you have the working codes now. Write code for creating random and heavy test cases and compare the output of your code and working code. It will be very easy to find error in this way rather than asking other people to read your complete code and point out the error.

5 Likes

The Problem Could be made really very nice if you just remove this condition " there must not be any ‘#’ symbols between the positions of the first and the second ‘#’ symbol he chooses, nor between the second and the third ", otherwise the problem is simply a piece of cake.

3 Likes

@pragrame ur editorials are awesome… In MAY13 cook-off where almost all problems were of dynamic programming and bitwise operators the explanation given by you were superb. Keep posting us these editorials… :) 1 Like

@pragrame the question was awesome…but most of the people(including me ) got WA at the time 0.00 so please post that test case…so that it will be very useful.

please can ny body suggest how to optimize the code . i m getting TLE

``````//Data Structure includes
#include<vector>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<deque>
#include<string>
#include<ctime>

//Other Includes
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cassert>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cctype>

using namespace std;

#define s(n)                                    scanf("%d",&n)
#define sl(n)                                   scanf("%lld",&n)
#define sf(n)                                   scanf("%lf",&n)
#define ss(n)                                   scanf("%s",n)
#define p(n)                                    printf("%d\n",n)
#define pl(n)                                   printf("%lld\n",n)
#define maX(a,b)                                ((a)>(b)?(a):(b))
#define miN(a,b)                                ((a)<(b)?(a):(b))
#define abS(x)                                  ((x)<0?-(x):(x))
#define FOR(i,a,b)                              for(int i=a;i<b;i++)
#define mp                                      make_pair
#define FF                                      first
#define SS                                      second
#define pb                                      push_back
#define SZ(v)                                   ((int)(v.size()))
#define all(x)                                  x.begin(),x.end()
#define INF                                     (int)1e9
#define LINF                                    (long long)1e18
#define EPS                                     1e-9
#define MOD                                     ((1 << 30)-1))
#define MSZ                                     80

typedef long long LL;
typedef pair<int,int> PII;
typedef pair<float,float> PFF;
typedef pair<LL,LL> PLL;
typedef vector<int> VI;
int countchar;
int check(int left,int right)
{
int max=0;

FOR(j,0,26)
{
if(left==0)
{
if(countchar[right][j]>max)
max=countchar[right][j];
}
else
{
if(countchar[right][j]-countchar[left-1][j]>max)
max=countchar[right][j]-countchar[left-1][j];
}

}
return max;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen ("input.txt", "r", stdin);
//freopen ("output.txt", "w", stdout);
#endif
int t;
s(t);
while(t--)
{
char a;
int b;
ss(a);
int length=strlen(a);
int k=0;
FOR(i,0,length)
{
if(a[i]=='#')
{
b[k++]=i;  //count pos of hash
}

}
FOR(i,0,length)
{
FOR(j,0,26)
{
countchar[i][j]=0;
}
}
FOR(i,0,length)
{
if(a[i]!='#')
{
FOR(j,i,length)
countchar[j][a[i]-'a']++; //count char count till index i
}
}
int max=0;
FOR(i,0,k-2)
{
if(i==0 && b==0)
continue;
int p1=check(0,b[i]-1);  //returmn count of char having max frequency
int p2=check(b[i]+1,b[i+1]-1);
int p3=check(b[i+1]+1,b[i+2]);
int p4=check(b[i+2]+1,length-1);
if(p1==0 || p2==0 || p3==0 || p4==0)
continue;
if(p1+p2+p3+p4>max)
max=p1+p2+p3+p4;
}
if(max==0)
p(max);
else
p(max+3);

}
return 0;
}``````

http://www.codechef.com/viewsolution/2281355 can anyone tell why i am gettin WA 1 Like

Hi,

Apologies for opening an old thread but i really liked the question and have been trying to get this one. My code is constantly getting WA for this question. I have tried the examples given along with some testcases mentioned in the comments, but their answer is coming fine.

http://www.codechef.com/viewsolution/3799371

Can someone please take a look and give any inputs he may want to share? It would be really helpful for me.

Thanks

The setter’s and tester’s solution links seem to be broken.

4 Likes

@admin why not you show at which case the answer is wrong or exceeding time limit or other case … after the contest is over…
I think this should not have any problem and I believe it will be good for a programmer like me having little bit error many times…and had to think everything in a new way…

1 Like

The sub-sequence that you are talking about (eg. a#b#c#d ) is the W string. Whereas a#b#c#d#e is just a string that contains W strings in it. Which does not make it a W string.

It would not only become really nice, but also really hard Who knows, maybe it’ll appear in future as a challenge question!

4 Likes

Thanks Its great to have this feedback. I personally am a fan of @anton_lunyov, who has mastered setting, testing, and editorial-writing.

@nihalpi1, solution links do tend to be broken at the initial point of uploading Editorials. The general aim is to roll out the Editorial as soon as possible after the contest. The Setter and Tester links will work in a day or two.

Actually ,when I read the problem statement for the first time, i do skipped that condition and coded the whole logic but then saw that it was not working for the third case.Wasted about 3 hours :’(