ASYA2 editorial

level : simple-easy ;
we can map string to bitset<26> and with a 2 loop O(N1N2(bitset_oper)) for each bitset of N1 and bitset of N2 get OR | to
such bitset and if now bitset.count==26 increase result (1 sec)
my code :slight_smile: --> https://www.codechef.com/status/ASYA2,ahmadreza1994

3 Likes

here is my code with complexity O(26*N1)

 #include<stdio.h>
int n1,j,k,i,l,n2,s1[10000][26],prev[26],s2[26],min;
long int w;
int main()
{
	char a;
	scanf("%d%d",&n1,&n2);
	scanf("%c",&a);
	for(i=0;i<n1;i++)
	{
		scanf("%c",&a);
		while(a!=10){
		s1[i][a-65]=1;	
		scanf("%c",&a);
	}
	}
		for(i=1;i<=n2;i++)
	{
		scanf("%c",&a);
		while(a!=10){
		if(prev[a-65]!=i)
		{
		 prev[a-65]=i;		
		s2[a-65]++;}	
		scanf("%c",&a);
	}
	}
	for(i=0;i<n1;i++)
	{
		k=0;min=n2;j=0;
		for(l=0;l<26;l++){
		if(!s1[i][l]&&s2[l]>=1){
		k++;
		if(min>s2[l])
		min=s2[l];
	    }
	    else if (s1[i][l])
	    j++;
	    }
		if(k+j==26)
		w+=min;
	}
	printf("%ld\n",w); return 0;
}

I was unable to submit my code yesterday so i have uploaded my solution, Don’t know whether it is right or wrong. I think it is right but please help me if it is wrong.

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