Question

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,

S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

void getstring(char str1[],char str2[],char hash[])
{

    int len=strlen(str1);
    int i=0,j;              //i keeps the previous index
    int min=INT_MAX;
    int cnt=0;  
    for(j=0;j<len;j++)
        {
            if(hash[str1[j]]==-1)   //means uninitialized, will be checked for first time only
            {

                cnt++;              
                hash[str1[j]]=j;                    

                if(cnt==1)
                i=j;

                int y=check(hash,str2); //checking for the characters
                //printf("y=%d",y);
                if(y)       //means all the characters are present
                {
                    if( (j-i+1)<min)
                    {
                    min=j-i+1;
                    I=i;
                    J=j;
                    }               


                }


               }
                else if(hash[str1[j]]>=0)
                {


                    hash[str1[j]]=j;



                    if(check(hash,str2) )
                    {                   

                    if( hash[str1[i]]==hash[str1[j]] )      //it means that we are moving the first
                    {                       //index only
                    i=getmin(hash,str2);    //get minimum index of the element              

                        if( (j-i+1)<min)
                        {
                            min=j-i+1;
                            I=i;
                            J=j;
                        }


                    }
                    //else
                    }
                    else
                    {
                    if( hash[str1[i]]==hash[str1[j]] )
                    i=hash[str1[i]];                        
                    }

                }//end of else-if
                }//end of for    
}

I have made the code for it using hash i.e. i am keeping the index values of the characters of the string T in the hash and using two indexes, as soon as i get any character ahead same as the character at the lower index, i first check the length and then updates the index.

This approach would take O(nk) in worst case.

n - is the number of characters in S
k - is the number of characters in T

Is there any approach which will take O(n) time for this problem?

Answer 1

So make a pass of S keeping track of when you last saw each letter in T.

At each point the farthest of the last seen letters will delimit the left edge of a window (with the current point being the right edge).

Of these windows, simply keep track of the smallest one seen so far. At the end of the algorithm this will be the smallest window overall.