Find length of smallest window that contains all the characters of a string in another string

Question

Recently i have been interviewed. I didn\'t do well cause i got stuck at the following question

suppose a sequence is given : A D C B D A B C D A C D and search seq

Answer 1

Here's my solution. It follows one of the pattern matching solutions. Please comment/correct me if I'm wrong.

Given the input string as in the question A D C B D A B C D A C D. Let's first compute the indices where A occurs. Assuming a zero based index this should be [0,5,9].

Now the pseudo code is as follows.

    Store the indices of A in a list say *orders*.// orders=[0,5,9]
    globalminStart, globalminEnd=0,localMinStart=0,localMinEnd=0;
    for (index: orders)
     {
       int i =index;
       Stack chars=new Stack();// to store the characters
      i=localminStart;
     while(i< length of input string)
       { 
           if(str.charAt(i)=='C') // we've already seen A, so we look for C
           st.push(str.charAt(i));
           i++;
           continue;
           else if(str.charAt(i)=='D' and st.peek()=='C')
           localminEnd=i; // we have a match! so assign value of i to len
           i+=1;
           break;
           else if(str.charAt(i)=='A' )// seen the next A
           break;
    }
     if (globalMinEnd-globalMinStart

P.S: this is pseudocode and a rough idea. Id be happy to correct it and understand if there's something wrong.

AFAIC Time complexity -O(n). Space complexity O(n)