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<localMinEnd-localMinStart)
     {
       globalMinEnd=localMinEnd;
       globalMinStart=localMinStart;
     }
   }

    return [globalMinstart,globalMinEnd]
    }

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)

Answer 2

Here is my O(m*n) algorithm in Java:

class ShortestWindowAlgorithm {

    Multimap<Character, Integer> charToNeedleIdx; // Character -> indexes in needle, from rightmost to leftmost | Multimap is a class from Guava
    int[] prefixesIdx; // prefixesIdx[i] -- rightmost index in the hay window that contains the shortest found prefix of needle[0..i]
    int[] prefixesLengths; // prefixesLengths[i] -- shortest window containing needle[0..i]

    public int shortestWindow(String hay, String needle) {
        init(needle);
        for (int i = 0; i < hay.length(); i++) {
            for (int needleIdx : charToNeedleIdx.get(hay.charAt(i))) {
                if (firstTimeAchievedPrefix(needleIdx) || foundShorterPrefix(needleIdx, i)) {
                    prefixesIdx[needleIdx] = i;
                    prefixesLengths[needleIdx] = getPrefixNewLength(needleIdx, i);
                    forgetOldPrefixes(needleIdx);
                }
            }
        }
        return prefixesLengths[prefixesLengths.length - 1];
    }

    private void init(String needle) {
        charToNeedleIdx = ArrayListMultimap.create();
        prefixesIdx = new int[needle.length()];
        prefixesLengths = new int[needle.length()];
        for (int i = needle.length() - 1; i >= 0; i--) {
            charToNeedleIdx.put(needle.charAt(i), i);
            prefixesIdx[i] = -1;
            prefixesLengths[i] = -1;
        }
    }

    private boolean firstTimeAchievedPrefix(int needleIdx) {
        int shortestPrefixSoFar = prefixesLengths[needleIdx];
        return shortestPrefixSoFar == -1 && (needleIdx == 0 || prefixesLengths[needleIdx - 1] != -1);
    }

    private boolean foundShorterPrefix(int needleIdx, int hayIdx) {
        int shortestPrefixSoFar = prefixesLengths[needleIdx];
        int newLength = getPrefixNewLength(needleIdx, hayIdx);
        return newLength <= shortestPrefixSoFar;
    }

    private int getPrefixNewLength(int needleIdx, int hayIdx) {
        return needleIdx == 0 ? 1 : (prefixesLengths[needleIdx - 1] + (hayIdx - prefixesIdx[needleIdx - 1]));
    }

    private void forgetOldPrefixes(int needleIdx) {
        if (needleIdx > 0) {
            prefixesLengths[needleIdx - 1] = -1;
            prefixesIdx[needleIdx - 1] = -1;
        }
    }
}

It works on every input and also can handle repeated characters etc.

Here are some examples:

public class StackOverflow {

    public static void main(String[] args) {
        ShortestWindowAlgorithm algorithm = new ShortestWindowAlgorithm();
        System.out.println(algorithm.shortestWindow("AXCXXCAXCXAXCXCXAXAXCXCXDXDXDXAXCXDXAXAXCD", "AACD")); // 6
        System.out.println(algorithm.shortestWindow("ADCBDABCDACD", "ACD")); // 3
        System.out.println(algorithm.shortestWindow("ADCBDABCD", "ACD")); // 4
    }