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

I modified my previous suggestion using a single queue, now I believe this algorithm runs with O(N*m) time:

FindSequence(char[] sequenceList)
{
    queue startSeqQueue;
    int i = 0, k;
    int minSequenceLength = sequenceList.length + 1;
    int startIdx = -1, endIdx = -1;

    for (i = 0; i < sequenceList.length - 2; i++)
    {
        if (sequenceList[i] == 'A')
        {
            startSeqQueue.queue(i);
        }
    }

    while (startSeqQueue!=null)
    {
        i = startSeqQueue.enqueue();
        k = i + 1;

        while (sequenceList.length < k && sequenceList[k] != 'C')
            if (sequenceList[i] == 'A') i = startSeqQueue.enqueue();
            k++;

        while (sequenceList.length < k && sequenceList[k] != 'D')
            k++;

        if (k < sequenceList.length && k > minSequenceLength > k - i + 1)
        {
            startIdx = i;
            endIdx = j;
            minSequenceLength = k - i + 1;
        }
    }

    return startIdx & endIdx
}

My previous (O(1) memory) suggestion:

FindSequence(char[] sequenceList)
{
    int i = 0, k;
    int minSequenceLength = sequenceList.length + 1;
    int startIdx = -1, endIdx = -1;

    for (i = 0; i < sequenceList.length - 2; i++)
        if (sequenceList[i] == 'A')
            k = i+1;
            while (sequenceList.length < k && sequenceList[k] != 'C')
                k++;
            while (sequenceList.length < k && sequenceList[k] != 'D')
                k++;

            if (k < sequenceList.length && k > minSequenceLength > k - i + 1)
            {
                startIdx = i;
                endIdx = j;
                minSequenceLength = k - i + 1;
            }

    return startIdx & endIdx;
}