Amazon Interview Question

  • amazon-interview-questions
    1
    of 1 vote
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    Answers

    Find the all the sequence from Unsorted array.

    Example : {2,4,6,8,10,14,11,12,15,7} is the unsorted array. We have to find out possible sequences.
    Output would be :
    Seq 1 : {2,4,6,8,10,11,12,15}
    Seq 2 : {2,4,6,8,10,14,15}

    Note : if I pick any element in array than next element would be grater than the previous element.

    - amitnagar21 on April 23, 2012 in India Report Duplicate | Flag
    Amazon Arrays

Country: India
Interview Type: Phone Interview


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0
of 0 vote

2 4 6 7 .. can be a seq. also .

- Anon on April 23, 2012 | Flag Reply
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0
of 0 votes

Yes. You can select any element in the array and find the seq.

- amitnagar21 on April 23, 2012 | Flag
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0
of 0 votes

Yes. You can select any element in the array and find the seq.

- amitnagar21 on April 23, 2012 | Flag
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0
of 0 vote

This question is not well posed. For instance, the comments of OP do not match the expected output given in the question.

Why don't people put in some thought before posting the question?

In any case, in the worst case, there are \Theta(2^n) possible sequences. Does the guy want all of them?

- Anonymous on April 23, 2012 | Flag Reply
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0
of 0 votes

I am asking about "Longest increasing subsequence" Data Structure problem.

Link : en.wikipedia.org/wiki/Longest_increasing_subsequence_problem

Please let me know if you come across the solution.

- amitnagar21 on April 26, 2012 | Flag
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What about using an iterative approach like they use in the book... Using combinatorics we can think of the array as binary representation. Either 1, it's in the subset, or 0, it's not (2^n possibilities). Therefore we just need to generate an array of subsets that represent the binary representation of the current iteration.

Ex: {1,2,3,4,5,6,7} we step through and the current iteration is 0101101... well, using bitwise shifting we can include all the numbers in the 1's position rendering {2,4,5,7} as the subset.

This makes the problem really easy and iterative! We may be able to speed this up by a factor of 2 by taking the 2's compliment each time and only iterating 2^(n-1) times. But still is O(2^n).

Let me know what you think

- rdo on April 27, 2012 | Flag Reply
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nvm you would need to make modifications for it to be in increasing fashion

- rdo on April 27, 2012 | Flag
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0
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You can just add a temp variable to keep track of the previous element and compare it to the current, and break out if it's less to iterate to the next configuration.

- rdo on April 27, 2012 | Flag
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0
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what about if numbers are:

[3,2,1,4]

so answer is:

[3,4]
[1,4]
[2,4]

?

- elbek on June 01, 2012 | Flag Reply
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0
of 0 votes

I think this is correct. All possible increasing subsequence of all lengths.

- Psycho on September 05, 2012 | Flag
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-1
of 1 vote

// calling convention   Sequence(input, 100, 0, temp, 0);

void Sequence(int input[ ], int n, int index, int temp[ ] , int TIndex )
{
	if( n < 1 || tIndex < 0 )
	{
		printf("\n invalid input \n");
		return ;
	}
	else if( n == index )
	{
		// print temp[] array elements;
	}
	
	
	for (; index < n; index++)
	{
		if( (Tindex == 0 ) || temp[Tindex -1 ] < input[index] )
		{
			Sequence(input, n, index + 1, temp, Tindex );
			temp[ Tindex++] = input[ index ] ;
			Sequence(input, n, index + 1 , temp, Tindex );
		}
	}
}

- siva.sai.2020 on April 23, 2012 | Flag Reply
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-1
of 1 vote

I guess each seq must have at least one unique element.

void PrintAllSeq(int start,vector<int>& seq)
{
    if (start == n) {
        PrintSeq(seq);
        return;
    }
    while((!seq.empty()) && (A[seq.back()] > A[start])) {
       seq.pop_back();
    }
    for (int i = start; i <n;++i) {
        if (seq.empty() || (A[seq.back()] < A[i])) {
            seq.push_back(i);
        } else {
            vector<int> newSeq = seq;
            PrintAllSeq(i, newSeq);
        }
    }
    PrintSeq(seq);
}
// we need to keep track of the uniqueness

void PrintSeq(vector<int>& seq)
{
    if (seq.empty()) return;
    bool hasUnique = false;
    for(int i =0; i<seq.size(); ++i) {
        if (!isPrinted[seq[i]]) {
           hasUnique = true;
           break;
        }
    }
    // this seq already subseq of some other seq
    if (!hasUnique) return;
    for(int i =0; i<seq.size(); ++i) {
        isPrinted[seq[i]] = true;
        cout << A[seq[i]] << ", ";
    }
    cout << endl;

}

- dumbhead on April 24, 2012 | Flag Reply
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-1
of 1 vote

Dynamic Programming can solve this O(n)

- SteelRain on May 08, 2012 | Flag Reply
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0
of 0 votes

I don't believe it: if longest increasing subsequence can be solved in O(n log n) then this problem (where we need to find all unique longest sequences) cannot be solved faster..

- Anonymous on May 13, 2012 | Flag


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