Moonfrog Labs Interview Experience | Set 3

Q1. Given a sequence of integers, find the longest increasing subsequence. Example:
arr = [1, 2, 5, 3, 7]
ans : [1, 2, 5, 7] or [1, 2, 3, 7]

arr = [4, 3, 1, 2]
ans: [1, 2].

Solution:

import java.util.Arrays;

/** @author hiccup  */
class LIS
{
    static int[] maxLIS;
    static int[] result;

    public static void getLCS(int[] arr)
    {
        if (null == arr || 0 == arr.length)
            return;

        maxLIS = new int[arr.length];
        /* At least LCS is 1 i.e. element is the
                   only one in given sequence */
        Arrays.fill(maxLIS, 1);


        /**
         *
         */
        for (int curIdx = 1; curIdx < arr.length; curIdx++)
        {
            for (int beginIdx = 0; beginIdx <= curIdx - 1; beginIdx++)
            {
                if (arr[curIdx] > arr[beginIdx])
                {
                    if (maxLIS[curIdx] < maxLIS[beginIdx] + 1)
                    {
                        //System.out.print(arr[curIdx] + "  ");
                        maxLIS[curIdx] = maxLIS[beginIdx] + 1;
                    }
                }
            }
        }

        int max = maxLIS[0];
        result = new int[arr.length];
        Arrays.fill(result, -1);

        int cpIdx = 0;
        for (int idx = 0; idx < maxLIS.length; idx++)
        {
            /* Put sequence at cpIdx   */
            if (-1 == result[maxLIS[idx] - 1])
            {
                result[cpIdx++] = arr[idx];
            }
        }

        /*  Print sequence       */
        for (int idx = 0; idx < result.length; idx++)
        {
            System.out.print(result[idx] + " ");
        }
    }

    public static void main(String[] args)
    {
        int[] arr = {1, 2, 5, 3, 7};
        LIS.getLCS(arr);
    }
}

[1, 2, 3, 3, 4]

—————————————————————————————————————————————–
Q2. Given a vectors of numbers of fixed length, for example:

v1 = [4, 3, 1, 2] v2 = [2, 4, 3, 5]

The relationship nested between two vectors is defined as follows:

if the corresponding entries of a vector are all smaller than the other vector, after rearranging entries of vector if needed, then first vector
is said to be nested in the other. Example

Not nested
v1 – [4, 3, 1, 2] v2 – [2, 4, 3, 5]
v2 – [2, 4, 3, 5] v1 – [4, 3, 1, 2]

After re-arranging:

Nested
v1 – [4, 3, 1, 2]
v2 – [5, 4, 2, 3]

Hence v1 is nested in v2.

Given a pair of such vectors , write a function as follows:

function isNested(Vec a, Vec b);

Result:
-1 if a is nested in b
1 if b is nested in a
0 if nesting is not possible.

———————————————————————————————————————-

Q3. Given a list of numbers in random order. Is it possible to pair all elements in the list in such a way that no two pairs share an element
and the sum of elements in a pair is divisible by 101. Example:

v1 [ 1, 100, 1]
Ans: No

v2 [1, 100, 100, 1] [2, 98, 101, 1]
Ans: Yes

v3 [1, 200, 100, 100, 2, 1]
Ans: yes

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