package problems

/**
 * https://codility.com/media/train/13-CaterpillarMethod.pdf
 *
 * The Caterpillar method is a likeable name for a popular means of solving algorithmic tasks.
 * The idea is to check elements in a way that’s reminiscent of movements of a caterpillar.
 * The caterpillar crawls through the array. We remember the front and back positions of the caterpillar,
 * and at every step either of them is moved forward.
 */
object CaterPillarMethod {
    /**
     * Let’s check whether a sequence a0, a1, . . . , an-1 (1 < ai <= 10^9)
     * contains a contiguous sub-sequence whose sum of elements equals s.
     */
    fun isThereASequenceOfSum(A: IntArray, givenSum: Int): Boolean {
        var runningSum = 0
        var front = 0
        //for each new back position
        for (back in A) {
            //keep moving the front until the runningSum <= givenSum
            //1. If we find a runningSum == givenSum then return true
            //2. If we find a runningSum > givenSum then stop the
            // front there, move back by 1 step and then move front
            while (front < A.size && (runningSum + A[front]) <= givenSum) {
                runningSum += A[front]
                front++
            }

            if (runningSum == givenSum) {
                return true
            }

            //we did not find a sum with given back, so remove current back from runningSum and move it by 1 step
            runningSum -= back
        }

        //we did not find a sub-sequence whose sum = givenSum
        return false
    }

    /**
     * https://codility.com/media/train/13-CaterpillarMethod.pdf
     * -------Problem----------
     * You are given n sticks (of lengths 1 <= a0 <= a1  ̨ ...  <= an-1 <= 10^9). The goal is to count the number of triangles
     * that can be constructed using these sticks. More precisely, we have to count the number of triplets at indices x<y<z,
     * such that ax+ay > az
     *
     * For every pair x,y we can find the largest stick z that can be used to construct the triangle.
     * Every stick k, less than that largest stick z (such that y < k < z), can also be used to build triangle,
     * because the condition ax + ay > ak will still be true. We can add up all these triangles at once to calculate
     * all triangles for given (x, y) using formula (largestZ - y) because each z after y can make up a triangle with
     * given (x, y). We repeat the same process for remaining (x, y)
     */
    fun countTrianglesOfSticks(A: IntArray): Int {
        var trianglesCount = 0
        //for every x
        for (x in 0 until A.size) {
            var z = x + 2 //x + 2 because x+1 will be y to achieve condition (x < y < z)
            //for every y in range [x+1, n]
            for (y in x+1 until A.size) {
                //find the largest z(or you can say all the possible z) that can make triangle with
                //given (x, y) (i.e, which satisfy condition Ax + Ay > Az
                while (z < A.size && (A[x] + A[y]) < A[z]) {
                    //current z can make a triangle with given (x, y) so move to next to see if that can
                    z++ //call this currentZ
                }

                //all z values in range [x+2, currentZ] can make triangles
                //NOTE: -1 in (z - y - 1) because the last z is invalid as the while loop broke due to that z
                //NOTE: we use this calculation for following y because this current z value can make triangles
                //with all the Ys < currentZ because of given assumption (1 <= a0 <= a1  ̨ ...  <= an-1 <= 10^9) so
                // if Ax + Ay > Az then it will definitely be true for all Ys < currentZ
                trianglesCount += z - y - 1
            }
        }

        return trianglesCount
    }

    /**
     * https://app.codility.com/programmers/lessons/15-caterpillar_method/count_distinct_slices/
     * CountDistinctSlices
     * Count the number of distinct slices (containing only unique numbers).
     * Programming language:
     * An integer M and a non-empty array A consisting of N non-negative integers are given. All integers in array A are less than or equal to M.
     * A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. The slice consists of the elements A[P], A[P + 1], ..., A[Q]. A distinct slice is a slice consisting of only unique numbers. That is, no individual number occurs more than once in the slice.
     * For example, consider integer M = 6 and array A such that:
     * A[0] = 3
     * A[1] = 4
     * A[2] = 5
     * A[3] = 5
     * A[4] = 2
     * There are exactly nine distinct slices: (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2), (3, 3), (3, 4) and (4, 4).
     * The goal is to calculate the number of distinct slices.
     * Write a function:
     * class Solution { public int solution(int M, int[] A); }
     * that, given an integer M and a non-empty array A consisting of N integers, returns the number of distinct slices.
     * If the number of distinct slices is greater than 1,000,000,000, the function should return 1,000,000,000.
     * For example, given integer M = 6 and array A such that:
     * A[0] = 3
     * A[1] = 4
     * A[2] = 5
     * A[3] = 5
     * A[4] = 2
     * the function should return 9, as explained above.
     * Write an efficient algorithm for the following assumptions:
     * N is an integer within the range [1..100,000];
     * M is an integer within the range [0..100,000];
     * each element of array A is an integer within the range [0..M].
     */
    fun countDistinctSlices(A: IntArray, M: Int): Int {
        val maxDistinctSlices = 1_000_000_000
        //array to keep check of encountered numbers. We will use it to check if we already encountered
        //a number previously in the given slice or not
        val isDiscovered = Array(M + 1) { false }
        var distinctSlicesCount = A.size //each number in A is also a distinct slice of size = 1 as per 0 ≤ P ≤ Q < N

        var front = 0
        for (back in 0 until A.size) {
            //we keep going for all numbers that are not discovered in this slice (back, front)
            while (front < A.size && !isDiscovered[A[front]]) {
                //count this front as we have explored it
                isDiscovered[A[front]] = true
                front++
            }

            //all numbers till front-1 can make a slice with given back so
            //NOTE: -1 because last front is what broke the loop so is not valid for given slice
            distinctSlicesCount += front - back - 1

            //we only need to count slices till MAX_DISTINCT_SLICES
            if (distinctSlicesCount >= maxDistinctSlices) {
                return maxDistinctSlices
            }

            //as now we are no longer going to consider the current back so we mark it as not discovered
            isDiscovered[A[back]] = false
        }

        return distinctSlicesCount
    }

    /**
     * https://app.codility.com/programmers/lessons/15-caterpillar_method/count_triangles/
     * -----------------CountTriangles------------------
     * Count the number of triangles that can be built from a given set of edges.
     * An array A consisting of N integers is given. A triplet (P, Q, R) is triangular if it is possible to build a
     * triangle with sides of lengths A[P], A[Q] and A[R]. In other words, triplet (P, Q, R) is triangular
     * if 0 ≤ P < Q < R < N and:
     * A[P] + A[Q] > A[R],
     * A[Q] + A[R] > A[P],
     * A[R] + A[P] > A[Q].
     * For example, consider array A such that:
     * A[0] = 10    A[1] = 2    A[2] = 5
     * A[3] = 1     A[4] = 8    A[5] = 12
     * There are four triangular triplets that can be constructed from elements of this array,
     * namely (0, 2, 4), (0, 2, 5), (0, 4, 5), and (2, 4, 5).
     * Write a function:
     * class Solution { public int solution(int[] A); }
     * that, given an array A consisting of N integers, returns the number of triangular triplets in this array.
     * For example, given array A such that:
     * A[0] = 10    A[1] = 2    A[2] = 5
     * A[3] = 1     A[4] = 8    A[5] = 12
     * the function should return 4, as explained above.
     * Write an efficient algorithm for the following assumptions:
     * N is an integer within the range [0..1,000];
     * each element of array A is an integer within the range [1..1,000,000,000].
     */
    fun countTrianglesOfEdges(A: IntArray): Int {
        if (A.size < 3) return 0

        //NOTE: I am going to solve it same way as countTrianglesOfSticks() above.
        //first sort array as in countTrianglesOfSticks() problem, the array was sorted.
        A.sort()

        //now that array is sorted we can use the same technique we used in method countTrianglesOfSticks()
        //except that the condition for triangle is different this time. All other logic will be same
        var trianglesCount = 0
        for (p in 0 until A.size) {
            //because q will take the value (p+1) and p < q < r so r = q + 1 = p + 2
            var r = p + 2
            for (q in p+1 until A.size) {
                while (r < A.size && isTriangularTriplet(A[p], A[q], A[r])) {
                    r++
                }

                trianglesCount += r - q - 1
            }
        }

        return trianglesCount
    }

    private fun isTriangularTriplet(p: Int, q: Int, r: Int): Boolean {
        val condition1 = p.toLong() + q > r
        val condition2 = q.toLong() + r > p
        val condition3 = r.toLong() + p > q

        return condition1 && condition2 && condition3
    }

    /**
     * https://app.codility.com/programmers/lessons/15-caterpillar_method/abs_distinct/
     *
     * AbsDistinct
     * Compute number of distinct absolute values of sorted array elements.
     * Programming language:  Spoken language:
     * A non-empty array A consisting of N numbers is given. The array is sorted in non-decreasing order. The absolute distinct count of this array is the number of distinct absolute values among the elements of the array.
     * For example, consider array A such that:
     * A[0] = -5
     * A[1] = -3
     * A[2] = -1
     * A[3] =  0
     * A[4] =  3
     * A[5] =  6
     * The absolute distinct count of this array is 5, because there are 5 distinct absolute values among the elements of this array, namely 0, 1, 3, 5 and 6.
     * Write a function:
     * class Solution { public int solution(int[] A); }
     * that, given a non-empty array A consisting of N numbers, returns absolute distinct count of array A.
     * For example, given array A such that:
     * A[0] = -5
     * A[1] = -3
     * A[2] = -1
     * A[3] =  0
     * A[4] =  3
     * A[5] =  6
     * the function should return 5, as explained above.
     * Write an efficient algorithm for the following assumptions:
     * N is an integer within the range [1..100,000];
     * each element of array A is an integer within the range [−2,147,483,648..2,147,483,647];
     * array A is sorted in non-decreasing order.
     *
     */
    fun absDistinct(A: IntArray): Int {
        //NOTE: Any easy solution would be to sort the array by absolute values and the duplicates will
        //become adjacent values so can be easily discarded in one go or we can pass the array to Set class
        //to get distinct values but I want to solve it using CaterPillar method
        var count = 0
        var start = 0
        var end = A.lastIndex
        while (start <= end) {
            //count the current value as distinct because on iteration we will have
            //a distinct value. We will skip the duplicates using index increments in
            //the logic below
            count++

            //we are only interested in absolute values
            val startValue = Math.abs(A[start].toLong())
            val endValue = Math.abs(A[end].toLong())

            when {
                startValue == endValue -> {
                    //both start and end values are equal, duplicate encountered. We have considered 1 value above
                    //into our count and in case of duplicates only 1 value should be considered as we are counting
                    //distinct values.

                    //now that current startValue (which is also endValue as both are equal in this case)
                    //is discovered/counted, now remove duplicates on both start and end sides of the array

                    //discard start side duplicates
                    while (start <= end && Math.abs(A[start].toLong()) == endValue) {
                        start++
                    }

                    //discard end side duplicates
                    while (end >= start && Math.abs(A[end].toLong()) == endValue) {
                        end--
                    }
                }
                //start value > end value so move start by 1 element
                startValue > endValue -> {
                    //skip start side duplicates
                    while (start <= end && startValue == Math.abs(A[start].toLong())) {
                        start++
                    }
                }
                else -> { // endValue > startValue, move end by 1 element
                    //skip start side duplicates
                    while (end >= start && endValue == Math.abs(A[end].toLong())) {
                        end--
                    }
                }
            }
        }

        return count
    }

    /**
     * https://app.codility.com/programmers/lessons/15-caterpillar_method/min_abs_sum_of_two/
     * -------------MinAbsSumOfTwo-------------------
     * Find the minimal absolute value of a sum of two elements.
     * Programming language:
     * Let A be a non-empty array consisting of N integers.
     * The abs sum of two for a pair of indices (P, Q) is the absolute value |A[P] + A[Q]|, for 0 ≤ P ≤ Q < N.
     * For example, the following array A:
     * A[0] =  1
     * A[1] =  4
     * A[2] = -3
     * has pairs of indices (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).
     * The abs sum of two for the pair (0, 0) is A[0] + A[0] = |1 + 1| = 2.
     * The abs sum of two for the pair (0, 1) is A[0] + A[1] = |1 + 4| = 5.
     * The abs sum of two for the pair (0, 2) is A[0] + A[2] = |1 + (−3)| = 2.
     * The abs sum of two for the pair (1, 1) is A[1] + A[1] = |4 + 4| = 8.
     * The abs sum of two for the pair (1, 2) is A[1] + A[2] = |4 + (−3)| = 1.
     * The abs sum of two for the pair (2, 2) is A[2] + A[2] = |(−3) + (−3)| = 6.
     * Write a function:
     * class Solution { public int solution(int[] A); }
     * that, given a non-empty array A consisting of N integers, returns the minimal abs sum of two for any pair of indices in this array.
     * For example, given the following array A:
     * A[0] =  1
     * A[1] =  4
     * A[2] = -3
     * the function should return 1, as explained above.
     * Given array A:
     * A[0] = -8
     * A[1] =  4
     * A[2] =  5
     * A[3] =-10
     * A[4] =  3
     * the function should return |(−8) + 5| = 3.
     * Write an efficient algorithm for the following assumptions:
     * N is an integer within the range [1..100,000];
     * each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].
     *
     * ----------Solution------------
     * There are two cases in which we will get a min sum
     *
     * 1. A sum of same or very close positive and negative numbers. (like (4, -4) or (-3, 4) etc.) because in
     *      this case we will get a sum that is very small due to close values and opposite signs
     * 2. A sum of the absolute smallest number with itself like (e.g, (1, 1)) as in this case the total sum
     *      will be small as the number is smallest
     *
     *  So the smallest sum will be from either of the above two cases. If we sort the array then we will get
     *  negative numbers on start and positive numbers on the end and smallest number in between somewhere. Then we can use
     *  caterpillar method to find the smallest sum.
     *
     *  Idea behind code:
     *  Idea is to start from both ends (negative extreme end and positive extreme end) while keeping track of min sum
     *  and try to converge on the smallest possible absolute number (case-2). This way we will be able to cover both of the cases
     *  mentioned above
     */

    data class MyName(var name: String, var name2: String) {}

    fun printStudents(vararg students: MyName) {
        for (s in students) println(s)
    }

    fun minAbsSumOfTwo(A: IntArray): Int {
        val m = MyName("", "")
        printStudents(m)
        
        //pick sum of first number with itself just to initialize the sum with some valid value
        var minSum = Math.abs(A[0] + A[0])
        //sort the array so that negative and positive numbers are on opposite ends.
        A.sort()

        //start from both ends and try to converge to smallest absolute number
        var start = 0
        var end = A.lastIndex
        while (start <= end) {
            //only keep the min of the two
            minSum = Math.min(minSum, Math.abs(A[start] + A[end]))

            //as we want to converge to smallest absolute number so move the greater of start/end
            if (Math.abs(A[start]) > Math.abs(A[end])) {
                start++
            } else {
                end--
            }
        }

        return minSum
    }
}
package problems

/**
 * https://codility.com/media/train/13-CaterpillarMethod.pdf
 *
 * The Caterpillar method is a likeable name for a popular means of solving algorithmic tasks.
 * The idea is to check elements in a way that’s reminiscent of movements of a caterpillar.
 * The caterpillar crawls through the array. We remember the front and back positions of the caterpillar,
 * and at every step either of them is moved forward.
 */
object CaterPillarMethod {
    /**
     * Let’s check whether a sequence a0, a1, . . . , an-1 (1 < ai <= 10^9)
     * contains a contiguous sub-sequence whose sum of elements equals s.
     */
    fun isThereASequenceOfSum(A: IntArray, givenSum: Int): Boolean {
        var runningSum = 0
        var front = 0
        //for each new back position
        for (back in A) {
            //keep moving the front until the runningSum <= givenSum
            //1. If we find a runningSum == givenSum then return true
            //2. If we find a runningSum > givenSum then stop the
            // front there, move back by 1 step and then move front
            while (front < A.size && (runningSum + A[front]) <= givenSum) {
                runningSum += A[front]
                front++
            }

            if (runningSum == givenSum) {
                return true
            }

            //we did not find a sum with given back, so remove current back from runningSum and move it by 1 step
            runningSum -= back
        }

        //we did not find a sub-sequence whose sum = givenSum
        return false
    }

    /**
     * https://codility.com/media/train/13-CaterpillarMethod.pdf
     * -------Problem----------
     * You are given n sticks (of lengths 1 <= a0 <= a1  ̨ ...  <= an-1 <= 10^9). The goal is to count the number of triangles
     * that can be constructed using these sticks. More precisely, we have to count the number of triplets at indices x<y<z,
     * such that ax+ay > az
     *
     * For every pair x,y we can find the largest stick z that can be used to construct the triangle.
     * Every stick k, less than that largest stick z (such that y < k < z), can also be used to build triangle,
     * because the condition ax + ay > ak will still be true. We can add up all these triangles at once to calculate
     * all triangles for given (x, y) using formula (largestZ - y) because each z after y can make up a triangle with
     * given (x, y). We repeat the same process for remaining (x, y)
     */
    fun countTrianglesOfSticks(A: IntArray): Int {
        var trianglesCount = 0
        //for every x
        for (x in 0 until A.size) {
            var z = x + 2 //x + 2 because x+1 will be y to achieve condition (x < y < z)
            //for every y in range [x+1, n]
            for (y in x+1 until A.size) {
                //find the largest z(or you can say all the possible z) that can make triangle with
                //given (x, y) (i.e, which satisfy condition Ax + Ay > Az
                while (z < A.size && (A[x] + A[y]) < A[z]) {
                    //current z can make a triangle with given (x, y) so move to next to see if that can
                    z++ //call this currentZ
                }

                //all z values in range [x+2, currentZ] can make triangles
                //NOTE: -1 in (z - y - 1) because the last z is invalid as the while loop broke due to that z
                //NOTE: we use this calculation for following y because this current z value can make triangles
                //with all the Ys < currentZ because of given assumption (1 <= a0 <= a1  ̨ ...  <= an-1 <= 10^9) so
                // if Ax + Ay > Az then it will definitely be true for all Ys < currentZ
                trianglesCount += z - y - 1
            }
        }

        return trianglesCount
    }

    /**
     * https://app.codility.com/programmers/lessons/15-caterpillar_method/count_distinct_slices/
     * CountDistinctSlices
     * Count the number of distinct slices (containing only unique numbers).
     * Programming language:
     * An integer M and a non-empty array A consisting of N non-negative integers are given. All integers in array A are less than or equal to M.
     * A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. The slice consists of the elements A[P], A[P + 1], ..., A[Q]. A distinct slice is a slice consisting of only unique numbers. That is, no individual number occurs more than once in the slice.
     * For example, consider integer M = 6 and array A such that:
     * A[0] = 3
     * A[1] = 4
     * A[2] = 5
     * A[3] = 5
     * A[4] = 2
     * There are exactly nine distinct slices: (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2), (3, 3), (3, 4) and (4, 4).
     * The goal is to calculate the number of distinct slices.
     * Write a function:
     * class Solution { public int solution(int M, int[] A); }
     * that, given an integer M and a non-empty array A consisting of N integers, returns the number of distinct slices.
     * If the number of distinct slices is greater than 1,000,000,000, the function should return 1,000,000,000.
     * For example, given integer M = 6 and array A such that:
     * A[0] = 3
     * A[1] = 4
     * A[2] = 5
     * A[3] = 5
     * A[4] = 2
     * the function should return 9, as explained above.
     * Write an efficient algorithm for the following assumptions:
     * N is an integer within the range [1..100,000];
     * M is an integer within the range [0..100,000];
     * each element of array A is an integer within the range [0..M].
     */
    fun countDistinctSlices(A: IntArray, M: Int): Int {
        val maxDistinctSlices = 1_000_000_000
        //array to keep check of encountered numbers. We will use it to check if we already encountered
        //a number previously in the given slice or not
        val isDiscovered = Array(M + 1) { false }
        var distinctSlicesCount = A.size //each number in A is also a distinct slice of size = 1 as per 0 ≤ P ≤ Q < N

        var front = 0
        for (back in 0 until A.size) {
            //we keep going for all numbers that are not discovered in this slice (back, front)
            while (front < A.size && !isDiscovered[A[front]]) {
                //count this front as we have explored it
                isDiscovered[A[front]] = true
                front++
            }

            //all numbers till front-1 can make a slice with given back so
            //NOTE: -1 because last front is what broke the loop so is not valid for given slice
            distinctSlicesCount += front - back - 1

            //we only need to count slices till MAX_DISTINCT_SLICES
            if (distinctSlicesCount >= maxDistinctSlices) {
                return maxDistinctSlices
            }

            //as now we are no longer going to consider the current back so we mark it as not discovered
            isDiscovered[A[back]] = false
        }

        return distinctSlicesCount
    }

    /**
     * https://app.codility.com/programmers/lessons/15-caterpillar_method/count_triangles/
     * -----------------CountTriangles------------------
     * Count the number of triangles that can be built from a given set of edges.
     * An array A consisting of N integers is given. A triplet (P, Q, R) is triangular if it is possible to build a
     * triangle with sides of lengths A[P], A[Q] and A[R]. In other words, triplet (P, Q, R) is triangular
     * if 0 ≤ P < Q < R < N and:
     * A[P] + A[Q] > A[R],
     * A[Q] + A[R] > A[P],
     * A[R] + A[P] > A[Q].
     * For example, consider array A such that:
     * A[0] = 10    A[1] = 2    A[2] = 5
     * A[3] = 1     A[4] = 8    A[5] = 12
     * There are four triangular triplets that can be constructed from elements of this array,
     * namely (0, 2, 4), (0, 2, 5), (0, 4, 5), and (2, 4, 5).
     * Write a function:
     * class Solution { public int solution(int[] A); }
     * that, given an array A consisting of N integers, returns the number of triangular triplets in this array.
     * For example, given array A such that:
     * A[0] = 10    A[1] = 2    A[2] = 5
     * A[3] = 1     A[4] = 8    A[5] = 12
     * the function should return 4, as explained above.
     * Write an efficient algorithm for the following assumptions:
     * N is an integer within the range [0..1,000];
     * each element of array A is an integer within the range [1..1,000,000,000].
     */
    fun countTrianglesOfEdges(A: IntArray): Int {
        if (A.size < 3) return 0

        //NOTE: I am going to solve it same way as countTrianglesOfSticks() above.
        //first sort array as in countTrianglesOfSticks() problem, the array was sorted.
        A.sort()

        //now that array is sorted we can use the same technique we used in method countTrianglesOfSticks()
        //except that the condition for triangle is different this time. All other logic will be same
        var trianglesCount = 0
        for (p in 0 until A.size) {
            //because q will take the value (p+1) and p < q < r so r = q + 1 = p + 2
            var r = p + 2
            for (q in p+1 until A.size) {
                while (r < A.size && isTriangularTriplet(A[p], A[q], A[r])) {
                    r++
                }

                trianglesCount += r - q - 1
            }
        }

        return trianglesCount
    }

    private fun isTriangularTriplet(p: Int, q: Int, r: Int): Boolean {
        val condition1 = p.toLong() + q > r
        val condition2 = q.toLong() + r > p
        val condition3 = r.toLong() + p > q

        return condition1 && condition2 && condition3
    }

    /**
     * https://app.codility.com/programmers/lessons/15-caterpillar_method/abs_distinct/
     *
     * AbsDistinct
     * Compute number of distinct absolute values of sorted array elements.
     * Programming language:  Spoken language:
     * A non-empty array A consisting of N numbers is given. The array is sorted in non-decreasing order. The absolute distinct count of this array is the number of distinct absolute values among the elements of the array.
     * For example, consider array A such that:
     * A[0] = -5
     * A[1] = -3
     * A[2] = -1
     * A[3] =  0
     * A[4] =  3
     * A[5] =  6
     * The absolute distinct count of this array is 5, because there are 5 distinct absolute values among the elements of this array, namely 0, 1, 3, 5 and 6.
     * Write a function:
     * class Solution { public int solution(int[] A); }
     * that, given a non-empty array A consisting of N numbers, returns absolute distinct count of array A.
     * For example, given array A such that:
     * A[0] = -5
     * A[1] = -3
     * A[2] = -1
     * A[3] =  0
     * A[4] =  3
     * A[5] =  6
     * the function should return 5, as explained above.
     * Write an efficient algorithm for the following assumptions:
     * N is an integer within the range [1..100,000];
     * each element of array A is an integer within the range [−2,147,483,648..2,147,483,647];
     * array A is sorted in non-decreasing order.
     *
     */
    fun absDistinct(A: IntArray): Int {
        //NOTE: Any easy solution would be to sort the array by absolute values and the duplicates will
        //become adjacent values so can be easily discarded in one go or we can pass the array to Set class
        //to get distinct values but I want to solve it using CaterPillar method
        var count = 0
        var start = 0
        var end = A.lastIndex
        while (start <= end) {
            //count the current value as distinct because on iteration we will have
            //a distinct value. We will skip the duplicates using index increments in
            //the logic below
            count++

            //we are only interested in absolute values
            val startValue = Math.abs(A[start].toLong())
            val endValue = Math.abs(A[end].toLong())

            when {
                startValue == endValue -> {
                    //both start and end values are equal, duplicate encountered. We have considered 1 value above
                    //into our count and in case of duplicates only 1 value should be considered as we are counting
                    //distinct values.

                    //now that current startValue (which is also endValue as both are equal in this case)
                    //is discovered/counted, now remove duplicates on both start and end sides of the array

                    //discard start side duplicates
                    while (start <= end && Math.abs(A[start].toLong()) == endValue) {
                        start++
                    }

                    //discard end side duplicates
                    while (end >= start && Math.abs(A[end].toLong()) == endValue) {
                        end--
                    }
                }
                //start value > end value so move start by 1 element
                startValue > endValue -> {
                    //skip start side duplicates
                    while (start <= end && startValue == Math.abs(A[start].toLong())) {
                        start++
                    }
                }
                else -> { // endValue > startValue, move end by 1 element
                    //skip start side duplicates
                    while (end >= start && endValue == Math.abs(A[end].toLong())) {
                        end--
                    }
                }
            }
        }

        return count
    }

    /**
     * https://app.codility.com/programmers/lessons/15-caterpillar_method/min_abs_sum_of_two/
     * -------------MinAbsSumOfTwo-------------------
     * Find the minimal absolute value of a sum of two elements.
     * Programming language:
     * Let A be a non-empty array consisting of N integers.
     * The abs sum of two for a pair of indices (P, Q) is the absolute value |A[P] + A[Q]|, for 0 ≤ P ≤ Q < N.
     * For example, the following array A:
     * A[0] =  1
     * A[1] =  4
     * A[2] = -3
     * has pairs of indices (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).
     * The abs sum of two for the pair (0, 0) is A[0] + A[0] = |1 + 1| = 2.
     * The abs sum of two for the pair (0, 1) is A[0] + A[1] = |1 + 4| = 5.
     * The abs sum of two for the pair (0, 2) is A[0] + A[2] = |1 + (−3)| = 2.
     * The abs sum of two for the pair (1, 1) is A[1] + A[1] = |4 + 4| = 8.
     * The abs sum of two for the pair (1, 2) is A[1] + A[2] = |4 + (−3)| = 1.
     * The abs sum of two for the pair (2, 2) is A[2] + A[2] = |(−3) + (−3)| = 6.
     * Write a function:
     * class Solution { public int solution(int[] A); }
     * that, given a non-empty array A consisting of N integers, returns the minimal abs sum of two for any pair of indices in this array.
     * For example, given the following array A:
     * A[0] =  1
     * A[1] =  4
     * A[2] = -3
     * the function should return 1, as explained above.
     * Given array A:
     * A[0] = -8
     * A[1] =  4
     * A[2] =  5
     * A[3] =-10
     * A[4] =  3
     * the function should return |(−8) + 5| = 3.
     * Write an efficient algorithm for the following assumptions:
     * N is an integer within the range [1..100,000];
     * each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].
     *
     * ----------Solution------------
     * There are two cases in which we will get a min sum
     *
     * 1. A sum of same or very close positive and negative numbers. (like (4, -4) or (-3, 4) etc.) because in
     *      this case we will get a sum that is very small due to close values and opposite signs
     * 2. A sum of the absolute smallest number with itself like (e.g, (1, 1)) as in this case the total sum
     *      will be small as the number is smallest
     *
     *  So the smallest sum will be from either of the above two cases. If we sort the array then we will get
     *  negative numbers on start and positive numbers on the end and smallest number in between somewhere. Then we can use
     *  caterpillar method to find the smallest sum.
     *
     *  Idea behind code:
     *  Idea is to start from both ends (negative extreme end and positive extreme end) while keeping track of min sum
     *  and try to converge on the smallest possible absolute number (case-2). This way we will be able to cover both of the cases
     *  mentioned above
     */

    data class MyName(var name: String, var name2: String) {}

    fun printStudents(vararg students: MyName) {
        for (s in students) println(s)
    }

    fun minAbsSumOfTwo(A: IntArray): Int {
        val m = MyName("", "")
        printStudents(m)
        
        //pick sum of first number with itself just to initialize the sum with some valid value
        var minSum = Math.abs(A[0] + A[0])
        //sort the array so that negative and positive numbers are on opposite ends.
        A.sort()

        //start from both ends and try to converge to smallest absolute number
        var start = 0
        var end = A.lastIndex
        while (start <= end) {
            //only keep the min of the two
            minSum = Math.min(minSum, Math.abs(A[start] + A[end]))

            //as we want to converge to smallest absolute number so move the greater of start/end
            if (Math.abs(A[start]) > Math.abs(A[end])) {
                start++
            } else {
                end--
            }
        }

        return minSum
    }
}