Test results - Codility

Tasks Details

medium

1. NailingPlanks

Count the minimum number of nails that allow a series of planks to be nailed.

100%

You are given two non-empty arrays A and B consisting of N integers. These arrays represent N planks. More precisely, A[K] is the start and B[K] the end of the K−th plank.

Next, you are given a non-empty array C consisting of M integers. This array represents M nails. More precisely, C[I] is the position where you can hammer in the I−th nail.

We say that a plank (A[K], B[K]) is nailed if there exists a nail C[I] such that A[K] ≤ C[I] ≤ B[K].

The goal is to find the minimum number of nails that must be used until all the planks are nailed. In other words, you should find a value J such that all planks will be nailed after using only the first J nails. More precisely, for every plank (A[K], B[K]) such that 0 ≤ K < N, there should exist a nail C[I] such that I < J and A[K] ≤ C[I] ≤ B[K].

For example, given arrays A, B such that:

A[0] = 1 B[0] = 4 A[1] = 4 B[1] = 5 A[2] = 5 B[2] = 9 A[3] = 8 B[3] = 10

four planks are represented: [1, 4], [4, 5], [5, 9] and [8, 10].

Given array C such that:

C[0] = 4 C[1] = 6 C[2] = 7 C[3] = 10 C[4] = 2

if we use the following nails:

0, then planks [1, 4] and [4, 5] will both be nailed.

0, 1, then planks [1, 4], [4, 5] and [5, 9] will be nailed.

0, 1, 2, then planks [1, 4], [4, 5] and [5, 9] will be nailed.

0, 1, 2, 3, then all the planks will be nailed.

Thus, four is the minimum number of nails that, used sequentially, allow all the planks to be nailed.

Write a function:

class Solution { public int solution(int[] A, int[] B, int[] C); }

that, given two non-empty arrays A and B consisting of N integers and a non-empty array C consisting of M integers, returns the minimum number of nails that, used sequentially, allow all the planks to be nailed.

If it is not possible to nail all the planks, the function should return −1.

For example, given arrays A, B, C such that:

A[0] = 1 B[0] = 4 A[1] = 4 B[1] = 5 A[2] = 5 B[2] = 9 A[3] = 8 B[3] = 10 C[0] = 4 C[1] = 6 C[2] = 7 C[3] = 10 C[4] = 2

the function should return 4, as explained above.

Write an efficient algorithm for the following assumptions:

N and M are integers within the range [1..30,000];

each element of arrays A, B and C is an integer within the range [1..2*M];

A[K] ≤ B[K].

Solution

Programming language used Java 8

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 16:45:45 UTC, java, verify, result: Passed

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class Solution {
    public int solution(int[] A, int[] B, int[] C) {
        int[] ends = getPlankEnds(A, B, C);
        int[] starts = getPlankStarts(ends);
        int[] nails = getNailPositions(C, ends);

        int result = -1;
        Queue queue = new Queue(ends.length);
        for (int i = 1; i < ends.length; i++) {
            if (ends[i] > 0) {
                if (!queue.isEmpty() && ends[queue.last()] == -1) {
                     queue.removeLast();
                }
                ends[i] = -1;
                queue.addLast(i);
            } else if (queue.isEmpty()) {
                continue;
            }
            if (nails[i] != 0) {
                if (ends[queue.last()] == -1 || nails[i] < ends[queue.last()]) {
                    ends[queue.last()] = nails[i];
                    while (queue.size() > 1 && ends[queue.nextToLast()] > nails[i]) {
                        queue.removeNextToLast();
                    }
                }
            }
            if (starts[i] != 0) {
                int min = ends[queue.first()];
                if (min == -1) {
                    return -1;
                }
                if (starts[i] == queue.first()) {
                    queue.removeFirst();
                }
                if (result == -1 || min > result) {
                    result = min;
                }
            }
        }
        return result;
    }

    private int[] getPlankEnds(int[] A, int[] B, int[] C) {
        int[] planks = new int[2 * C.length + 1];
        for (int i = 0; i < A.length; i++) {
            if (planks[A[i]] == 0 || B[i] < planks[A[i]]) {
                planks[A[i]] = B[i];
            }
        }

        Queue stack = new Queue(2 * C.length);
        for (int i = 1; i < planks.length; i++) {
            if (planks[i] != 0) {
                while (!stack.isEmpty() && planks[i] <= planks[stack.last()]) {
                    stack.removeLast();
                }
                stack.addLast(i);
            }
        }

        int[] ends = new int[planks.length];
        while (!stack.isEmpty()) {
            int start = stack.removeLast();
            ends[start] = planks[start];
        }
        return ends;
    }

    private int[] getPlankStarts(int[] ends) {
		int[] starts = new int[ends.length];
		for (int i = 1; i < ends.length; i++) {
		    if (ends[i] > 0) {
		        starts[ends[i]] = i;
		    }
		}
        return starts;
    }

    private int[] getNailPositions(int[] C, int[] starts) {
        int[] nails = new int[starts.length];
        for (int i = 0; i < C.length; i++) {
            if (nails[C[i]] == 0) {
                nails[C[i]] = i + 1;
            }
        }
        return nails;
    }

    private static class Queue {
        private int[] array;
        private int start = 0;
        private int end = -1;
        private int size = 0;

        Queue(int capacity) {
            this.array = new int[capacity];
        }
        void addLast(int element) {
            size++;
            end++;
            if (end == array.length) {
                end = 0;
            }
            array[end] = element;
        }
        int removeLast() {
            size--;
            int result = array[end--];
            if (end < 0) {
                end = array.length - 1;
            }
            return result;
        }
        int removeFirst() {
            size--;
            int result = array[start++];
            if (start == array.length) {
                start = 0;
            }
            return result;
        }
        int removeNextToLast() {
            size--;
            int index = nextToLastIndex();
            int result = array[index];
            array[index] = array[end];
            end = index;
            return result;
        }
        int last() {
            return array[end];
        }
        int first() {
            return array[start];
        }
        int nextToLast() {
            return array[nextToLastIndex()];
        }
        boolean isEmpty() {
            return size == 0;
        }
        int size() {
            return size;
        }
        int nextToLastIndex() {
            return end > 0 ? end - 1 : array.length - 1;
        }
    }
}

Analysis

▶

example
example test

✔

Code: 16:45:50 UTC, java, final, score: 100


class Solution {
    public int solution(int[] A, int[] B, int[] C) {
        int[] ends = getPlankEnds(A, B, C);
        int[] starts = getPlankStarts(ends);
        int[] nails = getNailPositions(C, ends);

        int result = -1;
        Queue queue = new Queue(ends.length);
        for (int i = 1; i < ends.length; i++) {
            if (ends[i] > 0) {
                if (!queue.isEmpty() && ends[queue.last()] == -1) {
                     queue.removeLast();
                }
                ends[i] = -1;
                queue.addLast(i);
            } else if (queue.isEmpty()) {
                continue;
            }
            if (nails[i] != 0) {
                if (ends[queue.last()] == -1 || nails[i] < ends[queue.last()]) {
                    ends[queue.last()] = nails[i];
                    while (queue.size() > 1 && ends[queue.nextToLast()] > nails[i]) {
                        queue.removeNextToLast();
                    }
                }
            }
            if (starts[i] != 0) {
                int min = ends[queue.first()];
                if (min == -1) {
                    return -1;
                }
                if (starts[i] == queue.first()) {
                    queue.removeFirst();
                }
                if (result == -1 || min > result) {
                    result = min;
                }
            }
        }
        return result;
    }

    private int[] getPlankEnds(int[] A, int[] B, int[] C) {
        int[] planks = new int[2 * C.length + 1];
        for (int i = 0; i < A.length; i++) {
            if (planks[A[i]] == 0 || B[i] < planks[A[i]]) {
                planks[A[i]] = B[i];
            }
        }

        Queue stack = new Queue(2 * C.length);
        for (int i = 1; i < planks.length; i++) {
            if (planks[i] != 0) {
                while (!stack.isEmpty() && planks[i] <= planks[stack.last()]) {
                    stack.removeLast();
                }
                stack.addLast(i);
            }
        }

        int[] ends = new int[planks.length];
        while (!stack.isEmpty()) {
            int start = stack.removeLast();
            ends[start] = planks[start];
        }
        return ends;
    }

    private int[] getPlankStarts(int[] ends) {
		int[] starts = new int[ends.length];
		for (int i = 1; i < ends.length; i++) {
		    if (ends[i] > 0) {
		        starts[ends[i]] = i;
		    }
		}
        return starts;
    }

    private int[] getNailPositions(int[] C, int[] starts) {
        int[] nails = new int[starts.length];
        for (int i = 0; i < C.length; i++) {
            if (nails[C[i]] == 0) {
                nails[C[i]] = i + 1;
            }
        }
        return nails;
    }

    private static class Queue {
        private int[] array;
        private int start = 0;
        private int end = -1;
        private int size = 0;

        Queue(int capacity) {
            this.array = new int[capacity];
        }
        void addLast(int element) {
            size++;
            end++;
            if (end == array.length) {
                end = 0;
            }
            array[end] = element;
        }
        int removeLast() {
            size--;
            int result = array[end--];
            if (end < 0) {
                end = array.length - 1;
            }
            return result;
        }
        int removeFirst() {
            size--;
            int result = array[start++];
            if (start == array.length) {
                start = 0;
            }
            return result;
        }
        int removeNextToLast() {
            size--;
            int index = nextToLastIndex();
            int result = array[index];
            array[index] = array[end];
            end = index;
            return result;
        }
        int last() {
            return array[end];
        }
        int first() {
            return array[start];
        }
        int nextToLast() {
            return array[nextToLastIndex()];
        }
        boolean isEmpty() {
            return size == 0;
        }
        int size() {
            return size;
        }
        int nextToLastIndex() {
            return end > 0 ? end - 1 : array.length - 1;
        }
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O((N + M) * log(M))

▶

example
example test

✔

▶

extreme_single
single nail and single plank

✔

▶

extreme_point
nail is a point [1, 1]

✔

▶

few_nails_in_the_same_place
few nails are in the same place

✔

▶

random_small
random sequence, length = ~100

✔

▶

random_medium
random sequence, length = ~10,000

✔

▶

random_large
random sequence, length = ~30,000

✔

▶

extreme_large_planks
all large planks, length = ~30,000

✔

▶

large_point
all planks are points, length = ~30,000

✔