Test results - Codility

Tasks Details

medium

1. Peaks

Divide an array into the maximum number of same-sized blocks, each of which should contain an index P such that A[P - 1] < A[P] > A[P + 1].

36%

66%

A non-empty array A consisting of N integers is given.

A peak is an array element which is larger than its neighbors. More precisely, it is an index P such that 0 < P < N − 1, A[P − 1] < A[P] and A[P] > A[P + 1].

For example, the following array A:

A[0] = 1 A[1] = 2 A[2] = 3 A[3] = 4 A[4] = 3 A[5] = 4 A[6] = 1 A[7] = 2 A[8] = 3 A[9] = 4 A[10] = 6 A[11] = 2

has exactly three peaks: 3, 5, 10.

We want to divide this array into blocks containing the same number of elements. More precisely, we want to choose a number K that will yield the following blocks:

A[0], A[1], ..., A[K − 1],

A[K], A[K + 1], ..., A[2K − 1],
...

A[N − K], A[N − K + 1], ..., A[N − 1].

What's more, every block should contain at least one peak. Notice that extreme elements of the blocks (for example A[K − 1] or A[K]) can also be peaks, but only if they have both neighbors (including one in an adjacent blocks).

The goal is to find the maximum number of blocks into which the array A can be divided.

Array A can be divided into blocks as follows:

one block (1, 2, 3, 4, 3, 4, 1, 2, 3, 4, 6, 2). This block contains three peaks.

two blocks (1, 2, 3, 4, 3, 4) and (1, 2, 3, 4, 6, 2). Every block has a peak.

three blocks (1, 2, 3, 4), (3, 4, 1, 2), (3, 4, 6, 2). Every block has a peak. Notice in particular that the first block (1, 2, 3, 4) has a peak at A[3], because A[2] < A[3] > A[4], even though A[4] is in the adjacent block.

However, array A cannot be divided into four blocks, (1, 2, 3), (4, 3, 4), (1, 2, 3) and (4, 6, 2), because the (1, 2, 3) blocks do not contain a peak. Notice in particular that the (4, 3, 4) block contains two peaks: A[3] and A[5].

The maximum number of blocks that array A can be divided into is three.

Write a function:

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

that, given a non-empty array A consisting of N integers, returns the maximum number of blocks into which A can be divided.

If A cannot be divided into some number of blocks, the function should return 0.

For example, given:

A[0] = 1 A[1] = 2 A[2] = 3 A[3] = 4 A[4] = 3 A[5] = 4 A[6] = 1 A[7] = 2 A[8] = 3 A[9] = 4 A[10] = 6 A[11] = 2

the function should return 3, 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 [0..1,000,000,000].

Solution

Programming language used Java 8

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 10:32:16 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
     public static int solution(int[] A) {
            List<Integer> divList = getDivideSize(A.length);
            List<Integer> peakList = new ArrayList<>();

            for (int i = 0; i < A.length; i++) {
                if (i >= 1 && i < A.length - 1 && A[i] > A[i - 1] && A[i] > A[i + 1]) {
                    peakList.add(i);
                }
            }
            if(peakList.size() <= 1){
                return peakList.size();
            }

            // 1 2 3 4 6 12
            for (int div : divList) {
                int result = 1;
                for (int j = 0; j < peakList.size() - 1; j++) {
                    if (div * (j + 1) >= peakList.get(j) && div * (j + 1) < peakList.get(j + 1)) {
                        result ++;
                    } else {
                        break;
                    }
                    if(div * result == A.length){
                        return result;
                    }
                }
            }

            return 0;
        }

        public static List<Integer> getDivideSize(int N) {
            // write your code in Java SE 8
            List<Integer> divList = new ArrayList<>();
            for (int i = 1; i <= N; i++) {
                if (N % i == 0) {
                    divList.add(i);
                }
            }
            return divList;
}
}

Analysis

▶

example
example test

✔

Code: 10:32:21 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
     public static int solution(int[] A) {
            List<Integer> divList = getDivideSize(A.length);
            List<Integer> peakList = new ArrayList<>();

            for (int i = 0; i < A.length; i++) {
                if (i >= 1 && i < A.length - 1 && A[i] > A[i - 1] && A[i] > A[i + 1]) {
                    peakList.add(i);
                }
            }
            if(peakList.size() <= 1){
                return peakList.size();
            }

            // 1 2 3 4 6 12
            for (int div : divList) {
                int result = 1;
                for (int j = 0; j < peakList.size() - 1; j++) {
                    if (div * (j + 1) >= peakList.get(j) && div * (j + 1) < peakList.get(j + 1)) {
                        result ++;
                    } else {
                        break;
                    }
                    if(div * result == A.length){
                        return result;
                    }
                }
            }

            return 0;
        }

        public static List<Integer> getDivideSize(int N) {
            // write your code in Java SE 8
            List<Integer> divList = new ArrayList<>();
            for (int i = 1; i <= N; i++) {
                if (N % i == 0) {
                    divList.add(i);
                }
            }
            return divList;
}
}

Analysis

▶

example
example test

✔

Code: 10:32:24 UTC, java, final, score: 36

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
     public static int solution(int[] A) {
            List<Integer> divList = getDivideSize(A.length);
            List<Integer> peakList = new ArrayList<>();

            for (int i = 0; i < A.length; i++) {
                if (i >= 1 && i < A.length - 1 && A[i] > A[i - 1] && A[i] > A[i + 1]) {
                    peakList.add(i);
                }
            }
            if(peakList.size() <= 1){
                return peakList.size();
            }

            // 1 2 3 4 6 12
            for (int div : divList) {
                int result = 1;
                for (int j = 0; j < peakList.size() - 1; j++) {
                    if (div * (j + 1) >= peakList.get(j) && div * (j + 1) < peakList.get(j + 1)) {
                        result ++;
                    } else {
                        break;
                    }
                    if(div * result == A.length){
                        return result;
                    }
                }
            }

            return 0;
        }

        public static List<Integer> getDivideSize(int N) {
            // write your code in Java SE 8
            List<Integer> divList = new ArrayList<>();
            for (int i = 1; i <= N; i++) {
                if (N % i == 0) {
                    divList.add(i);
                }
            }
            return divList;
}
}

Analysis summary

The following issues have been detected: wrong answers.

Analysis

▶

example
example test

✔

▶

extreme_min
extreme min test

✔

▶

extreme_without_peaks
test without peaks

✔

▶

prime_length
test with prime sequence length

✘

WRONG ANSWER
got 0 expected 1

▶

anti_bin_search
anti bin_search test

✔

▶

simple1
simple test

✘

WRONG ANSWER
got 0 expected 4

▶

simple2
second simple test

✔

▶

medium_random
chaotic medium sequences, length = ~5,000

✘

WRONG ANSWER
got 0 expected 625

▶

medium_anti_slow
medium test anti slow solutions

✘

WRONG ANSWER
got 0 expected 1

▶

large_random
chaotic large sequences, length = ~50,000

✘

WRONG ANSWER
got 0 expected 5000

▶

large_anti_slow
large test anti slow solutions

✘

WRONG ANSWER
got 0 expected 30030

▶

extreme_max
extreme max test

✘

WRONG ANSWER
got 0 expected 25000