Test results - Codility

Tasks Details

medium

1. NumberOfDiscIntersections

Compute the number of intersections in a sequence of discs.

100%

We draw N discs on a plane. The discs are numbered from 0 to N − 1. An array A of N non-negative integers, specifying the radiuses of the discs, is given. The J-th disc is drawn with its center at (J, 0) and radius A[J].

We say that the J-th disc and K-th disc intersect if J ≠ K and the J-th and K-th discs have at least one common point (assuming that the discs contain their borders).

The figure below shows discs drawn for N = 6 and A as follows:

A[0] = 1 A[1] = 5 A[2] = 2 A[3] = 1 A[4] = 4 A[5] = 0

There are eleven (unordered) pairs of discs that intersect, namely:

discs 1 and 4 intersect, and both intersect with all the other discs;

disc 2 also intersects with discs 0 and 3.

Write a function:

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

that, given an array A describing N discs as explained above, returns the number of (unordered) pairs of intersecting discs. The function should return −1 if the number of intersecting pairs exceeds 10,000,000.

Given array A shown above, the function should return 11, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [0..100,000];

each element of array A is an integer within the range [0..2,147,483,647].

Solution

Programming language used Java 8

Time spent on task 4 minutes

Notes

not defined yet

Task timeline

Code: 15:34:47 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

// 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 int solution(int[] A) {
        // write your code in Java SE 8
        int l = A.length;  
        int[] arrayIn = new int[l];  
        int[] arrayOut = new int[l];  
        int inNumContext = 0;  
        int result = 0;  
          
        for(int i = 0; i < l; i ++) {  
            int in = (i - A[i]) < 0 ? 0 : (i - A[i]);  
            // take care the (A[i] + i) exceeds the value of MAX int   
            // which will become minus int  
            int out = (A[i] + i > l - 1 || A[i] + i < 0) ? (l - 1) : (A[i] + i);  
            arrayIn[in] ++;  
            arrayOut[out] ++;  
        }  
          
          
        for(int i = 0; i < l; i ++) {  
            if(arrayIn[i] != 0) {  
                // previous circles times new coming circles  
                result += inNumContext * arrayIn[i];  
                // new coming circles group with each other  
                result += arrayIn[i] * (arrayIn[i] - 1) / 2;  
                  
                if (result > 10000000) {  
                    return -1;  
                }  
                  
                // add coming circles to inNumContext  
                inNumContext += arrayIn[i];  
            }  
            // minus leaving circles from inNumContext  
            inNumContext -= arrayOut[i];  
        }   
          
        return result;
    }
}

Analysis

▶

example1
example test

✔

Code: 15:37:54 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

// 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 int solution(int[] A) {
        // write your code in Java SE 8
        int l = A.length;  
        int[] arrayIn = new int[l];  
        int[] arrayOut = new int[l];  
        int inNumContext = 0;  
        int result = 0;  
          
        for(int i = 0; i < l; i ++) {  
            int in = (i - A[i]) < 0 ? 0 : (i - A[i]);  
            // take care the (A[i] + i) exceeds the value of MAX int   
            // which will become minus int  
            int out = (A[i] + i > l - 1 || A[i] + i < 0) ? (l - 1) : (A[i] + i);  
            arrayIn[in] ++;  
            arrayOut[out] ++;  
        }  
          
          
        for(int i = 0; i < l; i ++) {  
            if(arrayIn[i] != 0) {  
                // previous circles times new coming circles  
                result += inNumContext * arrayIn[i];  
                // new coming circles group with each other  
                result += arrayIn[i] * (arrayIn[i] - 1) / 2;  
                  
                if (result > 10000000) {  
                    return -1;  
                }  
                  
                // add coming circles to inNumContext  
                inNumContext += arrayIn[i];  
            }  
            // minus leaving circles from inNumContext  
            inNumContext -= arrayOut[i];  
        }   
          
        return result;
    }
}

Analysis

▶

example1
example test

✔

Code: 15:38:12 UTC, java, final, score: 100

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

// 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 int solution(int[] A) {
        // write your code in Java SE 8
        int l = A.length;  
        int[] arrayIn = new int[l];  
        int[] arrayOut = new int[l];  
        int inNumContext = 0;  
        int result = 0;  
          
        for(int i = 0; i < l; i ++) {  
            int in = (i - A[i]) < 0 ? 0 : (i - A[i]);  
            // take care the (A[i] + i) exceeds the value of MAX int   
            // which will become minus int  
            int out = (A[i] + i > l - 1 || A[i] + i < 0) ? (l - 1) : (A[i] + i);  
            arrayIn[in] ++;  
            arrayOut[out] ++;  
        }  
          
          
        for(int i = 0; i < l; i ++) {  
            if(arrayIn[i] != 0) {  
                // previous circles times new coming circles  
                result += inNumContext * arrayIn[i];  
                // new coming circles group with each other  
                result += arrayIn[i] * (arrayIn[i] - 1) / 2;  
                  
                if (result > 10000000) {  
                    return -1;  
                }  
                  
                // add coming circles to inNumContext  
                inNumContext += arrayIn[i];  
            }  
            // minus leaving circles from inNumContext  
            inNumContext -= arrayOut[i];  
        }   
          
        return result;
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N*log(N)) or O(N)

▶

example1
example test

✔

▶

simple1

✔

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simple2

✔

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simple3

✔

▶

extreme_small
empty and [10]

✔

▶

small1

✔

▶

small2

✔

▶

small3

✔

▶

overflow
arithmetic overflow tests

✔

▶

medium1

✔

▶

medium2

✔

▶

medium3

✔

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medium4

✔

▶

10M_intersections
10.000.000 intersections

✔

▶

big1

✔

▶

big2

✔

▶

big3
[0]*50000

✔