Test results - Codility

Tasks Details

easy

1. PassingCars

Count the number of passing cars on the road.

100%

A non-empty array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road.

Array A contains only 0s and/or 1s:

0 represents a car traveling east,

1 represents a car traveling west.

The goal is to count passing cars. We say that a pair of cars (P, Q), where 0 ≤ P < Q < N, is passing when P is traveling to the east and Q is traveling to the west.

For example, consider array A such that:

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

We have five pairs of passing cars: (0, 1), (0, 3), (0, 4), (2, 3), (2, 4).

Write a function:

function solution(A);

that, given a non-empty array A of N integers, returns the number of pairs of passing cars.

The function should return −1 if the number of pairs of passing cars exceeds 1,000,000,000.

For example, given:

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

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 that can have one of the following values: 0, 1.

Solution

Programming language used JavaScript

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 16:36:59 UTC, js, verify, result: Passed

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// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(A) {
    // write your code in JavaScript (Node.js 8.9.4)

    // Idea here is to track how many cars going from Left-to-Right (East/0).
    // Then when we see a Right-to-Left car (West/1), we know how many Left-to-Right car it will pass later.
    let leftToRightCars = 0;
    let pathCrossCount = 0;
    for(let i = 0; i < A.length; ++i) {
        if(A[i] === 0) {
            ++leftToRightCars;
        } else { // This is a Left-to-Right car and will pass leftToRightCars amount of cars.
            pathCrossCount += leftToRightCars;
            // The function should return -1 if the number of pairs of passing cars exceeds 1,000,000,000.
            if (pathCrossCount > 1000000000) {
                return -1;
            }
        }
    }
    return pathCrossCount;
}

Analysis

▶

example
example test

✔

Code: 16:37:02 UTC, js, final, score: 100

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// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(A) {
    // write your code in JavaScript (Node.js 8.9.4)

    // Idea here is to track how many cars going from Left-to-Right (East/0).
    // Then when we see a Right-to-Left car (West/1), we know how many Left-to-Right car it will pass later.
    let leftToRightCars = 0;
    let pathCrossCount = 0;
    for(let i = 0; i < A.length; ++i) {
        if(A[i] === 0) {
            ++leftToRightCars;
        } else { // This is a Left-to-Right car and will pass leftToRightCars amount of cars.
            pathCrossCount += leftToRightCars;
            // The function should return -1 if the number of pairs of passing cars exceeds 1,000,000,000.
            if (pathCrossCount > 1000000000) {
                return -1;
            }
        }
    }
    return pathCrossCount;
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

▶

example
example test

✔

▶

single
single element

✔

▶

double
two elements

✔

▶

simple
simple test

✔

▶

small_random
random, length = 100

✔

▶

small_random2
random, length = 1000

✔

▶

medium_random
random, length = ~10,000

✔

▶

large_random
random, length = ~100,000

✔

▶

large_big_answer
0..01..1, length = ~100,000

✔

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large_alternate
0101..01, length = ~100,000

✔

▶

large_extreme
large test with all 1s/0s, length = ~100,000

✔