Tasks Details
easy
1.
PassingCars
Count the number of passing cars on the road.
Task Score
100%
Correctness
100%
Performance
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] = 1We 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] = 1the 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.
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used JavaScript
Time spent on task 2 minutes
Notes
not defined yet
Task timeline
Code: 07:52:30 UTC,
java,
autosave
Code: 07:52:36 UTC,
js,
autosave
function solution(A) {
// Implement your solution here
var current_sum = 0
var value_1_index = [], value_0_index = []
//紀錄兩個0中間經過的1
var value_0_interval = []
for(var i=0;i<A.length;i++){
if(A[i] == 0) {
value_0_index.push(i)
value_0_interval[value_0_index.length-1] = 0
}else {
value_1_index.push(i)
value_0_interval[value_0_index.length-1]++
}
}
var current_value_1 = value_1_index.length
for(var i=0;i<value_0_index.length;i++){
current_sum = current_sum + current_value_1
current_value_1 = current_value_1 - value_0_interval[i]
}
return (current_sum > 1000000000) ? -1 : current_sum
}
Code: 07:52:48 UTC,
js,
autosave
function solution(A) {
// Implement your solution here
var current_sum = 0
var value_1_index = [], value_0_index = []
//紀錄兩個0中間經過的1
var value_0_interval = []
for(var i=;i<A.length;i++){
if(A[i] == 0) {
value_0_index.push(i)
value_0_interval[value_0_index.length-1] = 0
}else {
value_1_index.push(i)
value_0_interval[value_0_index.length-1]++
}
}
var current_value_1 = value_1_index.length
for(var i=0;i<value_0_index.length;i++){
current_sum = current_sum + current_value_1
current_value_1 = current_value_1 - value_0_interval[i]
}
return (current_sum > 1000000000) ? -1 : current_sum
}
Code: 07:53:03 UTC,
js,
autosave
function solution(A) {
// Implement your solution here
var current_sum = 0
var value_1_index = [], value_0_index = []
//紀錄兩個0中間經過的1
var value_0_interval = []
for(var i=A.indexOf();i<A.length;i++){
if(A[i] == 0) {
value_0_index.push(i)
value_0_interval[value_0_index.length-1] = 0
}else {
value_1_index.push(i)
value_0_interval[value_0_index.length-1]++
}
}
var current_value_1 = value_1_index.length
for(var i=0;i<value_0_index.length;i++){
current_sum = current_sum + current_value_1
current_value_1 = current_value_1 - value_0_interval[i]
}
return (current_sum > 1000000000) ? -1 : current_sum
}
Code: 07:53:14 UTC,
js,
autosave
function solution(A) {
// Implement your solution here
var current_sum = 0
var value_1_index = [], value_0_index = []
//紀錄兩個0中間經過的1
var value_0_interval = []
for(var i=A.indexOf(1);i<A.length;i++){
if(A[i] == 0) {
value_0_index.push(i)
value_0_interval[value_0_index.length-1] = 0
}else {
value_1_index.push(i)
value_0_interval[value_0_index.length-1]++
}
}
var current_value_1 = value_1_index.length
for(var i=0;i<value_0_index.length;i++){
current_sum = current_sum + current_value_1
current_value_1 = current_value_1 - value_0_interval[i]
}
return (current_sum > 1000000000) ? -1 : current_sum
}
Code: 07:53:23 UTC,
js,
verify,
result: Passed
function solution(A) {
// Implement your solution here
var current_sum = 0
var value_1_index = [], value_0_index = []
//紀錄兩個0中間經過的1
var value_0_interval = []
for(var i=A.indexOf(0);i<A.length;i++){
if(A[i] == 0) {
value_0_index.push(i)
value_0_interval[value_0_index.length-1] = 0
}else {
value_1_index.push(i)
value_0_interval[value_0_index.length-1]++
}
}
var current_value_1 = value_1_index.length
for(var i=0;i<value_0_index.length;i++){
current_sum = current_sum + current_value_1
current_value_1 = current_value_1 - value_0_interval[i]
}
return (current_sum > 1000000000) ? -1 : current_sum
}
Analysis
Code: 07:53:29 UTC,
js,
verify,
result: Passed
function solution(A) {
// Implement your solution here
var current_sum = 0
var value_1_index = [], value_0_index = []
//紀錄兩個0中間經過的1
var value_0_interval = []
for(var i=A.indexOf(0);i<A.length;i++){
if(A[i] == 0) {
value_0_index.push(i)
value_0_interval[value_0_index.length-1] = 0
}else {
value_1_index.push(i)
value_0_interval[value_0_index.length-1]++
}
}
var current_value_1 = value_1_index.length
for(var i=0;i<value_0_index.length;i++){
current_sum = current_sum + current_value_1
current_value_1 = current_value_1 - value_0_interval[i]
}
return (current_sum > 1000000000) ? -1 : current_sum
}
Analysis
Code: 07:53:32 UTC,
js,
final,
score:
100
function solution(A) {
// Implement your solution here
var current_sum = 0
var value_1_index = [], value_0_index = []
//紀錄兩個0中間經過的1
var value_0_interval = []
for(var i=A.indexOf(0);i<A.length;i++){
if(A[i] == 0) {
value_0_index.push(i)
value_0_interval[value_0_index.length-1] = 0
}else {
value_1_index.push(i)
value_0_interval[value_0_index.length-1]++
}
}
var current_value_1 = value_1_index.length
for(var i=0;i<value_0_index.length;i++){
current_sum = current_sum + current_value_1
current_value_1 = current_value_1 - value_0_interval[i]
}
return (current_sum > 1000000000) ? -1 : current_sum
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N)
expand all
Correctness tests
1.
0.056 s
OK
2.
0.056 s
OK
1.
0.056 s
OK
2.
0.052 s
OK
3.
0.056 s
OK
4.
0.056 s
OK
1.
0.056 s
OK
1.
0.056 s
OK
1.
0.056 s
OK
expand all
Performance tests
1.
0.068 s
OK
1.
0.084 s
OK
1.
0.084 s
OK
2.
0.080 s
OK
1.
0.084 s
OK
2.
0.080 s
OK
1.
0.084 s
OK
2.
0.100 s
OK
3.
0.080 s
OK