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:
class Solution { public int solution(int[] 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.
// 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) {
System.out.println("A: " + Arrays.toString(A));
int[] prefixSums = new int[A.length];
prefixSums[0] = A[0];
int carsTravelingWest = A[0];
for (int i = 1; i < A.length; i++) {
prefixSums[i] = prefixSums[i - 1] + A[i];
if (A[i] == 1) {
carsTravelingWest += 1;
}
}
int numberOfPairsOfPassingCars = 0;
for (int i = 0; i < A.length; i++) {
if (A[i] == 0) {
numberOfPairsOfPassingCars += carsTravelingWest - prefixSums[i];
if (numberOfPairsOfPassingCars > 1000000000) {
numberOfPairsOfPassingCars = -1;
break;
}
}
}
System.out.println("Number of pairs of passing cars: " + numberOfPairsOfPassingCars);
return numberOfPairsOfPassingCars;
}
}
// 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) {
int[] prefixSums = new int[A.length];
prefixSums[0] = A[0];
int carsTravelingWest = A[0];
for (int i = 1; i < A.length; i++) {
prefixSums[i] = prefixSums[i - 1] + A[i];
if (A[i] == 1) {
carsTravelingWest += 1;
}
}
int numberOfPairsOfPassingCars = 0;
for (int i = 0; i < A.length; i++) {
if (A[i] == 0) {
numberOfPairsOfPassingCars += carsTravelingWest - prefixSums[i];
if (numberOfPairsOfPassingCars > 1000000000) {
numberOfPairsOfPassingCars = -1;
break;
}
}
}
System.out.println("Number of pairs of passing cars: " + numberOfPairsOfPassingCars);
return numberOfPairsOfPassingCars;
}
}
// 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) {
System.out.println("A: " + Arrays.toString(A));
int[] travellingToWestPassingCars = new int[A.length];
travellingToWestPassingCars[0] = A[0];
int totalCarsTravelingToWest = A[0];
for (int i = 1; i < A.length; i++) {
travellingToWestPassingCars[i] = travellingToWestPassingCars[i - 1] + A[i];
if (A[i] == 1) {
totalCarsTravelingToWest += 1;
}
}
int numberOfPairsOfPassingCars = 0;
for (int i = 0; i < A.length; i++) {
if (A[i] == 0) {
numberOfPairsOfPassingCars += totalCarsTravelingToWest - travellingToWestPassingCars[i];
if (numberOfPairsOfPassingCars > 1000000000) {
numberOfPairsOfPassingCars = -1;
break;
}
}
}
System.out.println("Number of pairs of passing cars: " + numberOfPairsOfPassingCars);
return numberOfPairsOfPassingCars;
}
}
// 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) {
int[] travellingToWestPassingCars = new int[A.length];
travellingToWestPassingCars[0] = A[0];
int totalCarsTravelingToWest = A[0];
for (int i = 1; i < A.length; i++) {
travellingToWestPassingCars[i] = travellingToWestPassingCars[i - 1] + A[i];
if (A[i] == 1) {
totalCarsTravelingToWest += 1;
}
}
int numberOfPairsOfPassingCars = 0;
for (int i = 0; i < A.length; i++) {
if (A[i] == 0) {
numberOfPairsOfPassingCars += totalCarsTravelingToWest - travellingToWestPassingCars[i];
if (numberOfPairsOfPassingCars > 1000000000) {
numberOfPairsOfPassingCars = -1;
break;
}
}
}
return numberOfPairsOfPassingCars;
}
}
// 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) {
int[] travellingToWestPassingCars = new int[A.length];
travellingToWestPassingCars[0] = A[0];
int totalCarsTravelingToWest = A[0];
for (int i = 1; i < A.length; i++) {
travellingToWestPassingCars[i] = travellingToWestPassingCars[i - 1] + A[i];
if (A[i] == 1) {
totalCarsTravelingToWest += 1;
}
}
int numberOfPairsOfPassingCars = 0;
for (int i = 0; i < A.length; i++) {
if (A[i] == 0) {
numberOfPairsOfPassingCars += totalCarsTravelingToWest - travellingToWestPassingCars[i];
if (numberOfPairsOfPassingCars > 1000000000) {
numberOfPairsOfPassingCars = -1;
break;
}
}
}
return numberOfPairsOfPassingCars;
}
}
// 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) {
int[] travellingToWestPassingCars = new int[A.length];
travellingToWestPassingCars[0] = A[0];
int totalCarsTravelingToWest = A[0];
for (int i = 1; i < A.length; i++) {
travellingToWestPassingCars[i] = travellingToWestPassingCars[i - 1] + A[i];
if (A[i] == 1) {
totalCarsTravelingToWest += 1;
}
}
int numberOfPairsOfPassingCars = 0;
for (int i = 0; i < A.length; i++) {
if (A[i] == 0) {
numberOfPairsOfPassingCars += totalCarsTravelingToWest - travellingToWestPassingCars[i];
if (numberOfPairsOfPassingCars > 1000000000) {
numberOfPairsOfPassingCars = -1;
break;
}
}
}
return numberOfPairsOfPassingCars;
}
}
The solution obtained perfect score.