In this Goodland Electricity HackerRank solution, Goodland is a country with a number of evenly spaced cities along a line. The distance between adjacent cities is unit. There is an energy infrastructure project planning meeting, and the government needs to know the fewest number of power plants needed to provide electricity to the entire list of cities. Determine that number. If it cannot be done, return -1.
You are given a list of city data. Cities that may contain a power plant have been labeled . Others not suitable for building a plant are labeled . Given a distribution range of , find the lowest number of plants that must be built such that all cities are served. The distribution range limits supply to cities where distance is less than k.
Example
Each city is unit distance from its neighbors, and we’ll use based indexing. We see there are cities suitable for power plants, cities and . If we build a power plant at , it can serve through because those endpoints are at a distance of and . To serve , we would need to be able to build a plant in city or . Since none of those is suitable, we must return -1. It cannot be done using the current distribution constraint.
Function Description
Complete the pylons function in the editor below.
pylons has the following parameter(s):
- int k: the distribution range
- int arr[n]: each city’s suitability as a building site
Returns
- int: the minimum number of plants required or -1
Input Format
The first line contains two space-separated integers and , the number of cities in Goodland and the plants’ range constant.
The second line contains space-separated binary integers where each integer indicates suitability for building a plant.
Constraints
- Each .
Subtask
- for of the maximum score.
Output Format
Print a single integer denoting the minimum number of plants that must be built so that all of Goodland’s cities have electricity. If this is not possible for the given value of , print .
Sample Input
STDIN Function
----- --------
6 2 arr[] size n = 6, k = 2
0 1 1 1 1 0 arr = [0, 1, 1, 1, 1, 0]
Sample Output
2
Explanation
Cities , , , and are suitable for power plants. Each plant will have a range of . If we build in cities cities, and , then all cities will have electricity.
Goodland Electricity HackerRank solution
I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.
Problem Solution in Python
def pylons(k, arr):
# Write your code here
index = 0
count = 0
while index < len(arr):
plant_not_found = True
for distance in range(k-1, -k, -1):
if (index + distance < len(arr)) and (index + distance >= 0) and (arr[index+distance] == 1):
count += 1
index += k + distance
plant_not_found = False
break
if plant_not_found:
return -1
return countProblem Solution in JavaScript
function pylons(k, arr) {
// Write your code here
let i=0,count=0;
while(i<arr.length){
let j=i+k-1;
while(j>i-k){
if(arr[j]==1){
count++;
i=j+k;
break;
}
j--;
if(j==i-k)
j--;
}
if(j<i-k)
return -1;
}
return count;
}Problem Solution in Java
public static int pylons(int k, List<Integer> arr) {
int plants = 0;
int index = k - 1;
int start = 0;
while(index < arr.size()) {
if(arr.get(index) == 1) {
start = index + 1;
plants++;
if((index + ((k - 1) * 2) + 1 > arr.size() - 1) && index + k < arr.size()){
if(arr.subList(arr.size() - k, arr.size()).contains(1)){
plants++;
break;
}else{
return -1;
}
}else{
index += ((k - 1) * 2) + 1;
}
}else if(index == start){
return -1;
}else{
index--;
}
}
return plants;
}