Hackerland Radio Transmitters HackerRank Solution

In this Hackerland Radio Transmitters HackerRank solution, Hackerland is a one-dimensional city with houses aligned at integral locations along a road. The Mayor wants to install radio transmitters on the roofs of the city’s houses. Each transmitter has a fixed range meaning it can transmit a signal to all houses within that number of units distance away.

Given a map of Hackerland and the transmission range, determine the minimum number of transmitters so that every house is within range of at least one transmitter. Each transmitter must be installed on top of an existing house.

Example

 antennae at houses  and  and  provide complete coverage. There is no house at location  to cover both  and . Ranges of coverage, are , , and .

Hackerland Radio Transmitters HackerRank solution

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Problem Solution in Python

def hackerlandRadioTransmitters(x, k):
    x.sort()
    count = 0
    while x:
        leftHouse = x[0]
        while x and leftHouse+k >= x[0]:
            transmitter=x.pop(0)
        while x and transmitter+k >= x[0]:
            x.pop(0)
        count += 1    
    return count

function hackerlandRadioTransmitters(x, k) {
    // Sort Array
    x.sort((a,b)=>a-b)
   
    let numTransmitters = 0
    let coveredPosition = 0
   
    for(let i=0;i<x.length;i++){
        // If current house is already covered -> skip
        if(x[i]<=coveredPosition) continue
       
        // Find for current i the next house which is outside of reach
        let j=i+1
        while(((x[i]+k)>=x[j])){
            j++
        }
        // Current x[j] is not in reach of x[i] anymore
        // -> Build transmitter on previous house
        numTransmitters++
        coveredPosition = x[j-1]+k
    }
    return numTransmitters
}

int hackerlandRadioTransmitters(vector<int> x, int k) {
    sort(x.begin(), x.end());
    int count=0;
    int i=0;
    while(i<x.size())
    {
        int j=i+1;
        while(x[i]+k>=x[j] &&j<x.size())
            j++;
        int m=j;
        j--;
        while(x[j]+k>=x[m] && m<x.size())
            m++;
        i=m;
        count++;
    }
    return count;
}

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