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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Flatland Space Stations
  5. Discussions

Flatland Space Stations

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948 Discussions

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  • krushnaingale201
     + 0 comments

    JAVA

    static int flatlandSpaceStations(int n, int[] c) {

        ArrayList<Integer> arrayList = new ArrayList<>();
    for(int i=0;i<n;i++){
        int k=0;
        int res = Integer.MAX_VALUE;
        while(k<c.length){
            int res1 = c[k]-i;
            if(res1<0){
                res1 = 0-res1;
            }
            if(res>res1){
                res = res1;
            }
            k++;
        }
        arrayList.add(res);
    }
    int max = 0;
    for(int i =0;i<arrayList.size();i++){
        if(arrayList.get(i)>max){
            max = arrayList.get(i);
        }
    }
    return max;
}

  • luka_robajac
     + 0 comments
    function flatlandSpaceStations($n, $m, $c) {
    sort($c);

    $max = $c[0];
    for($i = 0; $i < $m - 1; $i++) {
        $dist = floor(($c[$i + 1] - $c[$i]) / 2);
        if($dist > $max) {
            $max = $dist;
        }
    }
    
    $max = max($max, $n - 1 - $c[$m - 1]);
    
    return $max;
}

  • afzal_saiyed931
     + 0 comments
    def flatlandSpaceStations(n, m, c):
    if n == m:
        return 0
    c = sorted(c)
    start_distance = c[0]
    end_distance = (n-1) - c[-1]
    mid_distance = None
    for i in range(len(c)-1):
        gap = c[i+1] - c[i]
        distance = gap // 2
        if mid_distance is None or distance > mid_distance:
            mid_distance = distance
            
    return max(start_distance, mid_distance or 0, end_distance)

  • gabriel_atenci4
     + 0 comments

    js my solution... if (n == c.length) return 0;

    const lengthC = c.length - 1;

    c.sort((a, b) => a - b); console.log(c);

    let max = c[0]; for (let i = 0; i < lengthC; i++) { const tmp = Math.floor(c[i + 1] - (c[i + 1] + c[i]) / 2);

    if (tmp > max) max = tmp; }

    let lastOne = n - c[lengthC] - 1;

    if (lastOne > max) max = lastOne;

    console.log(max);

  • lsloan
     + 0 comments

    Python: My short solution…

    def flatlandSpaceStations(n, c):
    return max(
        (s := sorted(c))[0],  # dist.: 0 to 1st spaceport
        n - s[-1] - 1,  # dist.: last spaceport to n
        *map(lambda i: (s[i + 1] - s[i]) // 2,   # dist.: halfway between-
            range(len(s) - 1)))  # -each spaceport

    I updated my code to make it more readable. Functional solutions aren't always difficult to read.