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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Service Lane
  5. Discussions

Service Lane

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1299 Discussions

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  • rubinaparveen151
     + 0 comments
    def serviceLane(n, cases):
    # Write your code here
    result = []
    for i in range(len(cases)):
        width_max = min(width[cases[i][0]:cases[i][1]+1])
        result.append(width_max)
    return result

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/iuLCUdG77yQ You will need to update the serviLane function by adding the width vector, since it was forgotten.

    vector<int> serviceLane(int n, vector<vector<int>> cases, vector<int> width) {
    vector<int> result;
    for(auto cas: cases){
        int r = width[cas[0]];
        for(int i = cas[0]; i <= cas[1]; i++){
            r = min(r, width[i]);
        }
        result.push_back(r);
    }
    return result;
}

  • space59
     + 0 comments

    The example at the beginning uses 1-based indexing for the width array, whereas the sample later on uses 0-based indexing.

  • vudu44
     + 0 comments

    Typescript minimal solution (I had to fix the function since it missed the width param):

    function serviceLane(cases: number[][], width: number[]): number[] {
    const results = [].constructor(cases.length);
    cases.forEach((c, index) => {
        results[index] = width[c[0]];
        for (let i = c[0]; i <= c[1]; i++) {
            if (results[index] > width[i]) results[index] = width[i];
        }
    });
    return results;
}

  • BoredProgrammer
     + 0 comments
    def serviceLane(n, cases):
    l=[]
    for i in cases:
        l.append(min(width[i[0]:i[1]+1]))
    return l