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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Apple and Orange
  5. Discussions

Apple and Orange

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  • kelleher2
     + 0 comments

    function countApplesAndOranges(s, t, a, b, apples, oranges) { // Write your code here

    let appleCount = apples.filter(apple => {
    return (apple + a >= s) && (apple + a <= t)
})

let orangeCount = oranges.filter(orange => {
    return (orange + b >= s) && (orange + b <= t)
})    
console.log(appleCount.length)
console.log( orangeCount.length)

    }

  • valentinrodrigu2
     + 0 comments

    My solution, guys.

    void countApplesAndOranges(int s, int t, int a, int b, vector<int> apples, vector<int> oranges) {
  int napple = 0; 
  int norange = 0; 
  
  for(unsigned i = 0; i<apples.size(); i++){
    if((apples[i]+a) >= s && (apples[i]+a) <= t)
      napple += 1; 
  }
  for(unsigned i = 0; i<oranges.size(); i++){
    if((oranges[i]+b) >= s && (oranges[i]+b) <= t)
      norange += 1; 
  }
  
  cout<<napple<<endl;
  cout<<norange;
}

  • ddebajyati
     + 0 comments

    Look at my creative C++ solution :-)

    void countApplesAndOranges(int s, int t, int a, int b, vector<int> apples, vector<int> oranges) {
    std::pair<int, int> fruits_landed = {0,0};
    
    std::transform(apples.begin(), apples.end(), apples.begin(), [a](int pos)->int{
        return (pos + a);
    });
    std::transform(oranges.begin(), oranges.end(), oranges.begin(), [b](int pos)->int{
        return (pos + b);
    });
    
    fruits_landed.first = std::count_if(apples.begin(), apples.end(), [s,t](int pos)->bool{
        return (pos >= s && pos <= t);
    });
    fruits_landed.second = std::count_if(oranges.begin(), oranges.end(), [s,t](int pos)->bool{
        return (pos <= t && pos >= s);
    });
    
    std::cout << fruits_landed.first << endl << fruits_landed.second << endl;
}

    Here is an even better version - 

    void countApplesAndOranges(int s, int t, int a, int b, vector<int> apples, vector<int> oranges) {
    std::pair<int, int> fruits_landed = {0,0};
    std::function<bool(int)> condition = [s,t](int pos)->bool{ return (pos >= s && pos <= t);};
    
    std::transform(apples.begin(), apples.end(), apples.begin(), [a](int pos)->int{
        return (pos + a);
    });
    std::transform(oranges.begin(), oranges.end(), oranges.begin(), [b](int pos)->int{
        return (pos + b);
    });
    
    fruits_landed.first = std::count_if(apples.begin(), apples.end(), [condition](int pos)->bool{
        return condition(pos);
    });
    fruits_landed.second = std::count_if(oranges.begin(), oranges.end(), [condition](int pos)->bool{
        return condition(pos);
    });
    
    std::cout << fruits_landed.first << endl << fruits_landed.second << endl;
}

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can see the explanation here : https://youtu.be/YJVWinC21Fg

    void countApplesAndOranges(int s, int t, int a, int b, vector<int> apples, vector<int> oranges) {
    int orange = 0, apple = 0;
    for(int app: apples) if(app + a >= s && app + a <=t) apple++;
    for(int orr: oranges) if(orr + b >= s && orr + b <=t) orange++;
    cout << apple << endl << orange;
}

  • virajpatil009
     + 0 comments

    simple with es6 functions 

        console.log(apples.map(n=>n+a).filter(n=>n>=s&&n<=t).length)
    console.log(oranges.map(n=>n+b).filter(n=>n>=s&&n<=t).length)