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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Between Two Sets
  5. Discussions

Between Two Sets

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3032 Discussions

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  • abdlrhmn_l
     + 0 comments

    my simple javascript code, dont be overengineering

    function getTotalX(a, b) {
    // Write your code here
    let count = 0
    
    for (let i=a[a.length - 1];i<=b[0];i++) {
        const isFactorB = b.every(v => v % i === 0)
        const isAFactor = a.every(v => i % v === 0)
        
        if (isFactorB && isAFactor) {
            count++
        }
    }
    
    return count
}

  • vadlamani_bits
     + 0 comments
    def getTotalX(a, b):
    count=0
    bet_1=max(arr)
    bet_2=min(brr)
    
    for factor in range(bet_1, bet_2+1):
        if all(factor%i==0 for i in arr):
            if all(j%factor==0 for j in brr):
                count+=1
    return count

  • isacgabriel1095
     + 1 comment

    Javascript Code:

    function getTotalX(a, b) {
    // Write your code here
    let maxValue = Math.max(...a);
    let minValue = Math.min(...b);

    let i = 2;
    let elements = [];
    let element = maxValue;
    while(element<=minValue){ 
        let factors = a.filter(num=>element%num == 0); 
        if(factors.length == a.length){ 
            let isFactor = b.filter(num1=>num1%element == 0);
            if(isFactor.length == b.length){
                elements.push(element);
            }
        }
        element=maxValue*i; 
        i++; 
        
    }
    return elements.length;
}

  • surekha2019p
     + 0 comments

    JAVA8 CODE public static int gcd(int x, int y){ if(y==0) return x; return gcd(y, x%y); } public static int lcm(int x, int y){ return x*y/gcd(x,y); }

    public static int getTotalX(List<Integer> a, List<Integer> b) {
// Write your code here        
  int count =0;
  int Lcm = a.get(0);
  for(int i =0;i<a.size();i++){
    Lcm= lcm(Lcm,a.get(i));
  }
  int Gcd=b.get(0);
  for(int i =0;i<b.size();i++){
    Gcd=gcd(Gcd,b.get(i));
  }
  for(int i =Lcm;i<=Gcd;i+=Lcm){
    if(Gcd%i==0)
     count++;
  }
  return count;

}

    }

    }

    }

  • abhijithlalunni
     + 0 comments

    For Javascript, this is my solution. I don't think this is optimal solution. Coz it is having O(n^2).

    function getTotalX(a, b) {
    // Write your code here
    a.sort((a,b)=>(a-b));
    b.sort((a,b)=>(a-b));
    let count = 0;
    let startVal = a[a.length - 1];
    let endVal = b[0];
    while (startVal <= endVal) {
        let firstArr = 0;
        let secArr = 0;
        for (let num of a) {
            if (startVal % num === 0) {
                firstArr ++;
            }
        }
        for (let num of b) {
            if (num % startVal === 0) {
                secArr ++;
            }
        }
        if (firstArr === a.length && secArr === b.length) {
            count ++;
        }
        startVal ++;
    }
    return count;
}