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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Divisible Sum Pairs
  5. Discussions

Divisible Sum Pairs

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  • ayushkumarroy
     + 0 comments
    def divisibleSumPairs(n, k, ar):
    count = 0
    for i in range(n):
        for j in range(i+1,n):
            if (ar[i]+ar[j]) % k == 0:
                count+=1
    return count

  • sharp_thinker
     + 0 comments

    C++ Code here

    include

    using namespace std;

    int main (){

    int n, k, count = 0;
cin >> n >> k;
int array[n];
for (int i = 0; i < n; i++){
    cin >> array[i];
}
for (int i = 0; i < n; i++){
    for (int j = i+1; j < n; j++){
        if ((array[i] + array[j]) % k == 0){
            count++;
        }
    }
}

cout << count;

return 0;

    }

  • idttalbe_m
     + 0 comments
    Python 3 solution 
    def divisibleSumPairs(n, k, ar):
    # Write your code here
    couple = [sum([ar[i],ar[j]])%k == 0 for i in range(n) for j in range(i+1,n) ]
    return(sum(couple))

  • anurag_jain5
     + 0 comments

    PHP :

    function divisibleSumPairs($n, $k, $ar) {
    // Write your code here
    $count = 0;
    for($i = 0; $i < $n; $i++){
        for($j = $i+1; $j < $n; $j++){
            if(($ar[$i]+$ar[$j]) % $k == 0)
                $count++;
        }
    }
    return $count;
}

  • MrCyberlord
     + 0 comments

    Easy JavaScript Solution

    let counter = 0

for(var i=0;i<n-1;i++)
{
    for(var j=i+1;j<n;j++)
    {
        if((ar[i]+ar[j])%k==0)
        {
        counter++
        }
    }
}

return counter

    Please upvote if you like the solution

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