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

Divisible Sum Pairs

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2446 Discussions

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  • sam_a_d
     + 0 comments

    def divisibleSumPairs(n, k, ar):

    possible_comb = list()

for comb in itertools.combinations(ar, 2):
    if (comb[0] + comb[1]) % k == 0:
        possible_comb.append(comb)

return len(possible_comb)

  • thiago_sant3031
     + 0 comments

    Using Python

    from itertools import combinations

def divisibleSumPairs(n, k, ar):
    pairs = list(combinations(ar, 2))
    return len([p for p in pairs if sum(p) % k == 0])

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can whatch the explanation here: https://youtu.be/aDv1iUSgnd8

    Solution 1 : O(n^2)

    int divisibleSumPairs(int n, int k, vector<int> ar) {
    int result = 0;
    for(int i = 0; i < ar.size() - 1; i++){
        for(int j = i + 1; j < ar.size(); j++){
            if( (ar[i] + ar[j]) % k == 0) result++;
        }
    }
    return result;
}

    Solution 2 : O(n)

    int divisibleSumPairs(int n, int k, vector<int> ar) {
    map<int, int> mp;
    for(int i = 0; i < ar.size(); i++){
        mp[ar[i] % k]++;
    }
    int result = 0;
     for(int i = 0; i <= k/2; i++){
        if(i == 0 || i*2 == k) result += (mp[i] * (mp[i] - 1)) / 2;
        else result += mp[i] * mp[k - i]; 
    }
    return result;
}

  • petrenkosv
     + 0 comments

    C# solution

    public static int divisibleSumPairs(int n, int k, List<int> ar)
    {
        int result = 0;
        
        for(int i = 0; i < ar.Count; i++)
            for(int j = (i + 1); j < ar.Count; j++)
                if((ar[i] + ar[j]) % k == 0) result++;
        
        return result;
    }

  • highsoft_soi
     + 0 comments

    Perl:

    sub divisibleSumPairs {
    my ($n, $k, $s) = @_;

    my $res = 0;
    for (my $i = 0; $i < $n; $i++) {
        for (my $j = 0; $j < $n; $j++) {
            $res++ if ($i < $j && (($s->[$i] + $s->[$j]) % $k) == 0 );
        }
    }
    return $res;
}