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

Divisible Sum Pairs

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2443 Discussions

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  • 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;
}

  • eraymt21
     + 0 comments

    Hi here is my Python solution. I can take any suggestions .

    def divisibleSumPairs(n, k, ar):
# Write your code here
count=0
for index ,i in enumerate(ar):
    for index2 ,item in enumerate(ar):
        if(index!=index2):
             if ((i+item)%k==0):
                count=count+1

    print(int(count/2))
return int(count/2)

  • 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;
}

  • fkaanfirat
     + 1 comment
    def divisibleSumPairs(n, k, ar):
    # Write your code here
    count = 0
    for i in range(0, n) :
        
        for j in range(1, n): 
            if i < j and (ar[i] + ar[j]) % k == 0: 
                    count+= 1
                
    return count

  • kroizer21
     + 0 comments

    C# Solution: Explanation: we sort the array, and after the array is in a non descending order, we apply the two pointer method, "i" points to the current number and "j" starts from the end and works its way backwards until the number "ar[i] + ar[j] < k" (after that there is no reason to continue, because none of the pairs will be divisible by k)

            ar.Sort();
        int answer = 0;
        int i = 0;
        int j = n-1;
        while (i < j) {
            if ((ar[i] + ar[j])%k == 0) {
                answer++;
            }
            if (ar[i] + ar[j] < k || j==i+1) {
                i++;
                j=n-1;
            } else {
                j--; 
            }
        }
        return answer;