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

Divisible Sum Pairs

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2448 Discussions

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  • umairg404
     + 0 comments

    In C# 100%

    public static int divisibleSumPairs(int n, int k, List<int> ar)
    {
        int count = 0;
        int[] remainders = new int[k]; 
        
        for (int i = 0; i < n; i++)
        {
            int currentRemainder = ar[i] % k;
            if (currentRemainder < 0) currentRemainder += k;
            int complement = (k - currentRemainder) % k; 
            count += remainders[complement];
            remainders[currentRemainder]++;
        }
        
        return count;
    }

  • yuanhui5
     + 0 comments

    Python

    def divisibleSumPairs(n, k, ar):
    ar_mod = [i%k for i in ar]
    count_mod = {}
    for i in ar_mod:
        if i not in count_mod:
            count_mod[i] = 0
        count_mod[i] += 1
    
    count = 0
    
    for i in ar_mod:
        count_mod[i] -= 1
        
        if i != 0:
            sub = k-i
        else:
            sub = 0
        
        if sub in count_mod:
            count += count_mod[sub]
    
    return count

  • 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;
}

  • 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])