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

Divisible Sum Pairs

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2427 Discussions

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  • pitabashk196
     + 0 comments
    public static int divisibleSumPairs(int n, int k, List<Integer> ar) {
        
        Map<Integer,Integer> mp=new HashMap<>();
        int cnt=0;
        
        for(int i=0;i<ar.size();i++){
            int rem=ar.get(i)%k;
            int com=(k-rem==k?0:k-rem);
            if(mp.containsKey(com)){
                cnt+=mp.get(com);
                
            }
            mp.put(rem,mp.getOrDefault(rem,0)+1);
            
        }
        
        return cnt;

    }

}
****

  • ruisanchez93
     + 0 comments

    python

    def divisibleSumPairs(n, k, ar):
    count = 0
    for idx in range(len(ar)):
        for jdx in range(idx+1, len(ar)):
            if (ar[idx]+ar[jdx])%k==0:
                count+=1
    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;
}

  • pt023150
     + 0 comments

    java(8)

    public static int divisibleSumPairs(int n, int k, List<Integer> ar) {
    // Write your code here
    int ways = 0;
    for (int i = 0; i < ar.size(); i++) {

        for (int j = i + 1; j < ar.size(); j++) {
            if ((ar.get(i) + ar.get(j)) % k == 0) {
                ways++;
            }

        }

    }
    return ways;
}

  • nizarelghazal
     + 0 comments

    O(n+k)

    def divisibleSumPairs(n, k, ar):
    from collections import Counter
    modulo_count = Counter(num % k for num in ar)
    count = 0
    for i in range(k):
        j=-i % k
        
        num1 = modulo_count.get(i, 0)
        num2 = modulo_count.get(j, 0)
        
        if i>j or num1==0 or num2==0:
            continue

        if i==j:
            count+= num1*(num1-1)//2
        else:
            count+=num1*num2
    return count