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  1. Divisible Sum Pairs
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Divisible Sum Pairs

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209 Discussions

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  • pradymm
     + 0 comments
    public static int divisibleSumPairs(int n, int k, List<Integer> ar) {
// Write your code here
    Collections.sort(ar);
    int count = 0;
    for(int i = 0; i < n; i ++){
        for(int j = i +1; j < n; j ++){
            if ((ar.get(i) + ar.get(j)) % k == 0){
                count ++;
            }
        }
    }  
    return count;
}

  • yeremiaerensa10
     + 0 comments

    javascript

    function divisibleSumPairs(n, k, ar) {
    let totalPairs = 0;
    for (let iteration = 0; iteration < n; iteration++) {
        ar.forEach((element, index) => {
        if (index <= iteration) return;
        if ((ar[iteration] + element) % k === 0) totalPairs++;
        });
    }
    console.info(totalPairs)
    return totalPairs;

}

  • sawanpatel2508
     + 0 comments

    Typescript solution: I have made the algorithm to run o(n) operation using the hash map, and much better compared to o(n^2) if we use brut force method such as nested for loop. 

    function divisibleSumPairs(n: number, k: number, ar: number[]): number {
  // Write your code here
  const freq: { [key: number]: number } = {};
  let count = 0;
  ar.forEach((int) => {
    const remainder = int % k;
    const complement = (k - remainder) % k;
    if (complement in freq) {
      count += freq[complement];
    }
    if (remainder in freq) {
      freq[remainder]++;
    } else {
      freq[remainder] = 1;
    }
  });
  return count;
}

  • carlos_afv95
     + 1 comment

    C++:

    int divisibleSumPairs(int n, int k, vector<int> ar) {
    int count = 0;
    
    for (int i = 0; i < n; ++i)
    {
        for(int j = i; j < n; ++j)
        {   
            if(j != i)
            {
                if(IsDivisiblyByK(ar[i] + ar[j], k))
                {
                    count++;
                }
            }
        }
    }
    return count;
}

  • myclg7
     + 0 comments
    function divisibleSumPairs(n, k, ar) {
    
    let sum=0,count=0,i,j;
    
    for(i=0; i<n; i++){
        for(j=i+1; j<n; j++){
            sum=ar[i]+ar[j];
            if(sum%k==0){
                count++;
            }
        }
    }
 return count;
}