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  1. Divisible Sum Pairs
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Divisible Sum Pairs

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207 Discussions

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  • 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
     + 0 comments

    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;
}

  • rohan_baba888
     + 0 comments

    I belive so far this is the most simple to understand PYTHON solution without the use of any fancy functions but just if else and loops.

    sets=[]

for i in range(len(ar)):
    for j in range(len(ar)):
        if i<j:
            if (ar[i]+ar[j])%k==0:
                sets.append((ar[i],ar[j]))
return len(sets)

  • victor_reial
     + 0 comments

    Python best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    def divisible_sum_pairs(n: int, k: int, ar: list[int]):
    #Time complexity: O(n+k)
    #Space complexity (ignoring input): O(k)
    possible_remainders = {}
    total_pairs = 0

    for number in ar:
        remainder = number % k
        if remainder in possible_remainders:
            possible_remainders[remainder] += 1
        else:
            possible_remainders[remainder] = 1

    # For remainder 0 or k/2, the total pairs will be n choose 2
    if 0 in possible_remainders:
        total_pairs += int(possible_remainders[0] * (possible_remainders[0] - 1) / 2)
    k_is_pair = k % 2 == 0
    half_k = int(k / 2)
    if k_is_pair and (half_k in possible_remainders):
        total_pairs += int(
            possible_remainders[half_k] * (possible_remainders[half_k] - 1) / 2
        )

    # For the rest of the remainders, just need to multiply
    for remainder in range(1, int((k + 1) / 2)):
        if (remainder in possible_remainders) and (
            (k - remainder) in possible_remainders
        ):
            total_pairs += int(
                possible_remainders[remainder] * (possible_remainders[k - remainder])
            )

    return total_pairs