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

Divisible Sum Pairs

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2458 Discussions

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  • hristinaglavche1
     + 0 comments

    C# var remainderCounts = new int[k]; var count = 0;

    foreach (var num in ar) { var remainder = num % k; var complement = (k - remainder) % k; count += remainderCounts[complement]; remainderCounts[remainder]++; }

    return count;

  • tongyou1995
     + 0 comments

    Do not need the n in the function and using itertools to get all unique pairs:

    import os
from itertools import combinations

def divisibleSumPairs(n, k, ar):
    valid_pair = 0
    all_pairs = combinations(ar, 2)
    for pair in all_pairs:
        pair_sum = sum(pair)
        if pair_sum % k == 0:
            valid_pair += 1
    return valid_pair

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')
    
    first_multiple_input = input().rstrip().split()
    n = int(first_multiple_input[0])
    k = int(first_multiple_input[1])
    ar = list(map(int, input().rstrip().split()))
    result = divisibleSumPairs(n, k, ar)

    fptr.write(str(result) + '\n')
    fptr.close()

  • hujaakbar
     + 0 comments
    def divisibleSumPairs(n, k, ar):
    count = 0
    for index, i in enumerate(ar):
        for j in ar[index+1:]:
            if (i + j) % k == 0:
                count+=1        
    return count

  • muradbzv
     + 0 comments

    result=0 for num in range(n): for num1 in range(n): if num!=num1 and (ar[num]+ar[num1])%k==0: result=result+1 if result%2==0: return result//2 elif result%2==1: return result//2+1

  • mayankleo
     + 0 comments

    Simplest solution

    def divisibleSumPairs(n, k, ar):
    c=0
    for i in range(n):
        for j in range(i):
            if (ar[i]+ar[j])%k==0:
                c+=1
    return c