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

Divisible Sum Pairs

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2465 Discussions

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  • mohamed_alaa636
     + 0 comments

    condition should be i <= j not i < j inshallah, correct ?

  • tri_trinh207
     + 0 comments

    O(n) solution in JavaScript

    function divisibleSumPairs(n, k, ar) {
    let pairCount = 0;
    let remainders = new Array(k).fill(0);
    
    for(let i = 0; i < n; i++){
        let num = ar[i];
        let remainder = num % k;
        let compliment = (k - remainder)% k;
        
        pairCount += remainders[compliment];
        remainders[remainder]++;
    }
    return pairCount;
}

  • mustafaalp04q
     + 0 comments

    Is there any solution but O(n^2) ?

  • wmgarciaporcel
     + 0 comments

    This is my solution, with JavaScript:

    function divisibleSumPairs(n, k, ar) {
    // Write your code here
    let count = 0;
    while(ar.length>1){
        let i = ar.shift();
        for(let j=0; j<ar.length;j++){    
        let sum = i + ar[j];
        if(sum%k===0) count++;
        }
    }
    return count;
}

  • dhruvrparmar62
     + 0 comments

    def divisibleSumPairs(n, k, ar): freq = [0] * k count = 0

    for num in ar:
    mod = num % k
    complement = (k - num) % k
    count += freq[complement]
    freq[mod] += 1

return count