Divisible sum pairs
#beginners#programming#tutorial#javascript
Klecianny Melo

Klecianny Melo @kecbm

About: 💻 Software Engineer | SWE | ⚙️ JavaScript, .NET, React, Docker, Node.js, CSS | 📝 Author

Location:
Garanhuns, Pernambuco
Joined:
Apr 13, 2022

Divisible sum pairs

Publish Date: May 2 '24
17 8

Prepare your favorite cup of coffee, because we are about to enter the fantastic world of Divisible sum pairs.

The problem

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The solution

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To start our solution, let's define the divisibleSumPairs function that will receive the following parameters:

  • n: length of array arr;
  • k: integer divisor;
  • ar: input array. 
function divisibleSumPairs(n, k, ar) {}
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Next we will initialize the counter variable with the value 0:

let counter = 0;
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Now let's go through the ar array from the first element to the penultimate element:

for (let i = 0; i < n - 1; i++) {}
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Inside the first loop, we will start a second loop, to cycle through the elements of the ar array from the i + 1 element to the last element:

for (let j = i + 1; j < n; j++) {}
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After traversing the array ar with the two loops (integrating through the element i and the element j (i + 1)), we will check if the sum of the elements in the indices i and j of the array arr is divisible by k without remainder:

if ((ar[i] + ar[j]) % k === 0) {}
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If this condition is met, we will increase the value of the counter variable:

counter++;
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Thus, we will have the number of pairs that satisfy this condition counted in counter.

Finally, let's return the value of counter:

return counter;
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Final resolution

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After following the step by step we have our final resolution:

function divisibleSumPairs(n, k, ar) {
    let counter = 0;

    for (let i = 0; i < n - 1; i++) {
      for (let j = i + 1; j < n; j++) {
        if ((ar[i] + ar[j]) % k === 0) {
          counter++;
        }
      }
    }

    return counter;
}
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Share the code, spread knowledge and build the future! 😉

Images generated by DALL·E 3

Comments 8 total

  • Anuska Santos
    Anuska SantosMay 2, 2024

    Very interesting this solution, use two loops is 🤯

    • Klecianny Melo
      Klecianny MeloMay 2, 2024

      Thank you my dear! Two loops is the power 😎

  • Rayanny Bezerra
    Rayanny BezerraMay 2, 2024

    Nice algorithm for study

    • Klecianny Melo
      Klecianny MeloMay 2, 2024

      It's true Lala, thank you for your contribution! 🥰

  • 𒎏Wii 🏳️‍⚧️
    𒎏Wii 🏳️‍⚧️May 3, 2024

    Prepare your favorite cup of coffee, because we are about to enter the fantastic world of Divisible sum pairs.

    No! *prepares cup of tea* >:3

  • Tracy Gilmore
    Tracy GilmoreMay 13, 2024

    Hi Klecianny,
    I would like to offer two alternative implementations. Not better (probably worse), just different approaches. There is still a double pass through the array, just not as explicate as in your solution.

    Using reduce & filter methods

      function divisibleSumPairs(k, ar) {
    return ar.reduce(
      (acc, ar_i, i) =>
        acc + ar.slice(i + 1).filter(ar_j => !((ar_i + ar_j) % k)).length,
      0
    );
  }
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    Using recursion with a filter method

      function divisibleSumPairs(k, ar) {
    return _divisibleSumPairs(0, ...ar);

    function _divisibleSumPairs(count, ar_i, ...ar_) {
      return ar_.length
        ? _divisibleSumPairs(
            count + ar_.filter(ar_j => !((ar_i + ar_j) % k)).length,
            ...ar_
          )
        : count;
    }
  }
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    • Klecianny Melo
      Klecianny MeloJun 6, 2024

      Hello @tracygjg, how are you? Thank you very much for this solutions for the current problem! Programming is fascinating because we can solve the same problem in different ways. I look forward to your interactions in future articles! 😁

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