Sum up all even numbers in an array using JavaScript

Question

Sum up all even numbers in an array using JavaScript

I have an array of numbers: [20,40,60,60,20]. The task is to find pairs of two numbers whose sum is divisible by 60. In this array, there are 3 such pairs: (20,40), (40,20), and (60,60).

However, when I implemented the code to achieve this, it returned 4 instead of 3.

function countPairs (array) {
  let count = 0;
  for (let i = 0; i < array.length-1; i++) {
    for (let j = 0; j < array.length-1; j++) {
      let a = array[i];
      let b = array[j];
      if (checkPair(a, b) && notSameIndex(i, j)) {
        count++;
      }
    }
  }
  return count;
}

function checkPair (a, b) {
  return Number.isInteger((a + b)/60);
}

function notSameIndex (x, y) {
  return x !== y ? true : false;
}

Can you spot what's causing the discrepancy in the result?

javascript arrays duplicates nested-loops

Answer 1

Answer №1

Commence the second iteration from i+1 as you've already covered the ith index in the initial loop; therefore, traverse from i+1 to the array's length. Make sure to run both loops until i < array.length, considering that you initiate from index 0.

function tallyPlayList (array) {
  let count = 0;
  for (let i = 0; i < array.length; i++) {
    for (let j = i+1; j < array.length; j++) {
      let a = array[i];
      let b = array[j];
      if (validatePlayTime(a, b) && eliminateDuplicates(i, j)) {
        count++;
      }
    }
  }
  return count;
}

function validatePlayTime (a, b) {
  return Number.isInteger((a + b)/60);
}

function eliminateDuplicates (x, y) {
  return x !== y ? true : false;
}
console.log(tallyPlayList([20,40,60,60,20]))

Answer 2

Commence the second iteration from i+1 as you've already covered the ith index in the initial loop; therefore, traverse from i+1 to the array's length. Make sure to run both loops until i < array.length, considering that you initiate from index 0.

function tallyPlayList (array) {
  let count = 0;
  for (let i = 0; i < array.length; i++) {
    for (let j = i+1; j < array.length; j++) {
      let a = array[i];
      let b = array[j];
      if (validatePlayTime(a, b) && eliminateDuplicates(i, j)) {
        count++;
      }
    }
  }
  return count;
}

function validatePlayTime (a, b) {
  return Number.isInteger((a + b)/60);
}

function eliminateDuplicates (x, y) {
  return x !== y ? true : false;
}
console.log(tallyPlayList([20,40,60,60,20]))

Answer 3

Answer №2

You've encountered 2 issues in your code:

Ensure that you iterate through the entire array by using either

(let i = 0; i < array.length; i++)

or

(let i = 0; i <= array.length - 1; i++)

The inner loop should begin from the outer index + 1 to avoid checking every combination twice.

function countPlayList (array) {
  let count = 0;
  for (let i = 0; i < array.length - 1; i++) {
    for (let j = i + 1; j < array.length; j++) {
      let a = array[i];
      let b = array[j];
      if (checkPlayTime(a, b)) {
        count++;
      }
    }
  }
  return count;
}

function checkPlayTime (a, b) {
  return Number.isInteger((a + b)/60);
}

console.log(countPlayList([20,40,60,60,20]));

By the way, you can simplify return x !== y ? true : false; to just return x !== y;, as this value is already boolean.

Answer 4

You've encountered 2 issues in your code:

Ensure that you iterate through the entire array by using either

(let i = 0; i < array.length; i++)

or

(let i = 0; i <= array.length - 1; i++)

The inner loop should begin from the outer index + 1 to avoid checking every combination twice.

function countPlayList (array) {
  let count = 0;
  for (let i = 0; i < array.length - 1; i++) {
    for (let j = i + 1; j < array.length; j++) {
      let a = array[i];
      let b = array[j];
      if (checkPlayTime(a, b)) {
        count++;
      }
    }
  }
  return count;
}

function checkPlayTime (a, b) {
  return Number.isInteger((a + b)/60);
}

console.log(countPlayList([20,40,60,60,20]));

By the way, you can simplify return x !== y ? true : false; to just return x !== y;, as this value is already boolean.

Answer 5

Answer №3

You have the option to modify certain elements:

Start iterating from i + 1 to just before the length of the array,
Utilize a concatenated string value of the current pair,
Implement a Set to monitor visited pairs,
Adopt a different condition for checking sums and employ the remainder operator for obtaining the remaining value,
Include a check for already seen values,
Save an unseen pair in memory,
Finally, return the count of unique pairs.

function calculatePairCount(array) {
    let pairs = new Set;
        
    for (let i = 0; i < array.length - 1; i++) {
        for (let j = i + 1; j < array.length; j++) {
            let a = array[i],
                b = array[j],
                pair = [a, b].join('|');
                
            if ((a + b) % 60 === 0 && !pairs.has(pair)) {
                pairs.add(pair);
            }
        }
    }
    console.log(...pairs);
    return pairs.size;
}

console.log(calculatePairCount([20, 40, 60, 60, 20]));

Answer 6

You have the option to modify certain elements:

Start iterating from i + 1 to just before the length of the array,
Utilize a concatenated string value of the current pair,
Implement a Set to monitor visited pairs,
Adopt a different condition for checking sums and employ the remainder operator for obtaining the remaining value,
Include a check for already seen values,
Save an unseen pair in memory,
Finally, return the count of unique pairs.

function calculatePairCount(array) {
    let pairs = new Set;
        
    for (let i = 0; i < array.length - 1; i++) {
        for (let j = i + 1; j < array.length; j++) {
            let a = array[i],
                b = array[j],
                pair = [a, b].join('|');
                
            if ((a + b) % 60 === 0 && !pairs.has(pair)) {
                pairs.add(pair);
            }
        }
    }
    console.log(...pairs);
    return pairs.size;
}

console.log(calculatePairCount([20, 40, 60, 60, 20]));

Sum up all even numbers in an array using JavaScript

Answer №1

Answer №2

Answer №3

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