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JavaScript: Compute the sum of two parts and store into an array of size two

JavaScript Basic: Exercise-85 with Solution

Write a JavaScript code to divide a given array of positive integers into two parts. First element goes to first part, second element goes to second part, and third element goes to first part and so on. Now compute the sum of two parts and store into an array of size two.

Pictorial Presentation:

JavaScript: Compute the sum of two parts and store into an array of size two.

Sample Solution:

HTML Code:

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width">
  <title>Compute the sum of two parts and store into an array of size two</title>
</head>
<body>

</body>
</html>

JavaScript Code:

function alternate_Sums(arr) {
  var result = [0, 0];
  for(var i = 0; i < arr.length; i++) 
  {
    if(i % 2) result[1] += arr[i];
    else
      result[0] += arr[i];
  }
  return result
}

console.log(alternate_Sums([1, 3, 6, 2, 5, 10]))

Sample Output:

[12,15]

Flowchart:

Flowchart: JavaScript - Compute the sum of two parts and store into an array of size two

ES6 Version:

function alternate_Sums(arr) {
  const result = [0, 0];
  for(let i = 0; i < arr.length; i++) 
  {
    if(i % 2) result[1] += arr[i];
    else
      result[0] += arr[i];
  }
  return result
}

console.log(alternate_Sums([1, 3, 6, 2, 5, 10]))

Live Demo:

See the Pen javascript-basic-exercise-85 by w3resource (@w3resource) on CodePen.


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JavaScript: Tips of the Day

How to insert an item into an array at a specific index (JavaScript)?

What you want is the splice function on the native array object.

arr.splice(index, 0, item); will insert item into arr at the specified index (deleting 0 items first, that is, it's just an insert). In this example we will create an array and add an element to it into index 2:

var arr = [];
arr[0] = "Jani";
arr[1] = "Hege";
arr[2] = "Stale";
arr[3] = "Kai Jim";
arr[4] = "Borge";

console.log(arr.join());
arr.splice(2, 0, "Lene");
console.log(arr.join());

Ref: https://bit.ly/2BXbp04