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JavaScript: Check whether 1 appears in first or last position of a given array of integers

JavaScript Basic: Exercise-71 with Solution

Write a JavaScript program to check whether 1 appears in first or last position of a given array of integers. The array length must be greater or equal to 1.

Pictorial Presentation:

JavaScript: Check whether 1 appears in first or last position of a given array of integers.

Sample Solution:

HTML Code:

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width">
  <title>JavaScript program to check whether 1 appears in first or last position of a given array of integers. The array length must be greater or equal to 1.</title>
</head>
<body>

</body>
</html>

JavaScript Code:

function first_last_1(nums)
{
  var end_pos = nums.length - 1;
  return nums[0] == 1 || nums[end_pos] == 1;
}


console.log(first_last_1([1, 3, 5]));
console.log(first_last_1([1, 3, 5, 1]));
console.log(first_last_1([2, 4, 6]));

Sample Output:

true
true
false

Flowchart:

Flowchart: JavaScript - Check whether 1 appears in first or last position of a given array of integers

ES6 Version:

function first_last_1(nums)
{
  const end_pos = nums.length - 1;
  return nums[0] == 1 || nums[end_pos] == 1;
}


console.log(first_last_1([1, 3, 5]));
console.log(first_last_1([1, 3, 5, 1]));
console.log(first_last_1([2, 4, 6]));

Live Demo:

See the Pen JavaScript - Check whether 1 appears in first or last position of a given array of integers - basic-ex-71 by w3resource (@w3resource) on CodePen.


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JavaScript: Tips of the Day

Classes/function constructors

class Person {
  constructor() {
    this.name = 'Owen';
  }
}

Person = class AnotherPerson {
  constructor() {
    this.name = 'Eddie';
  }
};

const member = new Person();
console.log(member.name);

We can set classes equal to other classes/function constructors. In this case, we set Person equal to AnotherPerson. The name on this constructor is Eddie, so the name property on the new Person instance member is "Eddie".

Ref: https://bit.ly/3jFRBje