Determine whether there are a minimum of two elements in the array that are larger than zero - JavaScript/Typescript

Question

Determine whether there are a minimum of two elements in the array that are larger than zero - JavaScript/Typescript

Looking for an efficient way to determine if there are at least two values greater than 0 in an array and return true? Otherwise, return false. Here's a hypothetical but incorrect attempt using the example:

const x = [9, 1, 0];
const y = [0, 0, 0];
const z = [5, 0, 0];

const checkValues = (element) => element > 0 && somethingElseMaybe;

console.log(x.some(checkValues)); // expected output: true
console.log(y.some(checkValues)); // expected output: false
console.log(z.some(checkValues)); // expected output: false

javascript typescript

Answer 1

Answer №1

Utilize the filter() method to eliminate values that are below zero and then verify if the resulting array has a length of two or more

const checkGreater = arr => arr.filter(x => x > 0).length >= 2;

console.log(checkGreater([9, 1, 0])) //true
console.log(checkGreater([0, 0, 0])) //false
console.log(checkGreater([5, 0, 0])) //false

Answer 2

Utilize the filter() method to eliminate values that are below zero and then verify if the resulting array has a length of two or more

const checkGreater = arr => arr.filter(x => x > 0).length >= 2;

console.log(checkGreater([9, 1, 0])) //true
console.log(checkGreater([0, 0, 0])) //false
console.log(checkGreater([5, 0, 0])) //false

Answer 3

Answer №2

In order to prevent wasting effort, it is advisable to halt the checking process once the condition has been satisfied. I believe this solution aligns with your requirements.

function checkIfTwoGreaterThanZero(arr) { 
    let count = 0;
    for(let num of arr) {
        if(num > 0 && (++count > 1)) return true;
    }
    return false;
}

const array1 = [9, 1, 0];
const array2 = [0, 0, 0];
const array3 = [5, 0, 0];

console.log(checkIfTwoGreaterThanZero(array1)); // expected output: true
console.log(checkIfTwoGreaterThanZero(array2)); // expected output: false
console.log(checkIfTwoGreaterThanZero(array3)); // expected output: false

Answer 4

In order to prevent wasting effort, it is advisable to halt the checking process once the condition has been satisfied. I believe this solution aligns with your requirements.

function checkIfTwoGreaterThanZero(arr) { 
    let count = 0;
    for(let num of arr) {
        if(num > 0 && (++count > 1)) return true;
    }
    return false;
}

const array1 = [9, 1, 0];
const array2 = [0, 0, 0];
const array3 = [5, 0, 0];

console.log(checkIfTwoGreaterThanZero(array1)); // expected output: true
console.log(checkIfTwoGreaterThanZero(array2)); // expected output: false
console.log(checkIfTwoGreaterThanZero(array3)); // expected output: false

Answer 5

Answer №3

If you prefer not to loop through the entire array, you can utilize a loop and break out of it early once your condition is met. While using filter may be more elegant, in cases where the list is extremely large, there could be benefit in not iterating through the whole array.

I have deconstructed this function into its fundamental components and developed a curried version.

The main question at hand is "whether there are 2 or more values that are greater than 0 in the array." However, this can be simplified to "are there X or more values that fulfill the comparator"?

const a = [9, 1, 0];
const b = [0, 0, 0];
const c = [5, 0, 0];

const quantityCompare = compare => quantity => arr => {
  let count = 0;
  for (let i = 0; i < arr.length; i += 1) {
    if (compare(arr[i])) count += 1;
    if (count >= quantity) return true;
  }
  return false;
};

const twoElementsGreaterThanZero = quantityCompare(x => x > 0)(2);

console.log(twoElementsGreaterThanZero(a)); // true
console.log(twoElementsGreaterThanZero(b)); // false
console.log(twoElementsGreaterThanZero(c)); // false

Just for fun, another alternative is to utilize Array.some (similar to an Array.forEach with the ability to exit early):

const a = [9, 1, 0];
const b = [0, 0, 0];
const c = [5, 0, 0];

const quantityCompare = compare => quantity => arr => {
  let count = 0;
  return arr.some(el => {
    if (compare(el)) count += 1;
    if (count >= quantity) return true;
  })
}

const twoElementsGreaterThanZero = quantityCompare(x => x > 0)(2);

console.log(twoElementsGreaterThanZero(a)); // true
console.log(twoElementsGreaterThanZero(b)); // false
console.log(twoElementsGreaterThanZero(c)); // false

Answer 6

If you prefer not to loop through the entire array, you can utilize a loop and break out of it early once your condition is met. While using filter may be more elegant, in cases where the list is extremely large, there could be benefit in not iterating through the whole array.

I have deconstructed this function into its fundamental components and developed a curried version.

The main question at hand is "whether there are 2 or more values that are greater than 0 in the array." However, this can be simplified to "are there X or more values that fulfill the comparator"?

const a = [9, 1, 0];
const b = [0, 0, 0];
const c = [5, 0, 0];

const quantityCompare = compare => quantity => arr => {
  let count = 0;
  for (let i = 0; i < arr.length; i += 1) {
    if (compare(arr[i])) count += 1;
    if (count >= quantity) return true;
  }
  return false;
};

const twoElementsGreaterThanZero = quantityCompare(x => x > 0)(2);

console.log(twoElementsGreaterThanZero(a)); // true
console.log(twoElementsGreaterThanZero(b)); // false
console.log(twoElementsGreaterThanZero(c)); // false

Just for fun, another alternative is to utilize Array.some (similar to an Array.forEach with the ability to exit early):

const a = [9, 1, 0];
const b = [0, 0, 0];
const c = [5, 0, 0];

const quantityCompare = compare => quantity => arr => {
  let count = 0;
  return arr.some(el => {
    if (compare(el)) count += 1;
    if (count >= quantity) return true;
  })
}

const twoElementsGreaterThanZero = quantityCompare(x => x > 0)(2);

console.log(twoElementsGreaterThanZero(a)); // true
console.log(twoElementsGreaterThanZero(b)); // false
console.log(twoElementsGreaterThanZero(c)); // false

Answer 7

Answer №4

If you find yourself needing to perform tasks like this frequently, consider creating your own method for generating predicate functions. One approach is to create a main function that generates functions which return true or false based on whether an array meets a certain condition:

function minimumSatisfy(condition, minimum) {
  return function(array) {
    for (var i = 0, count = 0; i < array.length && count < minimum; i++) {
      if (condition(array[i]))
        count++;
    }
    return count >= minimum;
  }
}

To apply this concept to your specific scenario, create a custom function using the main function:

let twoGreaterThanZero = minimumSatisfy(value => value > 0, 2);

You can now use this new function to evaluate any array with the specified predicate:

if (twoGreaterThanZero(someArray)) {
  // ...
}

However, if you only need to perform this type of check in one part of your code, it may not be necessary to implement this solution.

Answer 8

If you find yourself needing to perform tasks like this frequently, consider creating your own method for generating predicate functions. One approach is to create a main function that generates functions which return true or false based on whether an array meets a certain condition:

function minimumSatisfy(condition, minimum) {
  return function(array) {
    for (var i = 0, count = 0; i < array.length && count < minimum; i++) {
      if (condition(array[i]))
        count++;
    }
    return count >= minimum;
  }
}

To apply this concept to your specific scenario, create a custom function using the main function:

let twoGreaterThanZero = minimumSatisfy(value => value > 0, 2);

You can now use this new function to evaluate any array with the specified predicate:

if (twoGreaterThanZero(someArray)) {
  // ...
}

However, if you only need to perform this type of check in one part of your code, it may not be necessary to implement this solution.

Answer 9

Answer №5

Give it a shot

var x = [7, 2, 1]
var y = [3, 4, 5]
var z= [6, 7, 8]

function checkDuplicateElements(arr, element) {
  return arr.indexOf(element) !== arr.lastIndexOf(element)
}

console.log(checkDuplicateElements(x, 1))
console.log(checkDuplicateElements(y, 3))
console.log(checkDuplicateElements(z, 8))

Answer 10

Give it a shot

var x = [7, 2, 1]
var y = [3, 4, 5]
var z= [6, 7, 8]

function checkDuplicateElements(arr, element) {
  return arr.indexOf(element) !== arr.lastIndexOf(element)
}

console.log(checkDuplicateElements(x, 1))
console.log(checkDuplicateElements(y, 3))
console.log(checkDuplicateElements(z, 8))

Answer 11

Answer №6

Give this a shot,

console.log(a.filter(s => s > 0).length >= 2); // output is true
console.log(b.filter(s => s > 0).length >= 2); // output is false
console.log(c.filter(s => s > 0).length >= 2); // output is false

Answer 12

Give this a shot,

console.log(a.filter(s => s > 0).length >= 2); // output is true
console.log(b.filter(s => s > 0).length >= 2); // output is false
console.log(c.filter(s => s > 0).length >= 2); // output is false

Determine whether there are a minimum of two elements in the array that are larger than zero - JavaScript/Typescript

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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