Let's solve freeCodeCamp's intermediate algorithm scripting challenge, 'Drop it'.

Starter Code

function dropElements(arr, func) {
  return arr;
}

dropElements([1, 2, 3], function(n) {return n < 3; });

Instructions

Given the array arr, iterate through and remove each element starting from the first element (the 0 index) until the function func returns true when the iterated element is passed through it.

Then return the rest of the array once the condition is satisfied, otherwise, arr should be returned as an empty array.

Test Cases

• dropElements([1, 2, 3, 4], function(n) {return n >= 3;}) should return [3, 4].
• dropElements([0, 1, 0, 1], function(n) {return n === 1;}) should return [1, 0, 1].
• dropElements([1, 2, 3], function(n) {return n > 0;}) should return [1, 2, 3].
• dropElements([1, 2, 3, 4], function(n) {return n > 5;}) should return [].
• dropElements([1, 2, 3, 7, 4], function(n) {return n > 3;}) should return [7, 4].
• dropElements([1, 2, 3, 9, 2], function(n) {return n > 2;}) should return [3, 9, 2].

Our Approach

After reading the instructions, starter code, and test cases more than once, this is what we're working with:

• Our function takes in two arguments, an array (arr), and a function (func). arr consists of numbers.

• We must return an array.

• We want to alter arr based on the constraints of func, then return arr.

I had to read the instructions a few times to get a better idea of what the challenge was actually asking for. We want to work with arr, and remove each index until we come across a truthy index. So let's explore one of the test cases first off:

dropElements([1, 2, 3, 4], function(n) {return n >= 3;}) // should return [3, 4]

So, based on the arguments, we want to evaluate arr with func and keep the values that are greater than or equal to 3. Seems some what straight forward. Let's look at the last test case provided though:

dropElements([1, 2, 3, 9, 2], function(n) {return n > 2;}) // should return [3, 9, 2].

We want values greater than two so why is [2] being returned? Have a read at the below instruction:

Then return the rest of the array once the condition is satisfied, otherwise, arr should be returned as an empty array.

So once we hit something truthy, we'll return all the following elements. This is a more clear example:

dropElements([8, 1, 2, 3], function(n) {return n > 3; });

// On arr[0], we have 8. 8 is greater than 3. Even though 1 and 2 are not greater than 3, arr[0] has already met the constraint from the function so we return the rest of the elements.

Alright, hopefully the instructions are making a little more sense now. Let's get into the code.

The first thing I will do is set a variable equal to arr's length so we know how many times to loop.

const arrLength = arr.length;

I think it is time for the for loop.

for (let i = 0; i < arrLength; i++) {}

// We'll start i at 0 since array's index begins at 0

The next thing we'll need is an if statement to see if each index meets the constraint provided in the second argument, func. func takes in one argument, n. We can use arr[0] as func's argument each loop to see if it will meet the constraint.

for (let i = 0; i < arrLength; i++) {
  if (!func(arr[0])) {
    // more code coming
  }
}

Above, I'm using arr[0] instead of arr[i] because of the code we will be putting inside the if statement. If func(arr[0]) is false, we will remove it, right? Enter the array method, shift(). pop() or push() are common but here we want to remove elements from the beginning of the array and that is what shift() does for us.

Array.shift() on MDN

for (let i = 0; i < arrLength; i++) {
  if (!func(arr[0])) {
    arr.shift();
  }
  else {
    return arr;
  }
}

So let's try this step by step with one of our test cases:

dropElements([1, 2, 3, 4], function(n) {return n >= 3;});

const arrLength = arr.length // 4
for (let i = 0; i < arrLength; i++) {
  if (!func(arr[0])) {
    arr.shift();
  }
  else {
    return arr;
  }
}

// Loop 1: arr[0] = 1; is not greater or equal to 3. shift() run
// Loop 2: [2,3,4]; arr[0] = 2; not greater or equal to 3. shift() run
// Loop 3: [3,4]; arr[0] = 3; 3 is greater than or equal. return arr

As always, return arr and that is it!

Our Solution

function dropElements(arr, func) {
  const arrLength = arr.length;
  for (let i = 0; i < arrLength; i++) {
    if (!func(arr[0])) {
      arr.shift();
    }
    else {
      return arr;
    }
  }
  return arr;
}

Links & Resources

'Drop it' Challenge on fCC

freeCodeCamp

Donate to FCC!

Solution on my GitHub

Thank you for reading!