Write a function that takes in an array of integers, and a string that will be either 'even' or 'odd'

Question

Im working on a problem in javascript where I am supposed to write a function that takes in an array of integers, and a string that will be either 'even' or 'odd'. The function will count how many times 4 even or 4 odd numbers show up in a row.

For example:

quadruples([3,2,2,4,8,5], 'even')  // 1
quadruples([2,4,6,8,10,5], 'even')  // 2
quadruples([2,4,6,8,10,5], 'odd')  // 0

so far this is where I am at:

function quadruples(givenArray, evenOrOdd) {
  let arr = []
  if(evenOrOdd == 'even') {
    if( i = 0; i < givenArray.length; i++) {
      
    }
};

I figured I need to run a for loop and then use a % operator but I am stuck on where to go from here.

Any help is appreciated!

Answer 1

You need dynamic programming for this with a local and global variable: [2, 4, 6, 8, 10, 5]

• 2 - even, count is 1, totalCount is 0
• 4 - even, count is 2, totalCount is 0
• 6 - even, count is 3, totalCount is 0
• 8 - even, count is 4, totalCount is 0
• 10 - even, count is 5, totalCount is 0
• 5 - odd, count is 5, increasing totalCount by 5 - 4 + 1 = 2, resetting count to 0

const quadruples = (givenArray, evenOrOdd) => {
  // never hardcode `magic numbers`, create constants for them
  const sequenceLength = 4

  // based on evenOrOdd calculating what the division by 2
  // will be if it is even, then 0, if it is odd, then 1
  const rest = evenOrOdd === 'even' ? 0 : 1

  // this will hold the total count of quadruples
  let totalCount = 0

  // this is the local count of contiguous elements
  let count = 0

  // looping over the array
  for (let i = 0; i <= givenArray.length; i += 1) {
    const el = givenArray[i]

    // if the element is not what we want
    if (i === givenArray.length || el % 2 !== rest) {
      // if the count is 4 or more, we add to totalCount the count
      // minus 4 and plus 1, meaning that if we have 4, it's 1 quadruple,
      // if it is 5, then it's 2 quadruples, etc.
      // Otherwise (count is less than 4) we add 0 (nothing)
      totalCount += count >= sequenceLength ? count - sequenceLength + 1 : 0

      // resetting the count to zero as we encountered the opposite
      // of what we are looking for (even/odd)
      count = 0

      // if the element is what we need (even or odd)
    } else {
      // increasing the count of how many we've seen by far
      count += 1
    }
  }

  // returning totalCount of quadruples
  return totalCount
}

console.log(quadruples([1, 3, 5, 7, 9, 11], 'odd')) // 3
console.log(quadruples([3, 2, 2, 4, 8, 5], 'even')) // 1
console.log(quadruples([2, 4, 6, 8, 10, 5], 'even')) // 2
console.log(quadruples([2, 4, 6, 8, 10, 5], 'odd')) // 0

Answer 2

I write this recursive.

console.log(quadruples([3, 2, 2, 4, 8, 5], 'even')); // 1
console.log(quadruples([2, 4, 6, 8, 10, 5], 'even')); // 2
console.log(quadruples([2, 4, 6, 8, 10, 5], 'odd')); // 0
console.log(quadruples([5, 4, 6, 8, 10, 5, 2, 2, 2, 2, 4, 4], 'even')); // 4

function quadruples(givenArray, evenOrOdd) {
  const maxSequence = 4;
  let result = 0;
  if (givenArray.length < maxSequence)
    return 0;
  for (let i = 0; i < maxSequence; i++) {
    if (givenArray[i] % 2 != (evenOrOdd == "even" ? 0 : 1)) {
      givenArray = givenArray.slice(i + 1);
      return (result += quadruples(givenArray, evenOrOdd));
    }
  }
  result++;
  givenArray = givenArray.slice(1);
  return (result += quadruples(givenArray, evenOrOdd));
}