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

#### Tags: arrays, for-loop, javascript, modulus

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
```

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!

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
}
}