Group the Numbers

// Given array of numbers, group them so that all evens are in left side and all odds are in right side

/* 
First Intuition: Use two pointers to partition the array like in quicksort

Time Complexity - O(n)
Space Complexity - O(1)
*/

function swap(array, i, j) {
	let temp = array[i]
	array[i] = array[j]
	array[j] = temp
}

function isEven(num) {
	return num % 2 === 0
}

function groupTheNumbers(array) {
	const len = array.length

	let evenPointer = -1,
		curr = 0

	for (curr; curr < len; curr++) {
		const element = array[curr]
		if (isEven(element)) {
			evenPointer++
			swap(array, evenPointer, curr)
		}
	}

	return array
}

console.log(groupTheNumbers([1, 2, 3, 5, -1, 0, 7, 8]))
console.log(groupTheNumbers([1, 2, 3, 4]))