# Scenario 

With **_Cereal crops_** like wheat or rice, before we can eat the grain kernel, we need to remove that inedible hull, or *to separate the wheat from the chaff*.
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# Task

**_Given_** a *sequence of n integers* , **_separate_** *the negative numbers (chaff) from positive ones (wheat).* 
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# Notes

* **_Sequence size_** is _at least_ **_3_**
* **_Return_** *a new sequence*, such that  **_negative numbers (chaff) come first, then positive ones (wheat)_**.
* In Java , *you're not allowed to modify the input Array/list/Vector*
* **_Have no fear_** , *it is guaranteed that there will be no zeroes* . 
* **_Repetition_** of numbers in *the input sequence could occur* , so **_duplications are included when separating_**.
* If a misplaced *positive* number is found in the front part of the sequence, replace it with the last misplaced negative number (the one found near the end of the input). The second misplaced positive number should be swapped with the second last misplaced negative number. *Negative numbers found at the head (begining) of the sequence* , **_should be kept in place_** .

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# Input >> Output Examples:

```
wheatFromChaff ({7, -8, 1 ,-2}) ==> return ({-2, -8, 1, 7}) 
```

## **_Explanation_**:

* **_Since_** `7 ` is a  **_positive number_** , it should not be located at the beginnig so it needs to be swapped with the **last negative number** `-2`.
____

```
wheatFromChaff ({-31, -5, 11 , -42, -22, -46, -4, -28 }) ==> return ({-31, -5,- 28, -42, -22, -46 , -4, 11})
```

## **_Explanation_**:

* **_Since_**, `{-31, -5} ` are  **_negative numbers_** *found at the head (begining) of the sequence* , *so we keep them in place* .
* Since `11` is a positive number, it's replaced by  the last negative which is `-28` , and so on till sepration is complete. 

____

```
wheatFromChaff ({-25, -48, -29, -25, 1, 49, -32, -19, -46, 1}) ==> return ({-25, -48, -29, -25, -46, -19, -32, 49, 1, 1})
```

## **_Explanation_**:

* **_Since_** `{-25, -48, -29, -25} ` are  **_negative numbers_** *found at the head (begining) of the input* , *so we keep them in place* .

* Since `1` is a positive number, it's replaced by  the last negative which is `-46` , and so on till sepration is complete. 

* Remeber, *duplications are included when separating* , that's why the number `1` appeared twice at the end of the output. 
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# Tune Your Code , There are 250 Assertions , 100.000 element For Each .

# Only O(N) Complexity Solutions Will pass . 
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Source: codewars | Difficulty: EASY
Tags: Fundamentals

Data structures

Starter code: def wheat_from_chaff(values):
	

Solutions:
def wheat_from_chaff(values):
	(i, j) = (0, len(values) - 1)
	while True:
		while i < j and values[i] < 0:
			i += 1
		while i < j and values[j] > 0:
			j -= 1
		if i >= j:
			return values
		(values[i], values[j]) = (values[j], values[i])


def wheat_from_chaff(values):
	(i, j) = (0, len(values) - 1)
	vals = values[:]
	while i < j:
		if vals[i] < 0:
			i += 1
		elif vals[j] > 0:
			j -= 1
		else:
			(vals[i], vals[j]) = (vals[j], vals[i])
			(i, j) = (i + 1, j - 1)
	return vals


from typing import List

def wheat_from_chaff(values: List[int]) -> List[int]:
	(i, latest_neg) = (0, len(values))
	while i < latest_neg:
		if values[i] > 0:
			try:
				latest_neg = next((j for j in range(latest_neg - 1, i, -1) if values[j] < 0))
			except StopIteration:
				break
			(values[i], values[latest_neg]) = (values[latest_neg], values[i])
		i += 1
	return values


def wheat_from_chaff(a):
	(r, i, l) = ([[i, j] for (i, j) in enumerate(a) if j < 0], 0, len(a))
	for i in range(len(a)):
		if a[i] > 0:
			if not r:
				break
			r_ = r.pop()
			if r_[0] < i:
				break
			(a[i], a[r_[0]]) = (r_[1], a[i])
	return a


def wheat_from_chaff(values):
	(i, j) = (0, len(values) - 1)
	while i < j:
		(a, b) = (values[i], values[j])
		if a > 0 and b < 0:
			(values[i], values[j]) = (values[j], values[i])
		elif b > 0:
			j -= 1
		else:
			i += 1
	return values


def wheat_from_chaff(values):
	values = values[:]
	(i, j) = (0, len(values) - 1)
	while i < j:
		while values[i] < 0:
			i += 1
		while values[j] > 0:
			j -= 1
		if i < j:
			(values[i], values[j]) = (values[j], values[i])
	return values


def wheat_from_chaff(values):
	end = len(values) - 1
	for (start, n) in enumerate(values):
		if start >= end:
			break
		if n > 0:
			while start < end and values[end] > 0:
				end -= 1
			(values[start], values[end]) = (values[end], values[start])
	return values


def wheat_from_chaff(values):
	results = values[:]
	head = 0
	tail = len(values) - 1
	while True:
		while results[head] <= 0 and head < tail:
			head += 1
		while results[tail] > 0 and tail > head:
			tail -= 1
		if head >= tail:
			break
		(results[head], results[tail]) = (results[tail], results[head])
	return results


def wheat_from_chaff(values):
	last_index = len(values) - 1
	for i in range(0, len(values)):
		if values[i] > 0:
			negative_index = None
			negative_value = None
			for ii in range(last_index, -1, -1):
				if values[ii] < 0:
					last_index = ii
					negative_index = ii
					negative_value = values[ii]
					break
			if negative_index == None:
				break
			elif negative_index > i:
				(values[i], values[ii]) = (values[ii], values[i])
			else:
				break
	return values


def wheat_from_chaff(values):
	negative = []
	positive = []
	while len(values) > 0:
		if values[0] < 0:
			negative.append(values.pop(0))
		elif values[-1] > 0:
			positive.append(values.pop(-1))
		else:
			positive.append(values.pop(0))
			negative.append(values.pop(-1))
	return negative + positive[::-1]