Write
```python
remove(text, what)
```
that takes in a string ```str```(```text``` in Python) and an object/hash/dict/Dictionary ```what``` and returns a string with the chars removed in ```what```.
For example:
```python
remove('this is a string',{'t':1, 'i':2}) == 'hs s a string'
# remove from 'this is a string' the first 1 't' and the first 2 i's.
remove('hello world',{'x':5, 'i':2}) == 'hello world'
# there are no x's or i's, so nothing gets removed
remove('apples and bananas',{'a':50, 'n':1}) == 'pples d bnns'
# we don't have 50 a's, so just remove it till we hit end of string.
```

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def remove(text, what):
	

Solutions:
def remove(text, what):
	for char in what:
		text = text.replace(char, '', what[char])
	return text


def remove(text, what):
	for (c, n) in what.items():
		text = text.replace(c, '', n)
	return text


def remove(text, what):
	for (char, count) in list(what.items()):
		text = ''.join(text.split(char, count))
	return text


def remove(text, what):
	(t, d) = (list(text), dict(what))
	for (i, c) in enumerate(text):
		n = d.get(c, 0)
		if n:
			d[c] -= 1
			t[i] = ''
	return ''.join(t)


def remove(text, what):
	for (c, v) in what.items():
		text = text.replace(c, '', v)
	return text


def remove(s, d):
	for key in d:
		s = s.replace(key, '', d[key])
	return s


def remove(text, what):
	for i in what.keys():
		text = text.replace(i, '', what[i])
	return text


def remove(text, what):
	for (k, v) in what.items():
		text = text.replace(k, '', v)
	return text