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.
```

Difficulty: introductory

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