# Instructing Children Programming – Take away Trailing Zeros From a String (strip, lstrip, rstrip)

Instructing Children Programming: Movies on Information Constructions and Algorithms

Given a constructive integer num represented as a string, return the integer num with out trailing zeros as a string.

Instance 1:
Enter: num = “51230100”
Output: “512301”
Clarification: Integer “51230100” has 2 trailing zeros, we take away them and return integer “512301”.

Instance 2:
Enter: num = “123”
Output: “123”
Clarification: Integer “123” has no trailing zeros, we return integer “123”.

Constraints:
1 <= num.size <= 1000
num consists of solely digits.
num doesn’t have any main zeros.

### Take away Trailing Zeros From a String (strip, lstrip, rstrip)

A string in Python is immutable – we can’t change it. Nonetheless, we are able to convert it to record, take away the trailing zeros and convert it again to string utilizing the be part of operate.

We repeat checking the rightmost character to see whether it is zero, use the pop to take away the final character. The time complexity is O(N) within the worst circumstances, all characters are eliminated. The house complexity is O(N) as we have to retailer a replica of the immutable string as a mutable record/array.

 ```1 2 3 4 5 6 7 8 ``` ```class Answer:     def removeTrailingZeros(self, num: str) -> str:         def f(x, t):             x = record(x)             whereas x and x[-1] == t:                 x.pop()             return "".be part of(x)         return f(num, '0')```
```class Answer:
def removeTrailingZeros(self, num: str) -> str:
def f(x, t):
x = record(x)
whereas x and x[-1] == t:
x.pop()
return "".be part of(x)
return f(num, '0')```

If we need to take away the main zeros, we’d want to make use of a deque (double-ended queue) to attain O(1) fixed time to push/pop on either side. The above code takes O(N) time. We are able to change to the next utilizing deque:

 ```1 2 3 4 5 6 7 8 ``` ```class Answer:     def removeTrailingZeros(self, num: str) -> str:         def f(x, t):             x = deque(record(x))             whereas x and x[-1] == t:                 x.pop()             return "".be part of(x)         return f(num, '0')```

```class Answer:
def removeTrailingZeros(self, num: str) -> str:
def f(x, t):
x = deque(record(x))
whereas x and x[-1] == t:
x.pop()
return "".be part of(x)
return f(num, '0')```

If we need to take away the main characters (strip left):

 ```1 2 3 4 5 ``` ```def lstrip(x, t):     x = deque(record(x))     whereas x and x[0] == t:         x.popleft()     return "".be part of(x)```

```def lstrip(x, t):
x = deque(record(x))
whereas x and x[0] == t:
x.popleft()
return "".be part of(x)```

In Python, we are able to use lstrip and rstrip to take away the main and trailing characters respectively. If the characters will not be specified, it’s white house.

 ```1 2 3 ``` ```class Answer:     def removeTrailingZeros(self, num: str) -> str:         return num.rstrip('0')```
```class Answer:
def removeTrailingZeros(self, num: str) -> str:
return num.rstrip('0')```

The strip operate removes each main and trailing so it’s the identical as:

 ```1 2 ``` ```def strip(s, x):     return s.lstrip(x).rstrip(x)```
```def strip(s, x):
return s.lstrip(x).rstrip(x)```

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