RemoveDigits | Topcoder Archive

Problem Statement

Given a non-negative integer n (n may have up to 50 digits) and another number m, remove exactly m digits from n so the remaining number is as big as possible. m will always be less than the number of digits in n.

Create a class RemoveDigits containing the method maxNumber which takes a String n and an int m and returns a String containing the biggest number when removing exactly m digits from n.

Definition

Class:: RemoveDigits
Method:: maxNumber
Parameters:: String, int
Returns:: String
Method signature:: String maxNumber(String n, int m)
(be sure your method is public)

Notes

n may contain leading zeros, see example 1.
The return value should contain exactly |n|-m digits, where |n| is the number of digits in n.

Constraints

n will contain between 1 and 50 characters, inclusive.
n will only contain digits ('0'-'9').
m will be between 0 and |n|-1, inclusive, where |n| is the number of digits in n.

Examples

```
"834391370"
```
```
2
```
```
Returns: "8491370"
```
We should try to get the highest digits in the front. It's not possible to get the 9 in the front since that would require removing 4 digits, so we have to be satisfied with the 8. We still can't get the 9 to the second place in the final number, so the 4 is the highest choice for this position and thus we remove the digit 3. Now we can get the 9 as the third digit from the left in the final number by removing the second digit 3. Now we have removed 2 digits, and the number remaining is 8491370.
```
"0021597002943020"
```
```
3
```
```
Returns: "2597002943020"
```
We start with removing the two leading zeros and then the first 1, thus getting 2597002943020.
```
"0000500390"
```
```
2
```
```
Returns: "00500390"
```

"91459800928398113866079548895152924499"

Returns: "9999995152924499"

```
"98"
```
```
1
```
```
Returns: "9"
```

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