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You are given a string number
representing a positive integer and a character digit
.
Return the resulting string after removing exactly one occurrence of digit
from number
such that the value of the resulting string in decimal form is maximized. The test cases are generated such that digit
occurs at least once in number
.
Example 1:
Input: number = "123", digit = "3"
Output: "12"
Explanation: There is only one '3' in "123". After removing '3', the result is "12".
Example 2:
Input: number = "1231", digit = "1"
Output: "231"
Explanation: We can remove the first '1' to get "231" or remove the second '1' to get "123".
Since 231 > 123, we return "231".
Example 3:
Input: number = "551", digit = "5"
Output: "51"
Explanation: We can remove either the first or second '5' from "551".
Both result in the string "51".
Constraints:
2 <= number.length <= 100
number
consists of digits from '1'
to '9'
.digit
is a digit from '1'
to '9'
.digit
occurs at least once in number
.string removeDigit(string n, char digit) {
for (int i = 0; i < n.size() - 1; ++i)
if (n[i] == digit && n[i + 1] > digit)
return n.substr(0, i) + n.substr(i + 1);
int last_d = n.rfind(digit);
return n.substr(0, last_d) + n.substr(last_d + 1);
}
class Solution {
public String removeDigit(String number, char digit) {
List<String> digits = new ArrayList<>();
for (int i = 0; i < number.length(); i++) {
if (number.charAt(i) == digit) {
String stringWithoutDigit = number.substring(0, i) + number.substring(i + 1);
digits.add(stringWithoutDigit);
}
}
Collections.sort(digits);
return digits.get(digits.size() - 1);
}
}
class Solution:
def removeDigit(self, number: str, digit: str) -> str:
# Initializing the last index as zero
last_index = 0
#iterating each number to find the occurences, \
# and to find if the number is greater than the next element \
for num in range(1, len(number)):
# Handling [case 1] and [case 2]
if number[num-1] == digit:
if int(number[num]) > int(number[num-1]):
return number[:num-1] + number[num:]
else:
last_index = num - 1
# If digit is the last number (last occurence) in the string [case 3]
if number[-1] == digit:
last_index = len(number) - 1
return number[:last_index] + number[last_index + 1:]
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