2259. Remove Digit From Number to Maximize Result 

Description

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.

Solutions

Solution 1: Brute Force Enumeration

We can enumerate all positions $\textit{i}$ in the string $\textit{number}$. If $\textit{number}[i] = \textit{digit}$, we take the prefix $\textit{number}[0:i]$ and the suffix $\textit{number}[i+1:]$ of $\textit{number}$ and concatenate them. This gives the result after removing $\textit{number}[i]$. We then take the maximum of all possible results.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $\textit{number}$.

Python3

class Solution:
    def removeDigit(self, number: str, digit: str) -> str:
        return max(
            number[:i] + number[i + 1 :] for i, d in enumerate(number) if d == digit
        )

Java

class Solution {
    public String removeDigit(String number, char digit) {
        String ans = "0";
        for (int i = 0, n = number.length(); i < n; ++i) {
            char d = number.charAt(i);
            if (d == digit) {
                String t = number.substring(0, i) + number.substring(i + 1);
                if (ans.compareTo(t) < 0) {
                    ans = t;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    string removeDigit(string number, char digit) {
        string ans = "0";
        for (int i = 0, n = number.size(); i < n; ++i) {
            char d = number[i];
            if (d == digit) {
                string t = number.substr(0, i) + number.substr(i + 1, n - i);
                if (ans < t) {
                    ans = t;
                }
            }
        }
        return ans;
    }
};

Go

func removeDigit(number string, digit byte) string {
	ans := "0"
	for i, d := range number {
		if d == rune(digit) {
			t := number[:i] + number[i+1:]
			if strings.Compare(ans, t) < 0 {
				ans = t
			}
		}
	}
	return ans
}

TypeScript

function removeDigit(number: string, digit: string): string {
    const n = number.length;
    let last = -1;
    for (let i = 0; i < n; ++i) {
        if (number[i] === digit) {
            last = i;
            if (i + 1 < n && number[i] < number[i + 1]) {
                break;
            }
        }
    }
    return number.substring(0, last) + number.substring(last + 1);
}

PHP

class Solution {
    /**
     * @param String $number
     * @param String $digit
     * @return String
     */
    function removeDigit($number, $digit) {
        $max = 0;
        for ($i = 0; $i < strlen($number); $i++) {
            if ($number[$i] == $digit) {
                $tmp = substr($number, 0, $i) . substr($number, $i + 1);
                if ($tmp > $max) {
                    $max = $tmp;
                }
            }
        }
        return $max;
    }
}

Solution 2: Greedy

We can enumerate all positions $\textit{i}$ in the string $\textit{number}$. If $\textit{number}[i] = \textit{digit}$, we record the last occurrence position of $\textit{digit}$ as $\textit{last}$. If $\textit{i} + 1 < \textit{n}$ and $\textit{number}[i] < \textit{number}[i + 1]$, then we can directly return $\textit{number}[0:i] + \textit{number}[i+1:]$ as the result after removing $\textit{number}[i]$. This is because if $\textit{number}[i] < \textit{number}[i + 1]$, removing $\textit{number}[i]$ will result in a larger number.

After the traversal, we return $\textit{number}[0:\textit{last}] + \textit{number}[\textit{last}+1:]$.

Python3

class Solution:
    def removeDigit(self, number: str, digit: str) -> str:
        last = -1
        n = len(number)
        for i, d in enumerate(number):
            if d == digit:
                last = i
                if i + 1 < n and d < number[i + 1]:
                    break
        return number[:last] + number[last + 1 :]

Java

class Solution {
    public String removeDigit(String number, char digit) {
        int last = -1;
        int n = number.length();
        for (int i = 0; i < n; ++i) {
            char d = number.charAt(i);
            if (d == digit) {
                last = i;
                if (i + 1 < n && d < number.charAt(i + 1)) {
                    break;
                }
            }
        }
        return number.substring(0, last) + number.substring(last + 1);
    }
}

C++

class Solution {
public:
    string removeDigit(string number, char digit) {
        int n = number.size();
        int last = -1;
        for (int i = 0; i < n; ++i) {
            char d = number[i];
            if (d == digit) {
                last = i;
                if (i + 1 < n && number[i] < number[i + 1]) {
                    break;
                }
            }
        }
        return number.substr(0, last) + number.substr(last + 1);
    }
};

Go

func removeDigit(number string, digit byte) string {
	last := -1
	n := len(number)
	for i := range number {
		if number[i] == digit {
			last = i
			if i+1 < n && number[i] < number[i+1] {
				break
			}
		}
	}
	return number[:last] + number[last+1:]
}