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Formatted question description: https://leetcode.ca/all/2259.html

2259. Remove Digit From Number to Maximize Result

  • Difficulty: Easy.
  • Related Topics: String, Greedy, Enumeration.
  • Similar Questions: Remove K Digits, Remove Vowels from a String, Second Largest Digit in a String.

Problem

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.

Solution (Java, C++, Python)

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

############

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);
    }
}

  • 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
        )

############

# 2259. Remove Digit From Number to Maximize Result
# https://leetcode.com/problems/remove-digit-from-number-to-maximize-result/

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


  • 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);
    }
};

  • 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:]
}

  • 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);
}


  • 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;
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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