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Formatted question description: https://leetcode.ca/all/2566.html
2566. Maximum Difference by Remapping a Digit
Description
You are given an integer num
. You know that Danny Mittal will sneakily remap one of the 10
possible digits (0
to 9
) to another digit.
Return the difference between the maximum and minimum values Danny can make by remapping exactly one digit in num
.
Notes:
- When Danny remaps a digit d1 to another digit d2, Danny replaces all occurrences of
d1
innum
withd2
. - Danny can remap a digit to itself, in which case
num
does not change. - Danny can remap different digits for obtaining minimum and maximum values respectively.
- The resulting number after remapping can contain leading zeroes.
- We mentioned "Danny Mittal" to congratulate him on being in the top 10 in Weekly Contest 326.
Example 1:
Input: num = 11891 Output: 99009 Explanation: To achieve the maximum value, Danny can remap the digit 1 to the digit 9 to yield 99899. To achieve the minimum value, Danny can remap the digit 1 to the digit 0, yielding 890. The difference between these two numbers is 99009.
Example 2:
Input: num = 90 Output: 99 Explanation: The maximum value that can be returned by the function is 99 (if 0 is replaced by 9) and the minimum value that can be returned by the function is 0 (if 9 is replaced by 0). Thus, we return 99.
Constraints:
1 <= num <= 108
Solutions
Solution 1: Greedy
First, we convert the number to a string $s$.
To get the minimum value, we just need to find the first digit $s[0]$ in the string $s$, and then replace all $s[0]$ in the string with $0$.
To get the maximum value, we need to find the first digit $s[i]$ in the string $s$ that is not $9$, and then replace all $s[i]$ in the string with $9$.
Finally, return the difference between the maximum and minimum values.
The time complexity is $O(\log n)$, and the space complexity is $O(\log n)$. Where $n$ is the size of the number $num$.
-
class Solution { public int minMaxDifference(int num) { String s = String.valueOf(num); int mi = Integer.parseInt(s.replace(s.charAt(0), '0')); for (char c : s.toCharArray()) { if (c != '9') { return Integer.parseInt(s.replace(c, '9')) - mi; } } return num - mi; } }
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class Solution { public: int minMaxDifference(int num) { string s = to_string(num); string t = s; char first = s[0]; for (char& c : s) { if (c == first) { c = '0'; } } int mi = stoi(s); for (int i = 0; i < t.size(); ++i) { if (t[i] != '9') { char second = t[i]; for (int j = i; j < t.size(); ++j) { if (t[j] == second) { t[j] = '9'; } } return stoi(t) - mi; } } return num - mi; } };
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class Solution: def minMaxDifference(self, num: int) -> int: s = str(num) mi = int(s.replace(s[0], '0')) for c in s: if c != '9': return int(s.replace(c, '9')) - mi return num - mi
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func minMaxDifference(num int) int { s := []byte(strconv.Itoa(num)) first := s[0] for i := range s { if s[i] == first { s[i] = '0' } } mi, _ := strconv.Atoi(string(s)) t := []byte(strconv.Itoa(num)) for i := range t { if t[i] != '9' { second := t[i] for j := i; j < len(t); j++ { if t[j] == second { t[j] = '9' } } mx, _ := strconv.Atoi(string(t)) return mx - mi } } return num - mi }
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function minMaxDifference(num: number): number { const s = num + ''; const min = Number(s.replace(new RegExp(s[0], 'g'), '0')); for (const c of s) { if (c !== '9') { return Number(s.replace(new RegExp(c, 'g'), '9')) - min; } } return num - min; }
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impl Solution { pub fn min_max_difference(num: i32) -> i32 { let s = num.to_string(); let min = s .replace(char::from(s.as_bytes()[0]), "0") .parse::<i32>() .unwrap(); for &c in s.as_bytes() { if c != b'9' { return s.replace(c, "9").parse().unwrap() - min; } } num - min } }