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
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
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;
}
}
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;
}
};
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
}
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;
}
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
}
}
int getLen(int num) {
int res = 0;
while (num) {
num /= 10;
res++;
}
return res;
}
int minMaxDifference(int num) {
int n = getLen(num);
int* nums = malloc(sizeof(int) * n);
int t = num;
for (int i = n - 1; i >= 0; i--) {
nums[i] = t % 10;
t /= 10;
}
int min = 0;
for (int i = 0; i < n; i++) {
min *= 10;
if (nums[i] != nums[0]) {
min += nums[i];
}
}
int max = 0;
int target = 10;
for (int i = 0; i < n; i++) {
max *= 10;
if (target == 10 && nums[i] != 9) {
target = nums[i];
}
max += nums[i] == target ? 9 : nums[i];
}
free(nums);
return max - min;
}