2165. Smallest Value of the Rearranged Number
Description
You are given an integer num. Rearrange the digits of num such that its value is minimized and it does not contain any leading zeros.
Return the rearranged number with minimal value.
Note that the sign of the number does not change after rearranging the digits.
Example 1:
Input: num = 310 Output: 103 Explanation: The possible arrangements for the digits of 310 are 013, 031, 103, 130, 301, 310. The arrangement with the smallest value that does not contain any leading zeros is 103.
Example 2:
Input: num = -7605 Output: -7650 Explanation: Some possible arrangements for the digits of -7605 are -7650, -6705, -5076, -0567. The arrangement with the smallest value that does not contain any leading zeros is -7650.
Constraints:
-1015 <= num <= 1015
Solutions
Solution 1: Counting
We first use an array $\textit{cnt}$ to record the number of occurrences of each digit in $\textit{num}$.
If $\textit{num}$ is negative, the digits should be arranged in descending order. Therefore, we traverse $\textit{cnt}$ from $9$ to $0$ and arrange the digits in descending order according to their occurrences.
If $\textit{num}$ is positive, we first find the first non-zero digit and place it in the first position, then arrange the remaining digits in ascending order.
The time complexity is $O(\log n)$, where $n$ is the size of the number $\textit{num}$. The space complexity is $O(1)$.
Python3
class Solution:
def smallestNumber(self, num: int) -> int:
neg = num < 0
num = abs(num)
cnt = [0] * 10
while num:
cnt[num % 10] += 1
num //= 10
ans = 0
if neg:
for i in reversed(range(10)):
for _ in range(cnt[i]):
ans *= 10
ans += i
return -ans
if cnt[0]:
for i in range(1, 10):
if cnt[i]:
ans = i
cnt[i] -= 1
break
for i in range(10):
for _ in range(cnt[i]):
ans *= 10
ans += i
return ans
Java
class Solution {
public long smallestNumber(long num) {
boolean neg = num < 0;
num = Math.abs(num);
int[] cnt = new int[10];
while (num > 0) {
++cnt[(int) (num % 10)];
num /= 10;
}
long ans = 0;
if (neg) {
for (int i = 9; i >= 0; --i) {
while (cnt[i] > 0) {
ans = ans * 10 + i;
--cnt[i];
}
}
return -ans;
}
if (cnt[0] > 0) {
for (int i = 1; i < 10; ++i) {
if (cnt[i] > 0) {
--cnt[i];
ans = i;
break;
}
}
}
for (int i = 0; i < 10; ++i) {
while (cnt[i] > 0) {
ans = ans * 10 + i;
--cnt[i];
}
}
return ans;
}
}
C++
class Solution {
public:
long long smallestNumber(long long num) {
bool neg = num < 0;
num = abs(num);
int cnt[10]{};
while (num > 0) {
++cnt[num % 10];
num /= 10;
}
long long ans = 0;
if (neg) {
for (int i = 9; i >= 0; --i) {
while (cnt[i] > 0) {
ans = ans * 10 + i;
--cnt[i];
}
}
return -ans;
}
if (cnt[0]) {
for (int i = 1; i < 10; ++i) {
if (cnt[i] > 0) {
--cnt[i];
ans = i;
break;
}
}
}
for (int i = 0; i < 10; ++i) {
while (cnt[i] > 0) {
ans = ans * 10 + i;
--cnt[i];
}
}
return ans;
}
};
Go
func smallestNumber(num int64) (ans int64) {
neg := num < 0
num = max(num, -num)
cnt := make([]int, 10)
for num > 0 {
cnt[num%10]++
num /= 10
}
if neg {
for i := 9; i >= 0; i-- {
for cnt[i] > 0 {
ans = ans*10 + int64(i)
cnt[i]--
}
}
return -ans
}
if cnt[0] > 0 {
for i := 1; i < 10; i++ {
if cnt[i] > 0 {
cnt[i]--
ans = int64(i)
break
}
}
}
for i := 0; i < 10; i++ {
for cnt[i] > 0 {
ans = ans*10 + int64(i)
cnt[i]--
}
}
return ans
}
TypeScript
function smallestNumber(num: number): number {
const neg = num < 0;
num = Math.abs(num);
const cnt = Array(10).fill(0);
while (num > 0) {
cnt[num % 10]++;
num = Math.floor(num / 10);
}
let ans = 0;
if (neg) {
for (let i = 9; i >= 0; i--) {
while (cnt[i] > 0) {
ans = ans * 10 + i;
cnt[i]--;
}
}
return -ans;
}
if (cnt[0] > 0) {
for (let i = 1; i < 10; i++) {
if (cnt[i] > 0) {
cnt[i]--;
ans = i;
break;
}
}
}
for (let i = 0; i < 10; i++) {
while (cnt[i] > 0) {
ans = ans * 10 + i;
cnt[i]--;
}
}
return ans;
}
Rust
impl Solution {
pub fn smallest_number(num: i64) -> i64 {
let mut neg = num < 0;
let mut num = num.abs();
let mut cnt = vec![0; 10];
while num > 0 {
cnt[(num % 10) as usize] += 1;
num /= 10;
}
let mut ans = 0;
if neg {
for i in (0..10).rev() {
while cnt[i] > 0 {
ans = ans * 10 + i as i64;
cnt[i] -= 1;
}
}
return -ans;
}
if cnt[0] > 0 {
for i in 1..10 {
if cnt[i] > 0 {
cnt[i] -= 1;
ans = i as i64;
break;
}
}
}
for i in 0..10 {
while cnt[i] > 0 {
ans = ans * 10 + i as i64;
cnt[i] -= 1;
}
}
ans
}
}
JavaScript
/**
* @param {number} num
* @return {number}
*/
var smallestNumber = function (num) {
const neg = num < 0;
num = Math.abs(num);
const cnt = Array(10).fill(0);
while (num > 0) {
cnt[num % 10]++;
num = Math.floor(num / 10);
}
let ans = 0;
if (neg) {
for (let i = 9; i >= 0; i--) {
while (cnt[i] > 0) {
ans = ans * 10 + i;
cnt[i]--;
}
}
return -ans;
}
if (cnt[0] > 0) {
for (let i = 1; i < 10; i++) {
if (cnt[i] > 0) {
cnt[i]--;
ans = i;
break;
}
}
}
for (let i = 0; i < 10; i++) {
while (cnt[i] > 0) {
ans = ans * 10 + i;
cnt[i]--;
}
}
return ans;
};