# 1842. Next Palindrome Using Same Digits

## Description

You are given a numeric string num, representing a very large palindrome.

Return the smallest palindrome larger than num that can be created by rearranging its digits. If no such palindrome exists, return an empty string "".

A palindrome is a number that reads the same backward as forward.

Example 1:

Input: num = "1221"
Output: "2112"
Explanation: The next palindrome larger than "1221" is "2112".


Example 2:

Input: num = "32123"
Output: ""
Explanation: No palindromes larger than "32123" can be made by rearranging the digits.


Example 3:

Input: num = "45544554"
Output: "54455445"
Explanation: The next palindrome larger than "45544554" is "54455445".


Constraints:

• 1 <= num.length <= 105
• num is a palindrome.

## Solutions

Solution 1: Find the Next Permutation of the First Half

According to the problem description, we only need to find the next permutation of the first half of the string, then traverse the first half and symmetrically assign values to the second half.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string.

• class Solution {
public String nextPalindrome(String num) {
char[] nums = num.toCharArray();
if (!nextPermutation(nums)) {
return "";
}
int n = nums.length;
for (int i = 0; i < n / 2; ++i) {
nums[n - 1 - i] = nums[i];
}
return String.valueOf(nums);
}

private boolean nextPermutation(char[] nums) {
int n = nums.length / 2;
int i = n - 2;
while (i >= 0 && nums[i] >= nums[i + 1]) {
--i;
}
if (i < 0) {
return false;
}
int j = n - 1;
while (j >= 0 && nums[i] >= nums[j]) {
--j;
}
swap(nums, i++, j);
for (j = n - 1; i < j; ++i, --j) {
swap(nums, i, j);
}
return true;
}

private void swap(char[] nums, int i, int j) {
char t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}

• class Solution {
public:
string nextPalindrome(string num) {
int n = num.size();
string nums = num.substr(0, n / 2);
if (!next_permutation(begin(nums), end(nums))) {
return "";
}
for (int i = 0; i < n / 2; ++i) {
num[i] = nums[i];
num[n - i - 1] = nums[i];
}
return num;
}
};

• class Solution:
def nextPalindrome(self, num: str) -> str:
def next_permutation(nums: List[str]) -> bool:
n = len(nums) // 2
i = n - 2
while i >= 0 and nums[i] >= nums[i + 1]:
i -= 1
if i < 0:
return False
j = n - 1
while j >= 0 and nums[j] <= nums[i]:
j -= 1
nums[i], nums[j] = nums[j], nums[i]
nums[i + 1 : n] = nums[i + 1 : n][::-1]
return True

nums = list(num)
if not next_permutation(nums):
return ""
n = len(nums)
for i in range(n // 2):
nums[n - i - 1] = nums[i]
return "".join(nums)


• func nextPalindrome(num string) string {
nums := []byte(num)
n := len(nums)
if !nextPermutation(nums) {
return ""
}
for i := 0; i < n/2; i++ {
nums[n-1-i] = nums[i]
}
return string(nums)
}

func nextPermutation(nums []byte) bool {
n := len(nums) / 2
i := n - 2
for i >= 0 && nums[i] >= nums[i+1] {
i--
}
if i < 0 {
return false
}
j := n - 1
for j >= 0 && nums[j] <= nums[i] {
j--
}
nums[i], nums[j] = nums[j], nums[i]
for i, j = i+1, n-1; i < j; i, j = i+1, j-1 {
nums[i], nums[j] = nums[j], nums[i]
}
return true
}

• function nextPalindrome(num: string): string {
const nums = num.split('');
const n = nums.length;
if (!nextPermutation(nums)) {
return '';
}
for (let i = 0; i < n >> 1; ++i) {
nums[n - 1 - i] = nums[i];
}
return nums.join('');
}

function nextPermutation(nums: string[]): boolean {
const n = nums.length >> 1;
let i = n - 2;
while (i >= 0 && nums[i] >= nums[i + 1]) {
i--;
}
if (i < 0) {
return false;
}
let j = n - 1;
while (j >= 0 && nums[i] >= nums[j]) {
j--;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
for (i = i + 1, j = n - 1; i < j; ++i, --j) {
[nums[i], nums[j]] = [nums[j], nums[i]];
}
return true;
}