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Problem solution · Java

Next Palindrome Using Same Digits

Next Palindrome Using Same Digits: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
58 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Next Palindrome Using Same Digits, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 58 lines of Java from the credited upstream file 1842.java.
  • The implementation visibly relies on sequence storage.
  • 5 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeNext Palindrome Using Same Digits · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public String nextPalindrome(String num) {    final int n = num.length();    int[] arr = new int[n / 2];     for (int i = 0; i < arr.length; ++i)      arr[i] = num.charAt(i) - '0';     if (!nextPermutation(arr))      return "";     StringBuilder sb = new StringBuilder();     for (final int a : arr)      sb.append(a);     if (n % 2 == 1)      return sb.toString() + num.charAt(n / 2) + sb.reverse().toString();    return sb.toString() + sb.reverse().toString();  }   private boolean nextPermutation(int[] nums) {    final int n = nums.length;     // From the back to the front, find the first num < nums[i + 1].    int i;    for (i = n - 2; i >= 0; --i)      if (nums[i] < nums[i + 1])        break;     if (i < 0)      return false;     // From the back to the front, find the first num > nums[i] and swap it with    // nums[i].    for (int j = n - 1; j > i; --j)      if (nums[j] > nums[i]) {        swap(nums, i, j);        break;      }     // Reverse nums[i + 1..n - 1].    reverse(nums, i + 1, n - 1);    return true;  }   private void reverse(int[] nums, int l, int r) {    while (l < r)      swap(nums, l++, r--);  }   private void swap(int[] nums, int i, int j) {    final int temp = nums[i];    nums[i] = nums[j];    nums[j] = temp;  }}

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