The Next Palindrome Using Same Digits problem gives you a numeric string of even length and asks for the next palindrome that is strictly greater and uses the same multiset of digits. This hard coding problem applies the Next Permutation concept to only the first half of the palindrome, since the second half mirrors the first.
Amazon asks this to test the combination of palindrome structure understanding with the Next Permutation algorithm. The key insight — that a palindrome is fully determined by its first half — reduces the problem to finding the next permutation of the first half and mirroring it. The two pointers and string interview pattern is applied with palindrome constraints.
Extract first half + apply next permutation + mirror. For a string of length 2n, extract the first n characters. Apply the Next Permutation algorithm to this half. Mirror it to get the full palindrome (first half + reverse(first half)). This gives the next palindrome using the same digits. If no next permutation exists (first half is in descending order), return "" (no solution).
Palindrome: "1221". First half: "12". Apply next permutation to "12": pivot at index 0 (1 < 2), swap with rightmost > 1 which is index 1 (2) → "21", no suffix to reverse. Next half = "21". Mirror: "21" + reverse("21") = "2112". Answer: "2112".
Palindrome construction problems often reduce to operations on just one half. If a palindrome of length 2n has the property P, you only need to work with the first n characters — the second half follows automatically. Combining this insight with the Next Permutation algorithm is a powerful technique. Practice "next greater palindrome with same digits," "largest palindrome from digits," and similar half-based construction problems to build this instinct.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Last Substring in Lexicographical Order | Hard | Solve | |
| Make String a Subsequence Using Cyclic Increments | Medium | Solve | |
| Valid Palindrome IV | Medium | Solve | |
| Magical String | Medium | Solve | |
| Minimum Length of String After Deleting Similar Ends | Medium | Solve |