The "Smallest Palindromic Rearrangement I" interview question asks you to take a string and determine if its characters can be rearranged into a palindrome. If they can, you need to find the lexicographically smallest such palindrome. A palindrome reads the same forwards and backwards. This "Smallest Palindromic Rearrangement I coding problem" combines frequency counting with lexicographical optimization.
Microsoft and Amazon ask this to test your understanding of palindromic properties and string construction. It requires you to know that for a palindrome to exist, at most one character can have an odd frequency. It also evaluates your ability to build a string by placing the smallest available characters at the "edges" (start and end) and working your way inward.
This follows the "Sorting, Counting Sort, and String interview pattern".
left_half + middle + right_half.String: aabbcc
a:2, b:2, c:2. All even. OK.left = "abc".reverse("abc") = "cba".
Result: "abccba".
String: aabbca:2, b:2, c:1. One odd ('c'). OK.The most frequent mistake is not using alphabetical order when building the left half, which fails the "lexicographically smallest" requirement. Another error is incorrectly handling the odd-frequency character or failing to realize that a palindrome cannot be formed if multiple characters have odd frequencies. Some candidates also forget to handle the reverse part of the string construction efficiently.
Always use a frequency array or a sorted map when lexicographical order is required. This ensures that you process characters in the correct order (a, b, c...) without extra sorting steps. This technique is a staple for many "Smallest Palindromic Rearrangement I interview question" variants.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Make String Anti-palindrome | Hard | Solve | |
| Sort Vowels in a String | Medium | Solve | |
| Rearrange Words in a Sentence | Medium | Solve | |
| Sorting the Sentence | Easy | Solve | |
| Maximum Number of Potholes That Can Be Fixed | Medium | Solve |