LeetCode #3517 — MEDIUM

Smallest Palindromic Rearrangement I

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a palindromic string s.

Return the lexicographically smallest palindromic permutation of s.

Example 1:

Input: s = "z"

Output: "z"

Explanation:

A string of only one character is already the lexicographically smallest palindrome.

Example 2:

Input: s = "babab"

Output: "abbba"

Explanation:

Rearranging "babab" → "abbba" gives the smallest lexicographic palindrome.

Example 3:

Input: s = "daccad"

Output: "acddca"

Explanation:

Rearranging "daccad" → "acddca" gives the smallest lexicographic palindrome.

Constraints:

1 <= s.length <= 10⁵
s consists of lowercase English letters.
s is guaranteed to be palindromic.

Step 01

Brute Force Baseline

Problem summary: You are given a palindromic string s. Return the lexicographically smallest palindromic permutation of s.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"z"

Example 2

"babab"

Example 3

"daccad"

Core Insight

What unlocks the optimal approach

Consider a palindrome as composed of two mirror-image halves.
Construct one half (using <code>s</code>), and then the other half is its reverse to obtain the lexicographically smallest permutation.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3517: Smallest Palindromic Rearrangement I
class Solution {
    public String smallestPalindrome(String s) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            cnt[c - 'a']++;
        }

        StringBuilder t = new StringBuilder();
        String ch = "";

        for (char c = 'a'; c <= 'z'; c++) {
            int idx = c - 'a';
            int v = cnt[idx] / 2;
            if (v > 0) {
                t.append(String.valueOf(c).repeat(v));
            }
            cnt[idx] -= v * 2;
            if (cnt[idx] == 1) {
                ch = String.valueOf(c);
            }
        }

        String ans = t.toString();
        ans = ans + ch + new StringBuilder(ans).reverse();
        return ans;
    }
}

// Accepted solution for LeetCode #3517: Smallest Palindromic Rearrangement I
func smallestPalindrome(s string) string {
	cnt := make([]int, 26)
	for i := 0; i < len(s); i++ {
		cnt[s[i]-'a']++
	}

	t := make([]byte, 0, len(s)/2)
	var ch byte
	for c := byte('a'); c <= 'z'; c++ {
		v := cnt[c-'a'] / 2
		for i := 0; i < v; i++ {
			t = append(t, c)
		}
		cnt[c-'a'] -= v * 2
		if cnt[c-'a'] == 1 {
			ch = c
		}
	}

	totalLen := len(t) * 2
	if ch != 0 {
		totalLen++
	}
	var sb strings.Builder
	sb.Grow(totalLen)

	sb.Write(t)
	if ch != 0 {
		sb.WriteByte(ch)
	}
	for i := len(t) - 1; i >= 0; i-- {
		sb.WriteByte(t[i])
	}
	return sb.String()
}

# Accepted solution for LeetCode #3517: Smallest Palindromic Rearrangement I
class Solution:
    def smallestPalindrome(self, s: str) -> str:
        cnt = Counter(s)
        t = []
        ch = ""
        for c in ascii_lowercase:
            v = cnt[c] // 2
            t.append(c * v)
            cnt[c] -= v * 2
            if cnt[c] == 1:
                ch = c
        ans = "".join(t)
        ans = ans + ch + ans[::-1]
        return ans

// Accepted solution for LeetCode #3517: Smallest Palindromic Rearrangement I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3517: Smallest Palindromic Rearrangement I
// class Solution {
//     public String smallestPalindrome(String s) {
//         int[] cnt = new int[26];
//         for (char c : s.toCharArray()) {
//             cnt[c - 'a']++;
//         }
// 
//         StringBuilder t = new StringBuilder();
//         String ch = "";
// 
//         for (char c = 'a'; c <= 'z'; c++) {
//             int idx = c - 'a';
//             int v = cnt[idx] / 2;
//             if (v > 0) {
//                 t.append(String.valueOf(c).repeat(v));
//             }
//             cnt[idx] -= v * 2;
//             if (cnt[idx] == 1) {
//                 ch = String.valueOf(c);
//             }
//         }
// 
//         String ans = t.toString();
//         ans = ans + ch + new StringBuilder(ans).reverse();
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3517: Smallest Palindromic Rearrangement I
function smallestPalindrome(s: string): string {
    const ascii_lowercase = 'abcdefghijklmnopqrstuvwxyz';
    const cnt = new Array<number>(26).fill(0);
    for (const chKey of s) {
        cnt[chKey.charCodeAt(0) - 97]++;
    }

    const t: string[] = [];
    let ch = '';
    for (let i = 0; i < 26; i++) {
        const c = ascii_lowercase[i];
        const v = Math.floor(cnt[i] / 2);
        t.push(c.repeat(v));
        cnt[i] -= v * 2;
        if (cnt[i] === 1) {
            ch = c;
        }
    }

    let ans = t.join('');
    ans = ans + ch + ans.split('').reverse().join('');
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Smallest Palindromic Rearrangement I

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries