LeetCode #409 — EASY

Longest Palindrome

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

The Problem

Problem Statement

Given a string s which consists of lowercase or uppercase letters, return the length of the longest palindrome that can be built with those letters.

Letters are case sensitive, for example, "Aa" is not considered a palindrome.

Example 1:

Input: s = "abccccdd"
Output: 7
Explanation: One longest palindrome that can be built is "dccaccd", whose length is 7.

Example 2:

Input: s = "a"
Output: 1
Explanation: The longest palindrome that can be built is "a", whose length is 1.

Constraints:

1 <= s.length <= 2000
s consists of lowercase and/or uppercase English letters only.

Step 01

Brute Force Baseline

Problem summary: Given a string s which consists of lowercase or uppercase letters, return the length of the longest palindrome that can be built with those letters. Letters are case sensitive, for example, "Aa" is not considered a palindrome.

Baseline thinking

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

Pattern signal: Hash Map · Greedy

Example 1

"abccccdd"

Example 2

"a"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #409: Longest Palindrome
class Solution {
    public int longestPalindrome(String s) {
        int[] cnt = new int[128];
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            ++cnt[s.charAt(i)];
        }
        int ans = 0;
        for (int v : cnt) {
            ans += v / 2 * 2;
        }
        ans += ans < n ? 1 : 0;
        return ans;
    }
}

// Accepted solution for LeetCode #409: Longest Palindrome
func longestPalindrome(s string) (ans int) {
	cnt := [128]int{}
	for _, c := range s {
		cnt[c]++
	}
	for _, v := range cnt {
		ans += v / 2 * 2
	}
	if ans < len(s) {
		ans++
	}
	return
}

# Accepted solution for LeetCode #409: Longest Palindrome
class Solution:
    def longestPalindrome(self, s: str) -> int:
        cnt = Counter(s)
        ans = sum(v // 2 * 2 for v in cnt.values())
        ans += int(ans < len(s))
        return ans

// Accepted solution for LeetCode #409: Longest Palindrome
use std::collections::HashMap;

impl Solution {
    pub fn longest_palindrome(s: String) -> i32 {
        let mut cnt = HashMap::new();
        for ch in s.chars() {
            *cnt.entry(ch).or_insert(0) += 1;
        }

        let mut ans = 0;
        for &v in cnt.values() {
            ans += (v / 2) * 2;
        }

        if ans < (s.len() as i32) {
            ans += 1;
        }

        ans
    }
}

// Accepted solution for LeetCode #409: Longest Palindrome
function longestPalindrome(s: string): number {
    const cnt: Record<string, number> = {};
    for (const c of s) {
        cnt[c] = (cnt[c] || 0) + 1;
    }
    let ans = Object.values(cnt).reduce((acc, v) => acc + Math.floor(v / 2) * 2, 0);
    ans += ans < s.length ? 1 : 0;
    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 + |\Sigma|)

Space

O(|\Sigma|)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.