LeetCode #3035 — MEDIUM

Maximum Palindromes After Operations

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 0-indexed string array words having length n and containing 0-indexed strings.

You are allowed to perform the following operation any number of times (including zero):

Choose integers i, j, x, and y such that 0 <= i, j < n, 0 <= x < words[i].length, 0 <= y < words[j].length, and swap the characters words[i][x] and words[j][y].

Return an integer denoting the maximum number of palindromes words can contain, after performing some operations.

Note: i and j may be equal during an operation.

Example 1:

Input: words = ["abbb","ba","aa"]
Output: 3
Explanation: In this example, one way to get the maximum number of palindromes is:
Choose i = 0, j = 1, x = 0, y = 0, so we swap words[0][0] and words[1][0]. words becomes ["bbbb","aa","aa"].
All strings in words are now palindromes.
Hence, the maximum number of palindromes achievable is 3.

Example 2:

Input: words = ["abc","ab"]
Output: 2
Explanation: In this example, one way to get the maximum number of palindromes is: 
Choose i = 0, j = 1, x = 1, y = 0, so we swap words[0][1] and words[1][0]. words becomes ["aac","bb"].
Choose i = 0, j = 0, x = 1, y = 2, so we swap words[0][1] and words[0][2]. words becomes ["aca","bb"].
Both strings are now palindromes.
Hence, the maximum number of palindromes achievable is 2.

Example 3:

Input: words = ["cd","ef","a"]
Output: 1
Explanation: In this example, there is no need to perform any operation.
There is one palindrome in words "a".
It can be shown that it is not possible to get more than one palindrome after any number of operations.
Hence, the answer is 1.

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 100
words[i] consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string array words having length n and containing 0-indexed strings. You are allowed to perform the following operation any number of times (including zero): Choose integers i, j, x, and y such that 0 <= i, j < n, 0 <= x < words[i].length, 0 <= y < words[j].length, and swap the characters words[i][x] and words[j][y]. Return an integer denoting the maximum number of palindromes words can contain, after performing some operations. Note: i and j may be equal during an operation.

Baseline thinking

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

Pattern signal: Array · Hash Map · Greedy

Example 1

["abbb","ba","aa"]

Example 2

["abc","ab"]

Example 3

["cd","ef","a"]

Core Insight

What unlocks the optimal approach

We can redistribute all the letters freely among the words.
Calculate the frequency of each letter and total the number of matching letter pairs that can be formed from the letters, i.e., <code>total = sum(freq[ch] / 2)</code> for all <code>'a' <= ch <= 'z'</code>.
We can greedily try making palindromes from <code>words[i]</code> with the smallest length to <code>words[i]</code> with the longest length.
For the current index, <code>i</code>, we try to make <code>words[i]</code> a palindrome. We need <code>len(words[i]) / 2</code> matching character pairs, and the letter in the middle (if it exists) can be freely chosen afterward.
We can check if we have enough pairs for index <code>i</code>; if we do, we increase the number of palindromes we can make and decrease the number of pairs we have. Otherwise, we end the loop at this index.
The answer is the number of palindromes we were able to make in the end.

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 #3035: Maximum Palindromes After Operations
class Solution {
    public int maxPalindromesAfterOperations(String[] words) {
        int s = 0, mask = 0;
        for (var w : words) {
            s += w.length();
            for (var c : w.toCharArray()) {
                mask ^= 1 << (c - 'a');
            }
        }
        s -= Integer.bitCount(mask);
        Arrays.sort(words, (a, b) -> a.length() - b.length());
        int ans = 0;
        for (var w : words) {
            s -= w.length() / 2 * 2;
            if (s < 0) {
                break;
            }
            ++ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3035: Maximum Palindromes After Operations
func maxPalindromesAfterOperations(words []string) (ans int) {
	var s, mask int
	for _, w := range words {
		s += len(w)
		for _, c := range w {
			mask ^= 1 << (c - 'a')
		}
	}
	s -= bits.OnesCount(uint(mask))
	sort.Slice(words, func(i, j int) bool {
		return len(words[i]) < len(words[j])
	})
	for _, w := range words {
		s -= len(w) / 2 * 2
		if s < 0 {
			break
		}
		ans++
	}
	return
}

# Accepted solution for LeetCode #3035: Maximum Palindromes After Operations
class Solution:
    def maxPalindromesAfterOperations(self, words: List[str]) -> int:
        s = mask = 0
        for w in words:
            s += len(w)
            for c in w:
                mask ^= 1 << (ord(c) - ord("a"))
        s -= mask.bit_count()
        words.sort(key=len)
        ans = 0
        for w in words:
            s -= len(w) // 2 * 2
            if s < 0:
                break
            ans += 1
        return ans

// Accepted solution for LeetCode #3035: Maximum Palindromes After Operations
// 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 #3035: Maximum Palindromes After Operations
// class Solution {
//     public int maxPalindromesAfterOperations(String[] words) {
//         int s = 0, mask = 0;
//         for (var w : words) {
//             s += w.length();
//             for (var c : w.toCharArray()) {
//                 mask ^= 1 << (c - 'a');
//             }
//         }
//         s -= Integer.bitCount(mask);
//         Arrays.sort(words, (a, b) -> a.length() - b.length());
//         int ans = 0;
//         for (var w : words) {
//             s -= w.length() / 2 * 2;
//             if (s < 0) {
//                 break;
//             }
//             ++ans;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3035: Maximum Palindromes After Operations
function maxPalindromesAfterOperations(words: string[]): number {
    let s: number = 0;
    let mask: number = 0;
    for (const w of words) {
        s += w.length;
        for (const c of w) {
            mask ^= 1 << (c.charCodeAt(0) - 'a'.charCodeAt(0));
        }
    }
    s -= (mask.toString(2).match(/1/g) || []).length;
    words.sort((a, b) => a.length - b.length);
    let ans: number = 0;
    for (const w of words) {
        s -= Math.floor(w.length / 2) * 2;
        if (s < 0) {
            break;
        }
        ans++;
    }
    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 log n)

Space

O(1)

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.

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.

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.