LeetCode #318 — MEDIUM

Maximum Product of Word Lengths

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

The Problem

Problem Statement

Given a string array words, return the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. If no such two words exist, return 0.

Example 1:

Input: words = ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".

Example 2:

Input: words = ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4
Explanation: The two words can be "ab", "cd".

Example 3:

Input: words = ["a","aa","aaa","aaaa"]
Output: 0
Explanation: No such pair of words.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: Given a string array words, return the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. If no such two words exist, return 0.

Baseline thinking

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

Pattern signal: Array · Bit Manipulation

Example 1

["abcw","baz","foo","bar","xtfn","abcdef"]

Example 2

["a","ab","abc","d","cd","bcd","abcd"]

Example 3

["a","aa","aaa","aaaa"]

Step 02

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 #318: Maximum Product of Word Lengths
class Solution {
    public int maxProduct(String[] words) {
        int n = words.length;
        int[] mask = new int[n];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (char c : words[i].toCharArray()) {
                mask[i] |= 1 << (c - 'a');
            }
            for (int j = 0; j < i; ++j) {
                if ((mask[i] & mask[j]) == 0) {
                    ans = Math.max(ans, words[i].length() * words[j].length());
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #318: Maximum Product of Word Lengths
func maxProduct(words []string) (ans int) {
	n := len(words)
	mask := make([]int, n)
	for i, s := range words {
		for _, c := range s {
			mask[i] |= 1 << (c - 'a')
		}
		for j, t := range words[:i] {
			if mask[i]&mask[j] == 0 {
				ans = max(ans, len(s)*len(t))
			}
		}
	}
	return
}

# Accepted solution for LeetCode #318: Maximum Product of Word Lengths
class Solution:
    def maxProduct(self, words: List[str]) -> int:
        mask = [0] * len(words)
        ans = 0
        for i, s in enumerate(words):
            for c in s:
                mask[i] |= 1 << (ord(c) - ord("a"))
            for j, t in enumerate(words[:i]):
                if (mask[i] & mask[j]) == 0:
                    ans = max(ans, len(s) * len(t))
        return ans

// Accepted solution for LeetCode #318: Maximum Product of Word Lengths
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn max_product(words: Vec<String>) -> i32 {
        let mut hm: HashMap<u32, usize> = HashMap::new();
        for word in words {
            let mut mask: u32 = 0;
            for c in word.bytes() {
                mask |= 1 << (c - b'a');
            }
            let size = hm.entry(mask).or_default();
            *size = word.len().max(*size);
        }
        let mut res = 0;
        for (&ka, &va) in &hm {
            for (&kb, &vb) in &hm {
                if ka & kb == 0 {
                    res = res.max(va * vb);
                }
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let words = vec_string!["abcw", "baz", "foo", "bar", "xtfn", "abcdef"];
    let res = 16;
    assert_eq!(Solution::max_product(words), res);
    let words = vec_string!["a", "ab", "abc", "d", "cd", "bcd", "abcd"];
    let res = 4;
    assert_eq!(Solution::max_product(words), res);
    let words = vec_string!["a", "aa", "aaa", "aaaa"];
    let res = 0;
    assert_eq!(Solution::max_product(words), res);
}

// Accepted solution for LeetCode #318: Maximum Product of Word Lengths
function maxProduct(words: string[]): number {
    const n = words.length;
    const mask: number[] = Array(n).fill(0);
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        for (const c of words[i]) {
            mask[i] |= 1 << (c.charCodeAt(0) - 'a'.charCodeAt(0));
        }
        for (let j = 0; j < i; ++j) {
            if ((mask[i] & mask[j]) === 0) {
                ans = Math.max(ans, words[i].length * words[j].length);
            }
        }
    }
    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^2 + L)

Space

O(n)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

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.