Leetcode•Mar 21, 2025
318. Maximum Product of Word Lengths
Hazrat Ali
Leetcode
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
/**
* Maximum Product of Word Lengths
*
* Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not
* share common letters. You may assume that each word will contain only lower case letters. If no such two words exist,
* return 0.
*
* Example 1:
*
* Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
* Output: 16
* Explanation: The two words can be "abcw", "xtfn".
*
* Example 2:
*
* Input: ["a","ab","abc","d","cd","bcd","abcd"]
* Output: 4
* Explanation: The two words can be "ab", "cd".
*
* Example 3:
*
* Input: ["a","aa","aaa","aaaa"]
* Output: 0
* Explanation: No such pair of words.
*/
Solution :
/**
* @param {string[]} words
* @return {number}
*/
const maxProduct = words => {
const n = words.length;
const values = Array(n).fill(0);
for (let i = 0; i < n; i++) {
for (let ch of words[i]) {
values[i] |= 1 << (ch.charCodeAt(0) - 'a'.charCodeAt(0));
}
}
let max = 0;
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
if ((values[i] & values[j]) === 0 && words[i].length * words[j].length > max) {
max = words[i].length * words[j].length;
}
}
}
return max;
};