LeetCode Maximum Product of Word Lengths

Description:

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:

Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".

Example 2:

Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".

Example 3:

Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.

Solution 1:WA

一开始的方法想错了，以为可以用并查集做，把凡是有相同char的word放到同一组中，然后在不同的组之间做乘法。

但是后来发现这个想法有bug，比如ab，bc，cd是同一组，但是ab和cd也可以做乘法。

因为并查集好久没写了，所以还是贴上来回忆一下。

<span style="font-size:18px;">import java.util.Arrays;public class Solution {public int maxProduct(String[] words) {int n = words.length;DisjointSet ds = new DisjointSet(n + 26);for (int i = 0; i < n; i++) {String str = words[i];for (int j = 0; j < str.length(); j++) {int ch = str.charAt(j) - 'a' + n;ds.union(ch, i);}}int arr[] = new int[26];for (int i = 0; i < n; i++) {int ch = ds.find(i) - n;System.out.printf("word %d belongs to %d \n", i, ch);arr[ch] = Math.max(arr[ch], words[i].length());}int max = 1;for (int i = 0; i < 26; i++)for (int j = i + 1; j < 26; j++)max = Math.max(max, arr[i] * arr[j]);for (int i = 0; i < 26; i++) {System.out.println((char) (i + 'a') + " " + arr[i]);}return max;}class DisjointSet {int n;int father[];DisjointSet(int n) {this.n = n;father = new int[n];for (int i = 0; i < n; i++)father[i] = i;}public int find(int x) {if (x == father[x])return x;father[x] = find(father[x]);return father[x];}public void union(int a, int b) {int ra = find(a);int rb = find(b);if (ra == rb)return;father[rb] = ra;}}}</span>

Solution 2:

突然发现既然所有的char都是在'a'-'z'之间，那么如果我将每个word转成26位的long，再做乘法运算就可以。

import java.util.Arrays;public class Solution {public int maxProduct(String[] words) {int n = words.length;long[] words_binary = new long[n];for (int i = 0; i < n; i++) {String str = words[i];for (int j = 0; j < str.length(); j++)words_binary[i] |= (1l << (str.charAt(j) - 'a'));}int max = 0;for (int i = 0; i < n; i++)for (int j = i + 1; j < n; j++)if ((words_binary[i] & words_binary[j]) == 0)max = Math.max(max, words[i].length() * words[j].length());return max;}public static void main(String[] args) {Solution s = new Solution();String[] arr = { "abcw", "baz", "foo", "bar", "xtfn", "abcdef" };// System.out.println(Long.toBinaryString(1l << 25));System.out.println(s.maxProduct(arr));}}