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Maximum Product of Word Lengths

描述

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.

分析

由于只有 26 个小写字母,所以,我们可以为数组中的每个word开辟一个长为 26 的布尔数组作为哈希表,然后用一个两重 for 循环,两两比较,如果不存在公共的字母,则计算二者的长度的乘积,取最大作为最终结果。时间复杂度O(26n^2),空间复杂度O(26n)

上面的方法可以进一步优化,即长度为 26 的布尔数组,小于 32 位,可以编码为一个整数,这样两个整数按位与,如果结果为 1,说明存在公共字母,如果结果为 0,说明不存在公共字母。时间复杂度O(n^2),空间复杂度O(n)

解法 1

// Maximum Product of Word Lengths
// Time Complexity: O(26n^2), Space Complexity: O(26n)
public class Solution {
public int maxProduct(String[] words) {
final int n = words.length;
final boolean[][] hashset = new boolean[n][ALPHABET_SIZE];

for (int i = 0; i < n; ++i) {
for (int j = 0; j < words[i].length(); ++j) {
hashset[i][words[i].charAt(j) - 'a'] = true;
}
}

int result = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
boolean hasCommon = false;
for (int k = 0; k < ALPHABET_SIZE; ++k) {
if (hashset[i][k] && hashset[j][k]) {
hasCommon = true;
break;
}
}
int tmp = words[i].length() * words[j].length();
if (!hasCommon && tmp > result) {
result = tmp;
}
}
}
return result;
}
private static final int ALPHABET_SIZE = 26;
}

解法 2

// Maximum Product of Word Lengths
// Time Complexity: O(n^2), Space Complexity: O(n)
public class Solution {
public int maxProduct(String[] words) {
final int n = words.length;
final int[] hashset = new int[n];

for (int i = 0; i < n; ++i) {
for (int j = 0; j < words[i].length(); ++j) {
hashset[i] |= 1 << (words[i].charAt(j) - 'a');
}
}

int result = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
int tmp = words[i].length() * words[j].length();
if ((hashset[i] & hashset[j]) == 0 && tmp > result) {
result = tmp;
}
}
}
return result;
}
}