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maxProduct.cpp
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maxProduct.cpp
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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.
class Solution {
public:
int maxProduct(vector<string>& words) {
vector<int> mask(words.size());
int result = 0;
for (int i=0; i<words.size(); ++i) {
for (char c : words[i])
mask[i] |= 1 << (c - 'a');
for (int j=0; j<i; ++j)
if (!(mask[i] & mask[j]))
result = max(result, int(words[i].size() * words[j].size()));
}
return result;
}
};
class Solution {
public:
int maxProduct(vector<string>& words) {
unordered_map<int,int> maxlen;
for (string word : words) {
int mask = 0;
for (char c : word)
mask |= 1 << (c - 'a');
maxlen[mask] = max(maxlen[mask], (int) word.size());
}
int result = 0;
for (auto a : maxlen)
for (auto b : maxlen)
if (!(a.first & b.first))
result = max(result, a.second * b.second);
return result;
}
};
class Solution {
public:
int maxProduct(vector<string>& words) {
unordered_map<int,int> maxlen;
int result = 0;
for (string word : words) {
int mask = 0;
for (char c : word)
mask |= 1 << (c - 'a');
maxlen[mask] = max(maxlen[mask], (int) word.size());
for (auto maskAndLen : maxlen)
if (!(mask & maskAndLen.first))
result = max(result, (int) word.size() * maskAndLen.second);
}
return result;
}
};