Maximum Product of Word Lengths – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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.
C++ Maximum Product of Word Lengths LeetCode Solution
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;
}
Java Maximum Product of Word Lengths LeetCode Solution
public static int maxProduct(String[] words) {
if (words == null || words.length == 0)
return 0;
int len = words.length;
int[] value = new int[len];
for (int i = 0; i < len; i++) {
String tmp = words[i];
value[i] = 0;
for (int j = 0; j < tmp.length(); j++) {
value[i] |= 1 << (tmp.charAt(j) - 'a');
}
}
int maxProduct = 0;
for (int i = 0; i < len; i++)
for (int j = i + 1; j < len; j++) {
if ((value[i] & value[j]) == 0 && (words[i].length() * words[j].length() > maxProduct))
maxProduct = words[i].length() * words[j].length();
}
return maxProduct;
}
Python 3 Maximum Product of Word Lengths LeetCode Solution
class Solution(object):
def maxProduct(self, words):
d = {}
for w in words:
mask = 0
for c in set(w):
mask |= (1 << (ord(c) - 97))
d[mask] = max(d.get(mask, 0), len(w))
return max([d[x] * d[y] for x in d for y in d if not x & y] or [0])
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