318. Maximum Product Of Word Lengths¶

Difficulty: Medium

318. Maximum Product of Word Lengths

Medium

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.

Solution¶

class Solution {
    public int maxProduct(String[] words) {
        int n = words.length;
        int freq[][] = new int[n][26];
        for (int i = 0; i < n; i++) {
            String s = words[i];
            for (int j = 0; j < s.length(); j++)
                freq[i][s.charAt(j) - 'a']++;
        }
        int maxi  = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                boolean isCommon = false;
                for (int k = 0; k < 26; k++) {
                    if (freq[i][k] > 0 && freq[j][k] > 0) {
                        isCommon = true;
                        break;
                    }
                }
                if (isCommon == false)
                    maxi = Math.max(maxi, words[i].length() * words[j].length());
            }
        }
        return maxi;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here