318. 最大单词长度乘积

题目描述

给你一个字符串数组 words ，找出并返回 length(words[i]) * length(words[j]) 的最大值，并且这两个单词不含有公共字母。如果不存在这样的两个单词，返回 0 。

示例 1：

输入：words = ["abcw","baz","foo","bar","xtfn","abcdef"]
输出：16 
解释：这两个单词为 "abcw", "xtfn"。

示例 2：

输入：words = ["a","ab","abc","d","cd","bcd","abcd"]
输出：4 
解释：这两个单词为 "ab", "cd"。

示例 3：

输入：words = ["a","aa","aaa","aaaa"]
输出：0 
解释：不存在这样的两个单词。

提示：

2 <= words.length <= 1000
1 <= words[i].length <= 1000
words[i] 仅包含小写字母

解法

方法一：位运算

题目需要我们找到两个没有公共字母的字符串，使得它们的长度乘积最大。我们可以将每个字符串用一个二进制数 $mask[i]$ 表示，这个二进制数的每一位表示字符串中是否含有某个字母。如果两个字符串没有公共字母，那么这两个字符串对应的二进制数的按位与的结果为 $0$，即 $mask[i] \& mask[j] = 0$。

我们遍历每个字符串，对于当前遍历到的字符串 $words[i]$，我们先算出对应的二进制数 $mask[i]$，然后再遍历 $j \in [0, i)$ 的所有字符串 $words[j]$，检查 $mask[i] \& mask[j] = 0$ 是否成立，如果成立就更新答案为 $\max(ans, |words[i]| \times |words[j]|)$。

遍历结束后，返回答案即可。

时间复杂度 $O(n^2 + L)$，空间复杂度 $O(n)$。其中 $n$ 是字符串数组 $words$ 的长度，而 $L$ 是字符串数组所有字符串的长度之和。

Python3JavaC++GoTypeScript

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        mask = [0] * len(words)
        ans = 0
        for i, s in enumerate(words):
            for c in s:
                mask[i] |= 1 << (ord(c) - ord("a"))
            for j, t in enumerate(words[:i]):
                if (mask[i] & mask[j]) == 0:
                    ans = max(ans, len(s) * len(t))
        return ans

class Solution {
    public int maxProduct(String[] words) {
        int n = words.length;
        int[] mask = new int[n];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (char c : words[i].toCharArray()) {
                mask[i] |= 1 << (c - 'a');
            }
            for (int j = 0; j < i; ++j) {
                if ((mask[i] & mask[j]) == 0) {
                    ans = Math.max(ans, words[i].length() * words[j].length());
                }
            }
        }
        return ans;
    }
}

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n = words.size();
        int mask[n];
        memset(mask, 0, sizeof(mask));
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (char& c : words[i]) {
                mask[i] |= 1 << (c - 'a');
            }
            for (int j = 0; j < i; ++j) {
                if ((mask[i] & mask[j]) == 0) {
                    ans = max(ans, (int) (words[i].size() * words[j].size()));
                }
            }
        }
        return ans;
    }
};

func maxProduct(words []string) (ans int) {
    n := len(words)
    mask := make([]int, n)
    for i, s := range words {
        for _, c := range s {
            mask[i] |= 1 << (c - 'a')
        }
        for j, t := range words[:i] {
            if mask[i]&mask[j] == 0 {
                ans = max(ans, len(s)*len(t))
            }
        }
    }
    return
}

function maxProduct(words: string[]): number {
    const n = words.length;
    const mask: number[] = Array(n).fill(0);
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        for (const c of words[i]) {
            mask[i] |= 1 << (c.charCodeAt(0) - 'a'.charCodeAt(0));
        }
        for (let j = 0; j < i; ++j) {
            if ((mask[i] & mask[j]) === 0) {
                ans = Math.max(ans, words[i].length * words[j].length);
            }
        }
    }
    return ans;
}

方法二

Python3JavaC++GoTypeScript

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        mask = defaultdict(int)
        ans = 0
        for s in words:
            a = len(s)
            x = 0
            for c in s:
                x |= 1 << (ord(c) - ord("a"))
            for y, b in mask.items():
                if (x & y) == 0:
                    ans = max(ans, a * b)
            mask[x] = max(mask[x], a)
        return ans

class Solution {
    public int maxProduct(String[] words) {
        Map<Integer, Integer> mask = new HashMap<>();
        int ans = 0;
        for (var s : words) {
            int a = s.length();
            int x = 0;
            for (char c : s.toCharArray()) {
                x |= 1 << (c - 'a');
            }
            for (var e : mask.entrySet()) {
                int y = e.getKey(), b = e.getValue();
                if ((x & y) == 0) {
                    ans = Math.max(ans, a * b);
                }
            }
            mask.merge(x, a, Math::max);
        }
        return ans;
    }
}

class Solution {
public:
    int maxProduct(vector<string>& words) {
        unordered_map<int, int> mask;
        int ans = 0;
        for (auto& s : words) {
            int a = s.size();
            int x = 0;
            for (char& c : s) {
                x |= 1 << (c - 'a');
            }
            for (auto& [y, b] : mask) {
                if ((x & y) == 0) {
                    ans = max(ans, a * b);
                }
            }
            mask[x] = max(mask[x], a);
        }
        return ans;
    }
};

func maxProduct(words []string) (ans int) {
    mask := map[int]int{}
    for _, s := range words {
        a := len(s)
        x := 0
        for _, c := range s {
            x |= 1 << (c - 'a')
        }
        for y, b := range mask {
            if x&y == 0 {
                ans = max(ans, a*b)
            }
        }
        mask[x] = max(mask[x], a)
    }
    return
}

function maxProduct(words: string[]): number {
    const mask: Map<number, number> = new Map();
    let ans = 0;
    for (const s of words) {
        const a = s.length;
        let x = 0;
        for (const c of s) {
            x |= 1 << (c.charCodeAt(0) - 'a'.charCodeAt(0));
        }
        for (const [y, b] of mask.entries()) {
            if ((x & y) === 0) {
                ans = Math.max(ans, a * b);
            }
        }
        mask.set(x, Math.max(mask.get(x) || 0, a));
    }
    return ans;
}

318. 最大单词长度乘积

题目描述

解法

方法一：位运算

方法二

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