953. Verifying an Alien Dictionary

Description

In an alien language, surprisingly, they also use English lowercase letters, but possibly in a different order. The order of the alphabet is some permutation of lowercase letters.

Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographically in this alien language.

 

Example 1:

Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
Output: true
Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.

Example 2:

Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
Output: false
Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.

Example 3:

Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false
Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character (More info).

 

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 20
• order.length == 26
• All characters in words[i] and order are English lowercase letters.

Solutions

Solution 1

Python3JavaC++GoTypeScriptRustC

class Solution:
    def isAlienSorted(self, words: List[str], order: str) -> bool:
        m = {c: i for i, c in enumerate(order)}
        for i in range(20):
            prev = -1
            valid = True
            for x in words:
                curr = -1 if i >= len(x) else m[x[i]]
                if prev > curr:
                    return False
                if prev == curr:
                    valid = False
                prev = curr
            if valid:
                return True
        return True

class Solution {
    public boolean isAlienSorted(String[] words, String order) {
        int[] m = new int[26];
        for (int i = 0; i < 26; ++i) {
            m[order.charAt(i) - 'a'] = i;
        }
        for (int i = 0; i < 20; ++i) {
            int prev = -1;
            boolean valid = true;
            for (String x : words) {
                int curr = i >= x.length() ? -1 : m[x.charAt(i) - 'a'];
                if (prev > curr) {
                    return false;
                }
                if (prev == curr) {
                    valid = false;
                }
                prev = curr;
            }
            if (valid) {
                break;
            }
        }
        return true;
    }
}

class Solution {
public:
    bool isAlienSorted(vector<string>& words, string order) {
        vector<int> m(26);
        for (int i = 0; i < 26; ++i) m[order[i] - 'a'] = i;
        for (int i = 0; i < 20; ++i) {
            int prev = -1;
            bool valid = true;
            for (auto& x : words) {
                int curr = i >= x.size() ? -1 : m[x[i] - 'a'];
                if (prev > curr) return false;
                if (prev == curr) valid = false;
                prev = curr;
            }
            if (valid) break;
        }
        return true;
    }
};

func isAlienSorted(words []string, order string) bool {
    m := make([]int, 26)
    for i, c := range order {
        m[c-'a'] = i
    }
    for i := 0; i < 20; i++ {
        prev := -1
        valid := true
        for _, x := range words {
            curr := -1
            if i < len(x) {
                curr = m[x[i]-'a']
            }
            if prev > curr {
                return false
            }
            if prev == curr {
                valid = false
            }
            prev = curr
        }
        if valid {
            break
        }
    }
    return true
}

function isAlienSorted(words: string[], order: string): boolean {
    const map = new Map();
    for (const c of order) {
        map.set(c, map.size);
    }
    const n = words.length;
    for (let i = 1; i < n; i++) {
        const s1 = words[i - 1];
        const s2 = words[i];
        const m = Math.min(s1.length, s2.length);
        let isEqual = false;
        for (let j = 0; j < m; j++) {
            if (map.get(s1[j]) > map.get(s2[j])) {
                return false;
            }
            if (map.get(s1[j]) < map.get(s2[j])) {
                isEqual = true;
                break;
            }
        }
        if (!isEqual && s1.length > s2.length) {
            return false;
        }
    }
    return true;
}

use std::collections::HashMap;
impl Solution {
    pub fn is_alien_sorted(words: Vec<String>, order: String) -> bool {
        let n = words.len();
        let mut map = HashMap::new();
        order.as_bytes().iter().enumerate().for_each(|(i, &v)| {
            map.insert(v, i);
        });
        for i in 1..n {
            let s1 = words[i - 1].as_bytes();
            let s2 = words[i].as_bytes();
            let mut is_equal = true;
            for i in 0..s1.len().min(s2.len()) {
                if map.get(&s1[i]) > map.get(&s2[i]) {
                    return false;
                }
                if map.get(&s1[i]) < map.get(&s2[i]) {
                    is_equal = false;
                    break;
                }
            }
            if is_equal && s1.len() > s2.len() {
                return false;
            }
        }
        true
    }
}

#define min(a, b) (((a) < (b)) ? (a) : (b))

bool isAlienSorted(char** words, int wordsSize, char* order) {
    int map[26] = {0};
    for (int i = 0; i < 26; i++) {
        map[order[i] - 'a'] = i;
    }
    for (int i = 1; i < wordsSize; i++) {
        char* s1 = words[i - 1];
        char* s2 = words[i];
        int n = strlen(s1);
        int m = strlen(s2);
        int len = min(n, m);
        int isEqual = 1;
        for (int j = 0; j < len; j++) {
            if (map[s1[j] - 'a'] > map[s2[j] - 'a']) {
                return 0;
            }
            if (map[s1[j] - 'a'] < map[s2[j] - 'a']) {
                isEqual = 0;
                break;
            }
        }
        if (isEqual && n > m) {
            return 0;
        }
    }
    return 1;
}

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