1023. Camelcase Matching

Description

Given an array of strings queries and a string pattern, return a boolean array answer where answer[i] is true if queries[i] matches pattern, and false otherwise.

A query word queries[i] matches pattern if you can insert lowercase English letters pattern so that it equals the query. You may insert each character at any position and you may not insert any characters.

Example 1:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
Output: [true,false,true,true,false]
Explanation: "FooBar" can be generated like this "F" + "oo" + "B" + "ar".
"FootBall" can be generated like this "F" + "oot" + "B" + "all".
"FrameBuffer" can be generated like this "F" + "rame" + "B" + "uffer".

Example 2:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
Output: [true,false,true,false,false]
Explanation: "FooBar" can be generated like this "Fo" + "o" + "Ba" + "r".
"FootBall" can be generated like this "Fo" + "ot" + "Ba" + "ll".

Example 3:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
Output: [false,true,false,false,false]
Explanation: "FooBarTest" can be generated like this "Fo" + "o" + "Ba" + "r" + "T" + "est".

Constraints:

1 <= pattern.length, queries.length <= 100
1 <= queries[i].length <= 100
queries[i] and pattern consist of English letters.

Solutions

Solution 1: Two Pointers

We can traverse every string in queries and check whether it matches pattern or not. If it matches, we add true to the answer array, otherwise we add false.

Next, we implement a function $check(s, t)$ to check whether the string $s$ matches the string $t$.

We can use two pointers $i$ and $j$ to traverse the two strings. If the characters pointed to by $i$ and $j$ are not the same and $s[i]$ is a lowercase letter, then we move the pointer $i$ to the next position.

If the pointer $i$ has reached the end of the string $s$ or the characters pointed to by $i$ and $j$ are not the same, we return false. Otherwise, we move both pointers $i$ and $j$ to the next position. When the pointer $j$ reaches the end of the string $t$, we need to check if the remaining characters in the string $s$ are all lowercase letters. If so, we return true, otherwise we return false.

Time complexity $(n \times m)$, where $n$ and $m$ are the length of the array queries and the string pattern respectively.

Python3JavaC++GoTypeScript

class Solution:
    def camelMatch(self, queries: List[str], pattern: str) -> List[bool]:
        def check(s, t):
            m, n = len(s), len(t)
            i = j = 0
            while j < n:
                while i < m and s[i] != t[j] and s[i].islower():
                    i += 1
                if i == m or s[i] != t[j]:
                    return False
                i, j = i + 1, j + 1
            while i < m and s[i].islower():
                i += 1
            return i == m

        return [check(q, pattern) for q in queries]

class Solution {
    public List<Boolean> camelMatch(String[] queries, String pattern) {
        List<Boolean> ans = new ArrayList<>();
        for (var q : queries) {
            ans.add(check(q, pattern));
        }
        return ans;
    }

    private boolean check(String s, String t) {
        int m = s.length(), n = t.length();
        int i = 0, j = 0;
        for (; j < n; ++i, ++j) {
            while (i < m && s.charAt(i) != t.charAt(j) && Character.isLowerCase(s.charAt(i))) {
                ++i;
            }
            if (i == m || s.charAt(i) != t.charAt(j)) {
                return false;
            }
        }
        while (i < m && Character.isLowerCase(s.charAt(i))) {
            ++i;
        }
        return i == m;
    }
}

class Solution {
public:
    vector<bool> camelMatch(vector<string>& queries, string pattern) {
        vector<bool> ans;
        auto check = [](string& s, string& t) {
            int m = s.size(), n = t.size();
            int i = 0, j = 0;
            for (; j < n; ++i, ++j) {
                while (i < m && s[i] != t[j] && islower(s[i])) {
                    ++i;
                }
                if (i == m || s[i] != t[j]) {
                    return false;
                }
            }
            while (i < m && islower(s[i])) {
                ++i;
            }
            return i == m;
        };
        for (auto& q : queries) {
            ans.push_back(check(q, pattern));
        }
        return ans;
    }
};

func camelMatch(queries []string, pattern string) (ans []bool) {
    check := func(s, t string) bool {
        m, n := len(s), len(t)
        i, j := 0, 0
        for ; j < n; i, j = i+1, j+1 {
            for i < m && s[i] != t[j] && (s[i] >= 'a' && s[i] <= 'z') {
                i++
            }
            if i == m || s[i] != t[j] {
                return false
            }
        }
        for i < m && s[i] >= 'a' && s[i] <= 'z' {
            i++
        }
        return i == m
    }
    for _, q := range queries {
        ans = append(ans, check(q, pattern))
    }
    return
}

function camelMatch(queries: string[], pattern: string): boolean[] {
    const check = (s: string, t: string) => {
        const m = s.length;
        const n = t.length;
        let i = 0;
        let j = 0;
        for (; j < n; ++i, ++j) {
            while (i < m && s[i] !== t[j] && s[i].codePointAt(0) >= 97) {
                ++i;
            }
            if (i === m || s[i] !== t[j]) {
                return false;
            }
        }
        while (i < m && s[i].codePointAt(0) >= 97) {
            ++i;
        }
        return i == m;
    };
    const ans: boolean[] = [];
    for (const q of queries) {
        ans.push(check(q, pattern));
    }
    return ans;
}

1023. Camelcase Matching

Description

Solutions

Solution 1: Two Pointers

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