README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1900-1999/1961.Check%20If%20String%20Is%20a%20Prefix%20of%20Array/README_EN.md	1234	Weekly Contest 253 Q1	Array Two Pointers String

1961. Check If String Is a Prefix of Array

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Description

Given a string s and an array of strings words, determine whether s is a prefix string of words.

A string s is a prefix string of words if s can be made by concatenating the first k strings in words for some positive k no larger than words.length.

Return true if s is a prefix string of words, or false otherwise.

 

Example 1:

Input: s = "iloveleetcode", words = ["i","love","leetcode","apples"]
Output: true
Explanation:
s can be made by concatenating "i", "love", and "leetcode" together.

Example 2:

Input: s = "iloveleetcode", words = ["apples","i","love","leetcode"]
Output: false
Explanation:
It is impossible to make s using a prefix of arr.

 

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 20
• 1 <= s.length <= 1000
• words[i] and s consist of only lowercase English letters.

Solutions

Solution 1: Traversal

We traverse the array $words$, using a variable $t$ to record the currently concatenated string. If the length of $t$ is greater than the length of $s$, it means that $s$ is not a prefix string of $words$, so we return $false$; if the length of $t$ is equal to the length of $s$, we return whether $t$ is equal to $s$.

At the end of the traversal, if the length of $t$ is less than the length of $s$, it means that $s$ is not a prefix string of $words$, so we return $false$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.

Python3

class Solution:
    def isPrefixString(self, s: str, words: List[str]) -> bool:
        n, m = len(s), 0
        for i, w in enumerate(words):
            m += len(w)
            if m == n:
                return "".join(words[: i + 1]) == s
        return False

Java

class Solution {
    public boolean isPrefixString(String s, String[] words) {
        StringBuilder t = new StringBuilder();
        for (var w : words) {
            t.append(w);
            if (t.length() > s.length()) {
                return false;
            }
            if (t.length() == s.length()) {
                return s.equals(t.toString());
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool isPrefixString(string s, vector<string>& words) {
        string t;
        for (auto& w : words) {
            t += w;
            if (t.size() > s.size()) {
                return false;
            }
            if (t.size() == s.size()) {
                return t == s;
            }
        }
        return false;
    }
};

Go

func isPrefixString(s string, words []string) bool {
	t := strings.Builder{}
	for _, w := range words {
		t.WriteString(w)
		if t.Len() > len(s) {
			return false
		}
		if t.Len() == len(s) {
			return t.String() == s
		}
	}
	return false
}

TypeScript

function isPrefixString(s: string, words: string[]): boolean {
    const t: string[] = [];
    const n = s.length;
    let m = 0;
    for (const w of words) {
        m += w.length;
        if (m > n) {
            return false;
        }
        t.push(w);
        if (m === n) {
            return s === t.join('');
        }
    }
    return false;
}