# 1961. Check If String Is a Prefix of Array

## 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 $s$.

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• 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;
}