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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 $s$.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
  }
  
  

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