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2788. Split Strings by Separator
Description
Given an array of strings words and a character separator, split each string in words by separator.
Return an array of strings containing the new strings formed after the splits, excluding empty strings.
Notes
separatoris used to determine where the split should occur, but it is not included as part of the resulting strings.- A split may result in more than two strings.
- The resulting strings must maintain the same order as they were initially given.
Example 1:
Input: words = ["one.two.three","four.five","six"], separator = "." Output: ["one","two","three","four","five","six"] Explanation: In this example we split as follows: "one.two.three" splits into "one", "two", "three" "four.five" splits into "four", "five" "six" splits into "six" Hence, the resulting array is ["one","two","three","four","five","six"].
Example 2:
Input: words = ["$easy$","$problem$"], separator = "$" Output: ["easy","problem"] Explanation: In this example we split as follows: "$easy$" splits into "easy" (excluding empty strings) "$problem$" splits into "problem" (excluding empty strings) Hence, the resulting array is ["easy","problem"].
Example 3:
Input: words = ["|||"], separator = "|" Output: [] Explanation: In this example the resulting split of "|||" will contain only empty strings, so we return an empty array [].
Constraints:
1 <= words.length <= 1001 <= words[i].length <= 20- characters in
words[i]are either lowercase English letters or characters from the string".,|$#@"(excluding the quotes) separatoris a character from the string".,|$#@"(excluding the quotes)
Solutions
Solution 1: Simulation
We traverse the string array $words$. For each string $w$, we use separator as the delimiter to split it. If the split string is not empty, we add it to the answer array.
The time complexity is $O(n \times m)$, and the space complexity is $O(m)$, where $n$ is the length of the string array $words$, and $m$ is the maximum length of the strings in the array $words$.
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import java.util.regex.Pattern; class Solution { public List<String> splitWordsBySeparator(List<String> words, char separator) { List<String> ans = new ArrayList<>(); for (var w : words) { for (var s : w.split(Pattern.quote(String.valueOf(separator)))) { if (s.length() > 0) { ans.add(s); } } } return ans; } } -
class Solution { public: vector<string> splitWordsBySeparator(vector<string>& words, char separator) { vector<string> ans; for (const auto& w : words) { istringstream ss(w); string s; while (getline(ss, s, separator)) { if (!s.empty()) { ans.push_back(s); } } } return ans; } }; -
class Solution: def splitWordsBySeparator(self, words: List[str], separator: str) -> List[str]: return [s for w in words for s in w.split(separator) if s] -
func splitWordsBySeparator(words []string, separator byte) (ans []string) { for _, w := range words { for _, s := range strings.Split(w, string(separator)) { if s != "" { ans = append(ans, s) } } } return } -
function splitWordsBySeparator(words: string[], separator: string): string[] { return words.flatMap(w => w.split(separator).filter(s => s.length > 0)); }