You are given a string
text of words that are placed among some number of spaces. Each word consists of one or more lowercase English letters and are separated by at least one space. It’s guaranteed that
text contains at least one word.
Rearrange the spaces so that there is an equal number of spaces between every pair of adjacent words and that number is maximized. If you cannot redistribute all the spaces equally, place the extra spaces at the end, meaning the returned string should be the same length as
Return the string after rearranging the spaces.
Input: text = " this is a sentence " Output: "this is a sentence" Explanation: There are a total of 9 spaces and 4 words. We can evenly divide the 9 spaces between the words: 9 / (4-1) = 3 spaces.
Input: text = " practice makes perfect" Output: "practice makes perfect " Explanation: There are a total of 7 spaces and 3 words. 7 / (3-1) = 3 spaces plus 1 extra space. We place this extra space at the end of the string.
Input: text = "hello world" Output: "hello world"
Input: text = " walks udp package into bar a" Output: "walks udp package into bar a "
Input: text = "a" Output: "a"
1 <= text.length <= 100
textconsists of lowercase English letters and
textcontains at least one word.
Splits the words by spaces and count how many spaces in the text. Base on the count of the space and the count of words, to add space equally between words, and if there is additional modulus space, add to the end.
class Solution: def reorderSpaces(self, text: str) -> str: words = text.split() spaces_count = 0 for c in text: if c == ' ': spaces_count += 1 if spaces_count == 0: return text if len(words) <= 1: return "".join(words) + ' ' * spaces_count space_width = spaces_count // (len(words) - 1) last_space_width = spaces_count % (len(words) - 1) result = "" for i, word in enumerate(words): if i != len(words) - 1: result += word + ' ' * space_width else: result += word result += ' ' * last_space_width return result
- Time Complexity: O(N).
- Space Complexity: O(N).