1 <= s.length <= 10⁵
s[i] is an English letter (uppercase or lowercase), digit, or space ' '.
There is at least one word in s.
s does not contain leading or trailing spaces.
All the words in s are guaranteed to be separated by a single space.

23:T41c,# Intuition\n\n\n# Approach\n\n\n# Complexity\n- Time complexity:\n\n\n- Space complexity:\n\n\n# Code\n```\nclass Solution {\n public void reverseWords(char[] s) {\n ArrayList l=new ArrayList<>();\n String h="";\n for(char r:s){\n if(r==\' \'){\n l.add(h);\n l.add(" ");\n h="";\n }\n else{\n h+=r+"";\n }\n }\n l.add(h);\n String k="";\n for(String h1:l){\n k+=h1;\n }\n String a[]=k.split(" ");\n String g="";\n for(int i=a.length-1;i>=0;i--){\n g+=a[i]+" ";\n }\n g=g.trim();\n int y=0;\n for(char t:g.toCharArray()){\n s[y++]=t;\n }\n \n }\n}\n```24:T54c,```C++ []\nclass Solution {\npublic:\n void reverseWords(vector& s) {\n\n int i = 0, j = 0;\n while (i < s.size())\n {\n j = i;\n while (j < s.size() && s[j] != \' \')\n {\n j++;\n }\n\n std::reverse(s.begin() + i, s.begin() + j);\n\n i = j+1;\n }\n std::reverse(s.begin(), s.end());\n }\n};\n```\n\n```Python3 []\nclass Solution:\n def reverseWords(self, s: List[str]) -> None:\n s[:] = \' \'.join(reversed(\'\'.join(s).split

Reverse Words in a String II

Problem Description

Problem Description

Answer Panel