Reverse Words in a String LeetCode Solution

Problem – Reverse Words in a String

Given an input string s, reverse the order of the words.

word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.

Return a string of the words in reverse order concatenated by a single space.

Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.

Example 1:

Input: s = "the sky is blue"
Output: "blue is sky the"

Example 2:

Input: s = "  hello world  "
Output: "world hello"
Explanation: Your reversed string should not contain leading or trailing spaces.

Example 3:

Input: s = "a good   example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.


  • 1 <= s.length <= 104
  • s contains English letters (upper-case and lower-case), digits, and spaces ' '.
  • There is at least one word in s.

Follow-up: If the string data type is mutable in your language, can you solve it in-place with O(1) extra space?

Reverse Words in a String LeetCode Solution in Java

public class Solution {
  public String reverseWords(String s) {
    if (s == null) return null;
    char[] a = s.toCharArray();
    int n = a.length;
    // step 1. reverse the whole string
    reverse(a, 0, n - 1);
    // step 2. reverse each word
    reverseWords(a, n);
    // step 3. clean up spaces
    return cleanSpaces(a, n);
  void reverseWords(char[] a, int n) {
    int i = 0, j = 0;
    while (i < n) {
      while (i < j || i < n && a[i] == ' ') i++; // skip spaces
      while (j < i || j < n && a[j] != ' ') j++; // skip non spaces
      reverse(a, i, j - 1);                      // reverse the word
  // trim leading, trailing and multiple spaces
  String cleanSpaces(char[] a, int n) {
    int i = 0, j = 0;
    while (j < n) {
      while (j < n && a[j] == ' ') j++;             // skip spaces
      while (j < n && a[j] != ' ') a[i++] = a[j++]; // keep non spaces
      while (j < n && a[j] == ' ') j++;             // skip spaces
      if (j < n) a[i++] = ' ';                      // keep only one space
    return new String(a).substring(0, i);
  // reverse a[] from a[i] to a[j]
  private void reverse(char[] a, int i, int j) {
    while (i < j) {
      char t = a[i];
      a[i++] = a[j];
      a[j--] = t;

Reverse Words in a String LeetCode Solution in C++

class Solution {

    // function to reverse any part of string from i to j (just one word or entire string)
    void reverseword(string &s, int i, int j){
          char t=s[i];
    void reverseWords(string &s) {
        int i=0, j=0;
        int l=0;
        int len=s.length();
        int wordcount=0;
            while(i<len && s[i] == ' ') i++;  // skip spaces in front of the word
            if(i==len) break;
            if(wordcount) s[j++]=' ';
            while(i<len && s[i] != ' ') {s[j]=s[i]; j++; i++;} 
            reverseword(s,l,j-1);                // reverse word in place
        s.resize(j);                           // resize result string
        reverseword(s,0,j-1);                  // reverse whole string

Reverse Words in a String LeetCode Solution in Python

class Solution(object):
    def reverseWords(self, s):
        arr = list(s)
        self.reverse_string(arr, 0, len(arr)-1)
        word = self.trim_sides(arr)
        res = self.trim_space(word)
        return ''.join(res)

    def reverse_string(self, arr, l, r):
        '''reverse a given string'''
        while l < r:
            arr[l], arr[r] = arr[r], arr[l]
            l += 1 ; r -= 1
        return arr
    def reverse_word(self, arr):
        '''reverse every words in a string'''
        l, r = 0 , 0
        while r < len(arr):
            while r < len(arr) and not arr[r].isspace(): r += 1
            self.reverse_string(arr, l, r-1)
            r += 1; l = r
        return arr
    def trim_sides(self, arr):
        '''str.strip() basically'''
        if ''.join(arr).isspace(): return []
        l , r = 0, len(arr) - 1
        while l < r and arr[l].isspace(): l += 1
        while l < r and arr[r].isspace(): r -= 1
        return arr[l:r+1]
    def trim_space(self, word):
        '''remove duplicating space in a word'''
        if not word: return []
        res = [word[0]]            
        for i in range(1, len(word)):
            if res[-1].isspace() and word[i].isspace(): continue
        return res
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