Program to reverse words in a given string in C++
Given a sentence in the form of string str, the task is to reverse each word of the given sentence in C++.
Examples:
Input: str = “the sky is blue”
Output: blue is sky the
Input: str = “I love programming”
Output: programming love I
Method 1: Using STL functions
- Reverse the given string str using STL function reverse().
- Iterate the reversed string and whenever a space is found reverse the word before that space using the STL function reverse().
Below is the implementation of the above approach:
CPP
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to reverse the given string string reverseString(string str) { // Reverse str using inbuilt function reverse(str.begin(), str.end()); // Add space at the end so that the // last word is also reversed str.insert(str.end(), ' ' ); int n = str.length(); int j = 0; // Find spaces and reverse all words // before that for ( int i = 0; i < n; i++) { // If a space is encountered if (str[i] == ' ' ) { reverse(str.begin() + j, str.begin() + i); // Update the starting index // for next word to reverse j = i + 1; } } // Remove spaces from the end of the // word that we appended str.pop_back(); // Return the reversed string return str; } // Driver code int main() { string str = "I like this code" ; // Function call string rev = reverseString(str); // Print the reversed string cout << rev; return 0; } |
code this like I
Time Complexity: O(n2)
Auxiliary Space: O(1)
Method 2: Without using the inbuilt function: We can create the reverse() function which is used to reverse the given string. Below are the steps:
- Initially reverse each word of the given string str.
- Now reverse the whole string to get the resultant string in desired order.
Below is the implementation of the above approach:
CPP
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function used to reverse a string // from index l to r void reversed(string& s, int l, int r) { while (l < r) { // Swap characters at l and r swap(s[l], s[r]); l++; r--; } } // Function to reverse the given string string reverseString(string str) { // Add space at the end so that the // last word is also reversed str.insert(str.end(), ' ' ); int n = str.length(); int j = 0; // Find spaces and reverse all words // before that for ( int i = 0; i < n; i++) { // If a space is encountered if (str[i] == ' ' ) { // Function call to our custom // reverse function() reversed(str, j, i - 1); // Update the starting index // for next word to reverse j = i + 1; } } // Remove spaces from the end of the // word that we appended str.pop_back(); // Reverse the whole string reversed(str, 0, str.length() - 1); // Return the reversed string return str; } // Driver code int main() { string str = "I like this code" ; // Function call string rev = reverseString(str); // Print the reversed string cout << rev; return 0; } |
code this like I
Time Complexity: O(n2)
Auxiliary Space: O(1)
Method – 3 : Without Using Extra Space
The above task can also be accomplished by splitting and directly swapping the string starting from the middle. As direct swapping is involved, less space is consumed too.
C++
// C++ code to reverse a string #include <bits/stdc++.h> using namespace std; // Reverse the string string RevString(string s[], int l) { // Check if number of words is even if (l % 2 == 0) { // Find the middle word int j = l / 2; // Starting from the middle // start swapping words at // jth position and l-1-j position while (j <= l - 1) { string temp; temp = s[l - j - 1]; s[l - j - 1] = s[j]; s[j] = temp; j += 1; } } // Check if number of words is odd else { // Find the middle word int j = (l / 2) + 1; // Starting from the middle start // swapping the words at jth // position and l-1-j position while (j <= l - 1) { string temp; temp = s[l - j - 1]; s[l - j - 1] = s[j]; s[j] = temp; j += 1; } } string S = s[0]; // Return the reversed sentence for ( int i = 1; i < 9; i++) { S = S + " " + s[i]; } return S; } // Driver code int main() { string s = "getting good at coding " "needs a lot of practice" ; string words[] = { "getting" , "good" , "at" , "coding" , "needs" , "a" , "lot" , "of" , "practice" }; cout << RevString(words, 9) << endl; return 0; } // This code is contributed by AJAY MAKVANA |
practice of lot a needs coding at good getting
Time Complexity: O(n)
Auxiliary Space: O(n)
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