Remove odd indexed characters from a given string
Given string str of size N, the task is to remove the characters present at odd indices (0-based indexing) of a given string.
Examples :
Input: str = “abcdef”
Output: ace
Explanation:
The characters ‘b’, ‘d’ and ‘f’ are present at odd indices, i.e. 1, 3 and 5 respectively. Therefore, they are removed from the string.Input: str = “geeks”
Output: ges
Approach: Follow the steps below to solve the problem:
- Initialize an empty string, say new_string, to store the result.
- Traverse the given string and for every index, check if it is even or not.
- If found to be true, append the characters at those indices to the string new_string.
- Finally, after complete traversal of the entire string, return the new_string.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to remove the odd // indexed characters from a given string string removeOddIndexCharacters(string s) { // Stores the resultant string string new_string = "" ; for ( int i = 0; i < s.length(); i++) { // If current index is odd if (i % 2 == 1) { // Skip the character continue ; } // Otherwise, append the // character new_string += s[i]; } // Return the result return new_string; } // Driver Code int main() { string str = "abcdef" ; // Function call cout << removeOddIndexCharacters(str); return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG { // Function to remove odd indexed // characters from a given string static String removeOddIndexCharacters( String s) { // Stores the resultant string String new_string = "" ; for ( int i = 0 ; i < s.length(); i++) { // If the current index is odd if (i % 2 == 1 ) // Skip the character continue ; // Otherwise, append the // character new_string += s.charAt(i); } // Return the modified string return new_string; } // Driver Code public static void main(String[] args) { String str = "abcdef" ; // Remove the characters which // have odd index str = removeOddIndexCharacters(str); System.out.print(str); } } |
Python3
# Python3 program to implement # the above approach # Function to remove the odd # indexed characters from a given string def removeOddIndexCharacters(s): # Stores the resultant string new_s = "" i = 0 while i < len (s): # If the current index is odd if (i % 2 = = 1 ): # Skip the character i + = 1 continue # Otherwise, append the # character new_s + = s[i] i + = 1 # Return the modified string return new_s # Driver Code if __name__ = = '__main__' : str = "abcdef" # Remove the characters which # have odd index str = removeOddIndexCharacters( str ) print ( str ) |
C#
// C# program to implement // the above approach using System; class GFG{ // Function to remove odd indexed // characters from a given string static string removeOddIndexCharacters( string s) { // Stores the resultant string string new_string = "" ; for ( int i = 0; i < s.Length; i++) { // If the current index is odd if (i % 2 == 1) // Skip the character continue ; // Otherwise, append the // character new_string += s[i]; } // Return the modified string return new_string; } // Driver Code public static void Main() { string str = "abcdef" ; // Remove the characters which // have odd index str = removeOddIndexCharacters(str); Console.Write(str); } } // This code is contributed by sanjoy_62 |
Javascript
// JavaScript program to implement // the above approach // Function to remove the odd // indexed characters from a given string function removeOddIndexCharacters(s) { // Stores the resultant string var new_string = "" ; for ( var i = 0; i < s.length; i++) { // If current index is odd if (i % 2 === 1) { // Skip the character continue ; } // Otherwise, append the // character new_string += s[i]; } // Return the result return new_string; } // Driver Code string str = "abcdef" ; // Function call console.log(removeOddIndexCharacters(str)); // This code is contributed by Abhijeet Kumar(abhijeet19403) |
Output:
ace
Time Complexity: O(N)
Auxiliary Space: O(N)
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