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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