Given string str, the task is to print the last character of each word in the string.
Examples:
Input: str = “Geeks for Geeks”
Output: s r sInput: str = “computer applications”
Output: r s
Approach: Append a space i.e. ” “ at the end of the given string so that the last word in the string is also followed by a space just like all the other words in the string. Now start traversing the string character by character, and print every character which is followed by a space.
Below is the implementation of the above approach:
// CPP implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to print the last character // of each word in the given string void printLastChar(string str)
{ // Now, last word is also followed by a space
str = str + " " ;
for ( int i = 1; i < str.length(); i++) {
// If current character is a space
if (str[i] == ' ' )
// Then previous character must be
// the last character of some word
cout << str[i - 1] << " " ;
}
} // Driver code int main()
{ string str = "Geeks for Geeks" ;
printLastChar(str);
} // This code is contributed by // Surendra_Gangwar |
// Java implementation of the approach class GFG {
// Function to print the last character
// of each word in the given string
static void printLastChar(String str)
{
// Now, last word is also followed by a space
str = str + " " ;
for ( int i = 1 ; i < str.length(); i++) {
// If current character is a space
if (str.charAt(i) == ' ' )
// Then previous character must be
// the last character of some word
System.out.print(str.charAt(i - 1 ) + " " );
}
}
// Driver code
public static void main(String s[])
{
String str = "Geeks for Geeks" ;
printLastChar(str);
}
} |
# Function to print the last character # of each word in the given string def printLastChar(string):
# Now, last word is also followed by a space
string = string + " "
for i in range ( len (string)):
# If current character is a space
if string[i] = = ' ' :
# Then previous character must be
# the last character of some word
print (string[i - 1 ], end = " " )
# Driver code string = "Geeks for Geeks"
printLastChar(string) # This code is contributed by Shrikant13 |
// C# implementation of the approach using System;
class GFG
{ // Function to print the last character
// of each word in the given string
static void printLastChar( string str)
{
// Now, last word is also followed by a space
str = str + " " ;
for ( int i = 1; i < str.Length; i++)
{
// If current character is a space
if (str[i] == ' ' )
// Then previous character must be
// the last character of some word
Console.Write(str[i - 1] + " " );
}
}
// Driver code
public static void Main()
{
string str = "Geeks for Geeks" ;
printLastChar(str);
}
} // This code is contributed by Ryuga |
<?php // PHP implementation of the approach // Function to print the last character // of each word in the given string function printLastChar( $str )
{ // Now, last word is also followed by a space
$str = $str . " " ;
for ( $i = 1; $i < strlen ( $str ); $i ++)
{
// If current character is a space
if (! strcmp ( $str [ $i ], ' ' ))
// Then previous character must be
// the last character of some word
echo ( $str [ $i - 1] . " " );
}
} // Driver code $str = "Geeks for Geeks" ;
printLastChar( $str );
// This code contributed by PrinciRaj1992 ?> |
<script> // JavaScript implementation of the approach
// Function to print the last character
// of each word in the given string
function printLastChar(str) {
// Now, last word is also followed by a space
str = str + " " ;
for ( var i = 1; i < str.length; i++) {
// If current character is a space
if (str[i] === " " )
// Then previous character must be
// the last character of some word
document.write(str[i - 1] + " " );
}
}
// Driver code
var str = "Geeks for Geeks" ;
printLastChar(str);
</script> |
s r s
Time complexity: O(n) where n is the length of the given string.
Auxiliary space: O(1)