Given an array of strings arr[] of size N where each string consists only of lowercase English letter. The task is to return the last palindromic string in the array.
Note: It guarantees that always one palindromic string is present.
Examples:
Input: arr[] = {“abc”, “car”, “ada”, “racecar”, “cool”}
Output: “racecar”
Explanation: The Last string that is palindromic is “racecar”.
Note that “ada” is also palindromic, but it is not the Last.Input: arr[] = {“def”, “aba”}
Output: “aba”
Approach: The solution is based on greedy approach. Check every string of an array from last if it is palindrome or not and also keep track of the last palindrome string. Follow the steps below to solve the problem:
- Initialize a string variable ans as an empty string.
-
Iterate over the range (N, 0] using the variable i and perform the following tasks:
- If arr[i] is a palindrome, then set the value of ans as arr[i].
- After performing the above steps, print the value of ans as the answer.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to check if given string // is Palindrome or not bool isPalindrome(string& s)
{ // Copy string s char into string a
string a = s;
reverse(s.begin(), s.end());
// Check if two string are equal or not
return s == a;
} // Function to return last Palindrome string string LastPalindrome(string arr[], int N)
{ // Loop to find the last palindrome string
for ( int i = N - 1; i >= 0; i--) {
// Checking if given string is
// palindrome or not
if (isPalindrome(arr[i])) {
// Return the answer
return arr[i];
}
}
} // Driver Code int main()
{ string arr[]
= { "abc" , "car" , "ada" , "racecar" ,
"cool" };
int N = sizeof (arr) / sizeof (arr[0]);
// Print required answer
cout << LastPalindrome(arr, N);
return 0;
} |
// Java program for the above approach class GFG {
// Function to check if given string
// is Palindrome or not
static boolean isPalindrome(String s)
{
// Copy string s char into string a
String a = s;
a = new StringBuffer(a).reverse().toString();
// Check if two string are equal or not
return s.equals(a);
}
// Function to return last Palindrome string
static String LastPalindrome(String arr[], int N) {
// Loop to find the last palindrome string
for ( int i = N - 1 ; i >= 0 ; i--) {
// Checking if given string is
// palindrome or not
if (isPalindrome(arr[i])) {
// Return the answer
return arr[i];
}
}
return "Hi" ;
}
// Driver Code
public static void main(String args[]) {
String arr[] = { "abc" , "car" , "ada" , "racecar" , "cool" };
int N = arr.length;
// Print required answer
System.out.println(LastPalindrome(arr, N));
}
} // This code is contributed by saurabh_jaiswal. |
# Python code for the above approach # Function to check if given string # is Palindrome or not def isPalindrome(s):
# find the length of a string
_len = len (s)
for i in range (_len / / 2 ):
# check if first and last string are same
if s[i] ! = s[_len - 1 - i]:
return 0
return 1
# Function to return last Palindrome string def LastPalindrome(arr, N):
# Loop to find the last palindrome string
for i in range (N - 1 , 0 , - 1 ):
# Checking if given string is
# palindrome or not
if isPalindrome(arr[i]):
# Return the answer
return arr[i]
# Driver Code arr = [ "abc" , "car" , "ada" , "racecar" , "cool" ]
N = len (arr)
# Print required answer print (LastPalindrome(arr, N))
# This code is contributed by gfgking |
// C# program for the above approach using System;
class GFG {
// Function to check if given string
// is Palindrome or not
static bool isPalindrome( string s)
{
// Copy string s char into string a
char [] a = s.ToCharArray();
Array.Reverse(a);
string p = new string (a);
//a = new StringBuffer(a).reverse().toString();
// Check if two string are equal or not
return s.Equals(p);
}
// Function to return last Palindrome string
static string LastPalindrome( string [] arr, int N) {
// Loop to find the last palindrome string
for ( int i = N - 1; i >= 0; i--) {
// Checking if given string is
// palindrome or not
if (isPalindrome(arr[i])) {
// Return the answer
return arr[i];
}
}
return "Hi" ;
}
// Driver Code
public static void Main() {
string []arr = { "abc" , "car" , "ada" , "racecar" , "cool" };
int N = arr.Length;
// Print required answer
Console.Write(LastPalindrome(arr, N));
}
} // This code is contributed by ukasp. |
<script> // JavaScript code for the above approach
// Function to check if given string
// is Palindrome or not
function isPalindrome(s)
{
// find the length of a string
let len = s.length;
for (let i = 0; i < len / 2; i++)
{
// check if first and last string are same
if (s[i] !== s[len - 1 - i]) {
return 0;
}
}
return 1;
}
// Function to return last Palindrome string
function LastPalindrome(arr, N) {
// Loop to find the last palindrome string
for (let i = N - 1; i >= 0; i--) {
// Checking if given string is
// palindrome or not
if (isPalindrome(arr[i])) {
// Return the answer
return arr[i];
}
}
}
// Driver Code
let arr = [ "abc" , "car" , "ada" , "racecar" ,
"cool" ];
let N = arr.length;
// Print required answer
document.write(LastPalindrome(arr, N));
// This code is contributed by Potta Lokesh
</script>
|
racecar
Time Complexity: O(N*W) where W is the maximum size of any string in arr[]
Auxiliary Space: O(1)