Given a floating-point number N, the task is to check whether it is palindrome or not.
Input: N = 123.321
Output: Yes
Input: N = 122.1
Output: No
Approach:
- First, convert the given floating-point number into a character array.
- Initialize the low to first index and high to the last index.
- While low < high:
- If the character at low is not equal to the character at high then exit and print “No”.
- If the character at low is equal to the character at high then continue after incrementing low and decrementing high…
- If the above loop completes successfully then print “Yes”.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if num is palindromebool isPalindrome(float num){ // Convert the given floating point number // into a string stringstream ss; ss << num; string s; ss >> s; // Pointers pointing to the first and // the last character of the string int low = 0; int high = s.size() - 1; while (low < high) { // Not a palindrome if (s[low] != s[high]) return false; // Update the pointers low++; high--; } return true;}// Driver codeint main(){ float n = 123.321f; if (isPalindrome(n)) cout << "Yes"; else cout << "No"; return 0;}// This code is contributed by Rajput-Ji |
Java
// Java implementation of the approachpublic class GFG { // Function that returns true if num is palindrome public static boolean isPalindrome(float num) { // Convert the given floating point number // into a string String s = String.valueOf(num); // Pointers pointing to the first and // the last character of the string int low = 0; int high = s.length() - 1; while (low < high) { // Not a palindrome if (s.charAt(low) != s.charAt(high)) return false; // Update the pointers low++; high--; } return true; } // Driver code public static void main(String args[]) { float n = 123.321f; if (isPalindrome(n)) System.out.print("Yes"); else System.out.print("No"); }} |
Python3
# Python3 implementation of the approach # Function that returns true if num is palindrome def isPalindrome(num) : # Convert the given floating point number # into a string s = str(num) # Pointers pointing to the first and # the last character of the string low = 0 high = len(s) - 1 while (low < high): # Not a palindrome if (s[low] != s[high]): return False # Update the pointers low += 1 high -= 1 return True# Driver code n = 123.321if (isPalindrome(n)): print("Yes") else: print("No") # This code is contributed by ihritik |
C#
// C# implementation of the approach using System;class GFG { // Function that returns true // if num is palindrome public static bool isPalindrome(float num) { // Convert the given floating point number // into a string string s = num.ToString(); // Pointers pointing to the first and // the last character of the string int low = 0; int high = s.Length - 1; while (low < high) { // Not a palindrome if (s[low] != s[high]) return false; // Update the pointers low++; high--; } return true; } // Driver code public static void Main() { float n = 123.321f; if (isPalindrome(n)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } }// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the approach// Function that returns true if num is palindromefunction isPalindrome(num){ // Convert the given floating point number // into a string var s = num.toString(); // Pointers pointing to the first and // the last character of the string var low = 0; var high = s.length - 1; while (low < high) { // Not a palindrome if (s[low] != s[high]) return false; // Update the pointers low++; high--; } return true;}// Driver codevar n = 123.321;if (isPalindrome(n)) document.write( "Yes");else document.write( "No");</script> |
Output:
Yes
Time Complexity: O(N).
Auxiliary Space: O(1).
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