Given an integer N, the task is to find the closest palindromic number which is smaller than N.
Examples:
Input: N = 4000
Output: 3993
Explanation:
3993 is the closest palindromic number to N(= 4000) which is also smaller than N(= 4000). Therefore, 3993 is the required answer.Input: N = 2889
Output: 2882
Approach: Follow the steps below to solve the problem:
- Decrement the value of N while N is not a palindromic number
- Finally, print the updated value of N which was found to be a Palindrome Number.
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 check if// a number is palindrome or notbool checkPalindrome(int N){ // Stores reverse of N int rev = 0; // Stores the value of N int temp = N; // Calculate reverse of N while (N != 0) { // Update rev rev = rev * 10 + N % 10; // Update N N = N / 10; } // Update N N = temp; // if N is equal to // rev of N if (N == rev) { return true; } return false;}// Function to find the closest// smaller palindromic number to Nint closestSmallerPalindrome(int N){ // Calculate closest smaller // palindromic number to N do { // Update N N--; } while (N >= 0 && !checkPalindrome(N)); return N;}// Driver Codeint main(){ int N = 4000; cout << closestSmallerPalindrome(N); return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;import java.lang.Math;class GFG { // Function to check whether // a number is palindrome or not static boolean checkPalindrome(int N) { // Stores reverse of N int rev = 0; // Stores the value of N int temp = N; // Calculate reverse of N while (N != 0) { // Update rev rev = rev * 10 + N % 10; // Update N N = N / 10; } // Update N N = temp; // if N is equal to // rev of N if (N == rev) { return true; } return false; } // Function to find the closest // smaller palindromic number to N static int closestSmallerPalindrome(int N) { // Calculate closest smaller // palindromic number to N do { // Update N N--; } while (N >= 0 && !checkPalindrome(N)); return N; } // Driver Code public static void main(String[] args) { int N = 4000; System.out.println( closestSmallerPalindrome(N)); }} |
Python3
# Python3 program to implement # the above approach # Function to check if# a number is palindrome or notdef checkPalindrome(N): # Stores reverse of N rev = 0 # Stores the value of N temp = N # Calculate reverse of N while (N != 0): # Update rev rev = rev * 10 + N % 10 # Update N N = N // 10 # Update N N = temp # If N is equal to # rev of N if (N == rev): return True return False # Function to find the closest# smaller palindromic number to Ndef closestSmallerPalindrome(N): # Calculate closest smaller # palindromic number to N while N >= 0 and not checkPalindrome(N): # Update N N -= 1 return N# Driver Codeif __name__ == '__main__': N = 4000 print(closestSmallerPalindrome(N))# This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to check whether// a number is palindrome or notstatic bool checkPalindrome(int N){ // Stores reverse of N int rev = 0; // Stores the value of N int temp = N; // Calculate reverse of N while (N != 0) { // Update rev rev = rev * 10 + N % 10; // Update N N = N / 10; } // Update N N = temp; // If N is equal to // rev of N if (N == rev) { return true; } return false;}// Function to find the closest// smaller palindromic number to Nstatic int closestSmallerPalindrome(int N){ // Calculate closest smaller // palindromic number to N do { // Update N N--; } while (N >= 0 && !checkPalindrome(N)); return N;}// Driver Codepublic static void Main(){ int N = 4000; Console.WriteLine(closestSmallerPalindrome(N));}}// This code is contributed by bgangwar59 |
Javascript
<script>// Javascript program to implement// the above approach // Function to check whether // a number is palindrome or not function checkPalindrome(N) { // Stores reverse of N let rev = 0; // Stores the value of N let temp = N; // Calculate reverse of N while (N != 0) { // Update rev rev = Math.floor(rev * 10 + N % 10); // Update N N = Math.floor( N / 10); } // Update N N = temp; // if N is equal to // rev of N if (N == rev) { return true; } return false; } // Function to find the closest // smaller palindromic number to N function closestSmallerPalindrome(N) { // Calculate closest smaller // palindromic number to N do { // Update N N--; } while (N >= 0 && !checkPalindrome(N)); return N; } // Driver Code let N = 4000; document.write(closestSmallerPalindrome(N));</script> |
Output:
3993
Time Complexity: O(N * log10N)
Auxiliary Space: O(log10N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!