Find all Palindrome Numbers in a given range

Problem Statement: Given a range of numbers, find all the palindrome numbers in the range.

Note: A palindromic number is a number that remains the same when its digits are reversed.OR a palindrome is a number that reads the same forward and backward Eg: 121,1221, 2552

Examples:

Example 1:
Input: min = 10 , max = 50
Output: 11 22 33 44 
Explanation: 11, 22, 33, 44 will remain the same when they read from forward or backward.

Example2:
Input: min = 100 , max = 150
Output: 101 111 121 131 141 
Explanation: 11, 22, 33, 44 will remain the same when they read from forward or backward.

Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Intuition: We run a for loop, and check for every number individually, if it is palindrome then we will print it.

Palindrome: We will reverse the number, if it is equal to the given number then it is palindrome otherwise it is not.

Approach:

Use a for loop and pass every element to the palindrome function.
Now make a palindrome function which takes an integer as a parameter.
Reverse the given number.
If the reversed number is equal to the given number then print it.

Reversing a number:

Make a result variable.
Take the last digit of the element.
Now multiple variable by 10 and add the last digit.
Repeat till element>0.

Code:

C++ Code

#include<iostream>
using namespace std;
 
// A function to check if n is palindrome
bool isPalindrome(int n)
{
    int reverse = 0;
    int temp = n;
    while(temp>0) {
        reverse = reverse*10 + temp%10;
        temp = temp/10;
    }
                 
            // If n and reverse are same,
            // then n is palindrome
    if(n==reverse)return true;
    return false;
}
 
int main()
{
    int min = 100;
    int max = 150;
    for(int i=min; i<=max; i++) {
        if(isPalindrome(i)) {
            cout<<i <<" ";
        }
    }    
    return 0;
}

Output: 101 111 121 131 141

Time Complexity: O(N)

Space Complexity: O(1)

Java Code

import java.util.*;
 
public class tuf {
      static boolean isPalindrome(int n)
        {
             
            int reverse = 0;
            int temp = n;
            while(temp>0) {
                reverse = reverse*10 + temp%10;
                temp = temp/10;
            }
                 
            // If n and reverse are same,
            // then n is palindrome
           if(n==reverse)return true;
           return false;
        }
       
        public static void main(String args[])
        {
            int min = 100;
            int max = 150;
            for(int i=min; i<=max; i++) {
                if(isPalindrome(i)) {
                    System.out.print(i + " ");
                }
            }
     
        }
}

Output: 101 111 121 131 141

Time Complexity: O(N)

Space Complexity: O(1)

Special thanks to Prashant Sahu for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article