Count three-digit numbers having difference X with its reverse

Given an integer X, the task is to count the total number of three-digit numbers having difference X with its reverse. If no such number exists, then print -1.

Examples:

Input: X = 792
Output : 10
Explanation :
901 – 109 = 792
911 – 119 = 792
921 – 129 = 792
931 – 139 = 792
941 – 149 = 792
951 – 159 = 792
961 – 169 = 792
971 – 179 = 792
981 – 189 = 792
991 – 199 = 792

Input: X = 0
Output: 90

Approach: The given problem can be solved based on the following observations:

Let N = rpq 
Therefore, N = 100r + 10q + p 
Therefore, reverse of N = 100p + 10q + r 
Therefore, the problem reduces to solving (100r + 10q + p) – (r + 10q + 100p) = X 
-> 99(r – p) = X 
-> r – p = X / 99
Therefore, if given X is a multiple of 99, then solution exists. 

Follow the steps below to solve the problem based on the above observations: 

• Check if X is multiple of 99 or not. If not found to be true, print -1 as no solution exists.
• Otherwise, calculate X / 99. Generate all pairs using digits [1, 9] and for each pair, check if their difference is equal to X / 99 or not.
• If found to be true for any pair, increase count by 10, as the middle digit can be permuted to place any value from the range [0, 9] for the obtained pair.
• Finally, print the value of the count obtained.

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 count three-digit
// numbers having difference x
// with its reverse
int Count_Number(int x)
{
    int ans = 0;

    // if x is not multiple of 99
    if (x % 99 != 0) {

        // No solution exists
        ans = -1;
    }
    else {

        int diff = x / 99;

        // Generate all possible pairs
        // of digits [1, 9]
        for (int i = 1; i < 10; i++) {
            for (int j = 1; j < 10; j++) {

                // If any pair is obtained
                // with difference x / 99
                if ((i - j) == diff) {

                    // Increase count
                    ans += 10;
                }
            }
        }
    }

    // Return the count
    return ans;
}

// Driver Code
int main()
{
    int x = 792;
    cout << Count_Number(x) << endl;

    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.Arrays;

class GFG{

// Function to count three-digit
// numbers having difference x
// with its reverse
static int Count_Number(int x)
{
    int ans = 0;

    // If x is not multiple of 99
    if (x % 99 != 0)
    {

        // No solution exists
        ans = -1;
    }
    else
    {
        int diff = x / 99;

        // Generate all possible pairs
        // of digits [1, 9]
        for(int i = 1; i < 10; i++)
        {
            for(int j = 1; j < 10; j++)
            {

                // If any pair is obtained
                // with difference x / 99
                if ((i - j) == diff)
                {

                    // Increase count
                    ans += 10;
                }
            }
        }
    }

    // Return the count
    return ans;
}

// Driver Code
public static void main (String[] args)
{
    int x = 792;

    System.out.println(Count_Number(x));
}
}

// This code is contributed by sanjoy_62

Python3

# Python3 program to implement
# the above approach

# Function to count three-digit
# numbers having difference x
# with its reverse
def Count_Number(x):

    ans = 0;

    # If x is not multiple
    # of 99
    if (x % 99 != 0):

        # No solution exists
        ans = -1;
    else:
        diff = x / 99;

        # Generate all possible pairs
        # of digits [1, 9]
        for i in range(1, 10):
            for j in range(1, 10):

                # If any pair is obtained
                # with difference x / 99
                if ((i - j) == diff):
                    # Increase count
                    ans += 10;

    # Return the count
    return ans;

# Driver Code
if __name__ == '__main__':

    x = 792;
    print(Count_Number(x));

# This code is contributed by shikhasingrajput

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to count three-digit
// numbers having difference x
// with its reverse
static int Count_Number(int x)
{
    int ans = 0;

    // If x is not multiple of 99
    if (x % 99 != 0)
    {

        // No solution exists
        ans = -1;
    }
    else
    {
        int diff = x / 99;

        // Generate all possible pairs
        // of digits [1, 9]
        for(int i = 1; i < 10; i++)
        {
            for(int j = 1; j < 10; j++)
            {

                // If any pair is obtained
                // with difference x / 99
                if ((i - j) == diff)
                {

                    // Increase count
                    ans += 10;
                }
            }
        }
    }

    // Return the count
    return ans;
}

// Driver Code
public static void Main()
{
    int x = 792;

    Console.WriteLine(Count_Number(x));
}
}

// This code is contributed by code_hunt

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to count three-digit
// numbers having difference x
// with its reverse
function Count_Number(x)
{
    let ans = 0;

    // If x is not multiple of 99
    if (x % 99 != 0)
    {

        // No solution exists
        ans = -1;
    }
    else
    {
        let diff = x / 99;

        // Generate all possible pairs
        // of digits [1, 9]
        for(let i = 1; i < 10; i++)
        {
            for(let j = 1; j < 10; j++)
            {

                // If any pair is obtained
                // with difference x / 99
                if ((i - j) == diff)
                {

                    // Increase count
                    ans += 10;
                }
            }
        }
    }

    // Return the count
    return ans;
}

// Driver code
let x = 792;

document.write(Count_Number(x));

// This code is contributed by splevel62

</script>

Output

10

Time Complexity: O(1)
Auxiliary Space: O(1)