Genere todos los números de N dígitos que tengan una diferencia absoluta como K entre dígitos adyacentes

Dados dos enteros N y K , la tarea es generar todos los enteros positivos con longitud N que tengan una diferencia absoluta de dígitos adyacentes igual a K .

Ejemplos:

Entrada: N = 4, K = 8
Salida: 1919, 8080, 9191
Explicación:
La diferencia absoluta entre cada dígito consecutivo de cada número es 8.
Por ejemplo: 8080 (abs(8-0) = 8, abs(0-8 ) = 8)

Entrada: N = 2, K = 2
Salida: 13, 24, 20, 35, 31, 46, 42, 57, 53, 68, 64, 79, 75, 86, 97
Explicación:
La diferencia absoluta entre cada dígito consecutivo de cada número es 1.
 

Enfoque: La idea es utilizar Backtracking . Iterar sobre los dígitos [1, 9] y para cada dígito del número de N dígitos que tenga una diferencia de dígito absoluto como K usando recursividad . A continuación se muestran los pasos:

1. Cree un vector ans[] para almacenar todos los números resultantes e itere recursivamente del 1 al 9 para generar números a partir de cada dígito. A continuación se muestran los casos: 
 

• Caso base: para todos los enteros de una sola longitud, es decir, N = 1 , agréguelos a ans[] .
• Llamada recursiva: si agregar el dígito K a los números al dígito de uno no excede 9 , llame recursivamente disminuyendo la N y actualice num a (10*num + num%10 + K) como se muestra a continuación:

if(num % 10 + K ≤ 9) { 
   función_recursiva(10 * num + (num % 10 + K), K, N – 1, y); 
}

• Si el valor de K es distinto de cero después de todas las llamadas recursivas y si num % 10 >= K , entonces vuelva a llamar recursivamente disminuyendo N y actualice num a (10*num + num%10 – K) como:

recursive_function(10 * num + num % 10 – K, K, N – 1, ans);

2. Después de todos los pasos anteriores, imprima los números almacenados en ans[] .

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that recursively finds the
// possible numbers and append into ans
void checkUntil(int num, int K,
                int N, vector<int>& ans)
{
   
    // Base Case
    if (N == 1)
    {
        ans.push_back(num);
        return;
    }
 
    // Check the sum of last digit and k
    // less than or equal to 9 or not
    if ((num % 10 + K) <= 9)
        checkUntil(10 * num
                       + (num % 10 + K),
                   K, N - 1, ans);
 
    // If k==0, then subtraction and
    // addition does not make any
    // difference
    // Hence, subtraction is skipped
    if (K) {
        if ((num % 10 - K) >= 0)
            checkUntil(10 * num
                           + num % 10 - K,
                       K, N - 1,
                       ans);
    }
}
 
// Function to call checkUntil function
// for every integer from 1 to 9
void check(int K, int N, vector<int>& ans)
{
    // check_util function recursively
    // store all numbers starting from i
    for (int i = 1; i <= 9; i++) {
        checkUntil(i, K, N, ans);
    }
}
 
// Function to print the all numbers
// which  satisfy the conditions
void print(vector<int>& ans)
{
    for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << ", ";
    }
}
 
// Driver Code
int main()
{
    // Given N and K
    int N = 4, K = 8;
 
    // To store the result
    vector<int> ans;
 
    // Function Call
    check(K, N, ans);
 
    // Print Resultant Numbers
    print(ans);
    return 0;
}

// Java program for
// the above approach
import java.util.*;
class GFG{
 
    // Function that recursively finds the
    // possible numbers and append into ans
    static void checkUntil(int num, int K, int N,
                           Vector<Integer> ans)
    {
       
        // Base Case
        if (N == 1)
        {
            ans.add(num);
            return;
        }
 
        // Check the sum of last digit and k
        // less than or equal to 9 or not
        if ((num % 10 + K) <= 9)
            checkUntil(10 * num + (num % 10 + K),
                       K, N - 1, ans);
 
        // If k==0, then subtraction and
        // addition does not make any
        // difference
        // Hence, subtraction is skipped
        if (K > 0)
        {
            if ((num % 10 - K) >= 0)
                checkUntil(10 * num + num % 10 - K,
                           K, N - 1, ans);
        }
    }
 
    // Function to call checkUntil function
    // for every integer from 1 to 9
    static void check(int K, int N, Vector<Integer> ans)
    {
       
        // check_util function recursively
        // store all numbers starting from i
        for (int i = 1; i <= 9; i++)
        {
            checkUntil(i, K, N, ans);
        }
    }
 
    // Function to print the all numbers
    // which  satisfy the conditions
    static void print(Vector<Integer> ans)
    {
        for (int i = 0; i < ans.size(); i++)
        {
            System.out.print(ans.get(i) + ", ");
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given N and K
        int N = 4, K = 8;
 
        // To store the result
        Vector<Integer> ans = new Vector<Integer>();;
 
        // Function Call
        check(K, N, ans);
 
        // Print Resultant Numbers
        print(ans);
    }
}
 
// This code is contributed by Amit Katiyar

# Python3 program for the above approach
 
# Function that recursively finds the
# possible numbers and append into ans
def checkUntil(num, K, N, ans):
     
    # Base Case
    if (N == 1):
        ans.append(num)
        return
 
    # Check the sum of last digit and k
    # less than or equal to 9 or not
    if ((num % 10 + K) <= 9):
        checkUntil(10 * num +
                 (num % 10 + K),
                 K, N - 1, ans)
 
    # If k==0, then subtraction and
    # addition does not make any
    # difference
    # Hence, subtraction is skipped
    if (K):
        if ((num % 10 - K) >= 0):
            checkUntil(10 * num +
                      num % 10 - K,
                     K, N - 1, ans)
     
# Function to call checkUntil function
# for every integer from 1 to 9
def check(K, N, ans):
     
    # check_util function recursively
    # store all numbers starting from i
    for i in range(1, 10):
        checkUntil(i, K, N, ans)
 
# Function to print the all numbers
# which satisfy the conditions
def print_list(ans):
     
    for i in range(len(ans)):
        print(ans[i], end = ", ")
 
# Driver Code
if __name__ == "__main__":
 
    # Given N and K
    N = 4
    K = 8;
 
    # To store the result
    ans = []
 
    # Function call
    check(K, N, ans)
 
    # Print resultant numbers
    print_list(ans)
 
# This code is contributed by chitranayal

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function that recursively finds the
// possible numbers and append into ans
static void checkUntil(int num, int K, int N,
                       List<int> ans)
{
 
    // Base Case
    if (N == 1)
    {
        ans.Add(num);
        return;
    }
 
    // Check the sum of last digit and k
    // less than or equal to 9 or not
    if ((num % 10 + K) <= 9)
        checkUntil(10 * num + (num % 10 + K),
                 K, N - 1, ans);
 
    // If k==0, then subtraction and
    // addition does not make any
    // difference
    // Hence, subtraction is skipped
    if (K > 0)
    {
        if ((num % 10 - K) >= 0)
            checkUntil(10 * num + num % 10 - K,
                     K, N - 1, ans);
    }
}
 
// Function to call checkUntil function
// for every integer from 1 to 9
static void check(int K, int N, List<int> ans)
{
 
    // check_util function recursively
    // store all numbers starting from i
    for(int i = 1; i <= 9; i++)
    {
        checkUntil(i, K, N, ans);
    }
}
 
// Function to print the all numbers
// which satisfy the conditions
static void print(List<int> ans)
{
    for(int i = 0; i < ans.Count; i++)
    {
        Console.Write(ans[i] + ", ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given N and K
    int N = 4, K = 8;
 
    // To store the result
    List<int> ans = new List<int>();;
 
    // Function call
    check(K, N, ans);
 
    // Print Resultant Numbers
    print(ans);
}
}
 
// This code is contributed by Amit Katiyar

<script>
 
// Javascript program to implement
// the above approach
 
    // Function that recursively finds the
    // possible numbers and append into ans
    function checkUntil(num, K, N,
                           ans)
    {
         
        // Base Case
        if (N == 1)
        {
            ans.push(num);
            return;
        }
   
        // Check the sum of last digit and k
        // less than or equal to 9 or not
        if ((num % 10 + K) <= 9)
            checkUntil(10 * num + (num % 10 + K),
                       K, N - 1, ans);
   
        // If k==0, then subtraction and
        // addition does not make any
        // difference
        // Hence, subtraction is skipped
        if (K > 0)
        {
            if ((num % 10 - K) >= 0)
                checkUntil(10 * num + num % 10 - K,
                           K, N - 1, ans);
        }
    }
   
    // Function to call checkUntil function
    // for every integer from 1 to 9
    function check(K, N, ans)
    {
         
        // check_util function recursively
        // store all numbers starting from i
        for (let i = 1; i <= 9; i++)
        {
            checkUntil(i, K, N, ans);
        }
    }
   
    // Function to print the all numbers
    // which  satisfy the conditions
    function print( ans)
    {
        for (let i = 0; i < ans.length; i++)
        {
            document.write(ans[i] + ", ");
        }
    }
 
// Driver Code
 
    // Given N and K
        let N = 4, K = 8;
   
        // To store the result
        let ans = []
   
        // Function Call
        check(K, N, ans);
   
        // Print Resultant Numbers
        print(ans);
     
</script>

1919, 8080, 9191,

Complejidad temporal: O(2 ^N )
Espacio auxiliar: O(N)