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Count of N digit numbers having absolute difference between adjacent digits as K

  • Last Updated : 16 Dec, 2021

Given two integers N and K. The task is to count all positive integers with length N having an absolute difference between adjacent digits equal to K.

Examples:

Input: N = 4, K = 8
Output: 3
Explanation: The absolute difference between every consecutive digit of each number is 8. Three possible numbers are 8080, 1919 and 9191.

Input: N = 2, K = 0
Output: 9
Explanation: 11, 22, 33, 44, 55, 66, 77, 88, 99. The absolute difference between every consecutive digit of each number is 0.

 

Approach: The approach is based on recursion. Iterate over digits [1, 9], and for each digit, count the N-digit number having a difference of absolute digit as K using recursion. Following cases arrive in the recursive function call.

  • Base Case: For all single-digit integers i.e. N = 1, increment answer count.
  • Recursive Call: If adding digit K to the one’s digit of the number formed till now (num) does not exceed 9, then recursively call by decreasing N and making num = (num*10 + num%10 + K).

if(num % 10 + K ≤ 9) 
   recursive_function(10 * num + (num % 10 + K), N – 1); 

  • If the value of K is non-zero after all the recursive calls and if num % 10 ≥ K, then again recursively call by decreasing the N and update num to (10*num + num%10 – K).

if(num % 10 ≥ K) 
    recursive_function(10 * num + num % 10 – K, N – 1)

Below is the implementation of the above approach.

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// To store the count of numbers
int countNums = 0;
 
// Function that recursively finds the
// possible numbers and append into ans
void checkUtil(int num, int K, int N)
{
    // Base Case
    if (N == 1) {
        countNums++;
        return;
    }
 
    // Check the sum of last digit and k
    // less than or equal to 9 or not
    if ((num % 10 + K) <= 9)
        checkUtil(10 * num +
                  (num % 10 + K), K, N - 1);
 
    // If K = 0, then subtraction and
    // addition does not make any
    // difference
    if (K) {
         
        // If unit's digit greater than K
        if ((num % 10 - K) >= 0)
            checkUtil(10 * num +
                      num % 10 - K, K, N - 1);
    }
}
 
// Function to call checkUtil function
// for every integer from 1 to 9
void check(int K, int N)
{
    // Loop to check for
    // all digits from 1 to 9
    for (int i = 1; i <= 9; i++) {
        checkUtil(i, K, N);
    }
}
 
// Driver Code
int main()
{
    // Given N and K
    int N = 4, K = 8;
    check(K, N);
 
    // Count total possible numbers
    cout << countNums << endl;
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// To store the count of numbers
static int countNums = 0;
 
// Function that recursively finds the
// possible numbers and append into ans
static void checkUtil(int num, int K, int N)
{
   
    // Base Case
    if (N == 1) {
        countNums++;
        return;
    }
 
    // Check the sum of last digit and k
    // less than or equal to 9 or not
    if ((num % 10 + K) <= 9)
        checkUtil(10 * num +
                  (num % 10 + K), K, N - 1);
 
    // If K = 0, then subtraction and
    // addition does not make any
    // difference
    if (K>0) {
         
        // If unit's digit greater than K
        if ((num % 10 - K) >= 0)
            checkUtil(10 * num +
                      num % 10 - K, K, N - 1);
    }
}
 
// Function to call checkUtil function
// for every integer from 1 to 9
static void check(int K, int N)
{
   
    // Loop to check for
    // all digits from 1 to 9
    for (int i = 1; i <= 9; i++) {
        checkUtil(i, K, N);
    }
}
 
// Driver Code
public static void main(String[] args)
{
   
    // Given N and K
    int N = 4, K = 8;
    check(K, N);
 
    // Count total possible numbers
    System.out.print(countNums +"\n");
}
}
 
// This code contributed by shikhasingrajput

Python3




# Python program for the above approach
 
# To store the count of numbers
countNums = 0;
 
# Function that recursively finds the
# possible numbers and append into ans
def checkUtil(num, K, N):
    global countNums;
     
    # Base Case
    if (N == 1):
        countNums += 1;
        return;
 
    # Check the sum of last digit and k
    # less than or equal to 9 or not
    if ((num % 10 + K) <= 9):
        checkUtil(10 * num + (num % 10 + K), K, N - 1);
 
    # If K = 0, then subtraction and
    # addition does not make any
    # difference
    if (K > 0):
 
        # If unit's digit greater than K
        if ((num % 10 - K) >= 0):
            checkUtil(10 * num + num % 10 - K, K, N - 1);
 
# Function to call checkUtil function
# for every integer from 1 to 9
def check(K, N):
   
    # Loop to check for
    # all digits from 1 to 9
    for i in range(1,10):
        checkUtil(i, K, N);
 
# Driver Code
if __name__ == '__main__':
    # Given N and K
    N = 4;
    K = 8;
    check(K, N);
 
    # Count total possible numbers
    print(countNums);
 
# This code is contributed by shikhasingrajput

C#




// C# program for the above approach
using System;
 
class GFG{
 
// To store the count of numbers
static int countNums = 0;
 
// Function that recursively finds the
// possible numbers and append into ans
static void checkUtil(int num, int K, int N)
{
     
    // Base Case
    if (N == 1)
    {
        countNums++;
        return;
    }
 
    // Check the sum of last digit and k
    // less than or equal to 9 or not
    if ((num % 10 + K) <= 9)
        checkUtil(10 * num +
                 (num % 10 + K), K, N - 1);
 
    // If K = 0, then subtraction and
    // addition does not make any
    // difference
    if (K > 0)
    {
         
        // If unit's digit greater than K
        if ((num % 10 - K) >= 0)
            checkUtil(10 * num +
                      num % 10 - K, K, N - 1);
    }
}
 
// Function to call checkUtil function
// for every integer from 1 to 9
static void check(int K, int N)
{
     
    // Loop to check for
    // all digits from 1 to 9
    for(int i = 1; i <= 9; i++)
    {
        checkUtil(i, K, N);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given N and K
    int N = 4, K = 8;
    check(K, N);
 
    // Count total possible numbers
    Console.Write(countNums + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
      // JavaScript code for the above approach
 
      // To store the count of numbers
      let countNums = 0;
 
      // Function that recursively finds the
      // possible numbers and append into ans
      function checkUtil(num, K, N)
      {
       
          // Base Case
          if (N == 1) {
              countNums++;
              return;
          }
 
          // Check the sum of last digit and k
          // less than or equal to 9 or not
          if ((num % 10 + K) <= 9)
              checkUtil(10 * num +
                  (num % 10 + K), K, N - 1);
 
          // If K = 0, then subtraction and
          // addition does not make any
          // difference
          if (K) {
 
              // If unit's digit greater than K
              if ((num % 10 - K) >= 0)
                  checkUtil(10 * num +
                      num % 10 - K, K, N - 1);
          }
      }
 
      // Function to call checkUtil function
      // for every integer from 1 to 9
      function check(K, N)
      {
       
          // Loop to check for
          // all digits from 1 to 9
          for (let i = 1; i <= 9; i++) {
              checkUtil(i, K, N);
          }
      }
 
      // Driver Code
 
      // Given N and K
      let N = 4, K = 8;
      check(K, N);
 
      // Count total possible numbers
      document.write(countNums + '<br>');
 
// This code is contributed by Potta Lokesh
  </script>
Output
3

Time Complexity: O(2N)
Auxiliary Space: O(1)

 


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