Given a number N, the task is to find the Nth number which has an absolute difference of 1 between every pair of its adjacent digits.
Examples:
Input : N = 5
Output : 5
Explanation:
The first 5 such numbers are 1,2,3,4 and 5.
Input : N = 15
Output : 23
Explanation:
The first 15 such numbers are 1,2,3,4,5,6,7,8,9,10,11,12,21,22 and 23.
Approach: In order to solve this problem we are using the Queue data structure.
- Prepare an empty Queue, and Enqueue all integers 1 to 9 in increasing order.
- Now perform the following operation N times.
- Dequeue and store in array arr which stores ith number of required type in arr[i].
- If (arr[i] % 10 != 0), then enqueue 10 * arr[i] + (arr[i] % 10) – 1.
- Enqueue 10 * arr[i] + (arr[i] % 10).
- If (arr[i] % 10 != 9), then enqueue 10 * arr[i] + (arr[i] % 10) + 1.
- Return arr[N] as the answer.
Below is the implementation of the given approach:
C++
// C++ Program to find Nth number with// absolute difference between all // adjacent digits at most 1.#include <bits/stdc++.h>using namespace std;// Return Nth number with// absolute difference between all // adjacent digits at most 1.void findNthNumber(int N){ // To store all such numbers long long arr[N + 1]; queue<long long> q; // Enqueue all integers from 1 to 9 // in increasing order. for (int i = 1; i <= 9; i++) q.push(i); // Perform the operation N times so that // we can get all such N numbers. for (int i = 1; i <= N; i++) { // Store the front element of queue, // in array and pop it from queue. arr[i] = q.front(); q.pop(); // If the last digit of dequeued integer is // not 0, then enqueue the next such number. if (arr[i] % 10 != 0) q.push(arr[i] * 10 + arr[i] % 10 - 1); // Enqueue the next such number q.push(arr[i] * 10 + arr[i] % 10); // If the last digit of dequeued integer is // not 9, then enqueue the next such number. if (arr[i] % 10 != 9) q.push(arr[i] * 10 + arr[i] % 10 + 1); } cout<<arr[N]<<endl;}// Driver Codeint main(){ int N = 21; findNthNumber(N); return 0;} |
Java
// Java program to find Nth number with// absolute difference between all // adjacent digits at most 1.import java.util.*;class GFG{// Return Nth number with// absolute difference between all // adjacent digits at most 1.static void findNthNumber(int N){ // To store all such numbers int []arr = new int[N + 1]; Queue<Integer> q = new LinkedList<>(); // Enqueue all integers from 1 to 9 // in increasing order. for(int i = 1; i <= 9; i++) q.add(i); // Perform the operation N times so // that we can get all such N numbers. for(int i = 1; i <= N; i++) { // Store the front element of queue, // in array and pop it from queue. arr[i] = q.peek(); q.remove(); // If the last digit of dequeued // integer is not 0, then enqueue // the next such number. if (arr[i] % 10 != 0) q.add(arr[i] * 10 + arr[i] % 10 - 1); // Enqueue the next such number q.add(arr[i] * 10 + arr[i] % 10); // If the last digit of dequeued // integer is not 9, then enqueue // the next such number. if (arr[i] % 10 != 9) q.add(arr[i] * 10 + arr[i] % 10 + 1); } System.out.println(arr[N]);}// Driver Codepublic static void main(String[] args){ int N = 21; findNthNumber(N);}}// This code is contributed by Amit Katiyar |
Python3
# Python 3 Program to find Nth number with# absolute difference between all # adjacent digits at most 1.# Return Nth number with# absolute difference between all # adjacent digits at most 1.def findNthNumber(N): # To store all such numbers arr = [0 for i in range(N + 1)] q = [] # Enqueue all integers from 1 to 9 # in increasing order. for i in range(1, 10, 1): q.append(i) # Perform the operation N times so that # we can get all such N numbers. for i in range(1, N+1, 1): # Store the front element of queue, # in array and pop it from queue. arr[i] = q[0] q.remove(q[0]) # If the last digit of dequeued integer is # not 0, then enqueue the next such number. if (arr[i] % 10 != 0): q.append(arr[i] * 10 + arr[i] % 10 - 1) # Enqueue the next such number q.append(arr[i] * 10 + arr[i] % 10) # If the last digit of dequeued integer is # not 9, then enqueue the next such number. if (arr[i] % 10 != 9): q.append(arr[i] * 10 + arr[i] % 10 + 1) print(arr[N])# Driver Codeif __name__ == '__main__': N = 21 findNthNumber(N)# This code is contributed by Samarth |
C#
// C# program to find Nth number with// absolute difference between all // adjacent digits at most 1.using System;using System.Collections.Generic;class GFG{// Return Nth number with// absolute difference between all // adjacent digits at most 1.static void findNthNumber(int N){ // To store all such numbers int []arr = new int[N + 1]; Queue<int> q = new Queue<int>(); // Enqueue all integers from 1 to 9 // in increasing order. for(int i = 1; i <= 9; i++) q.Enqueue(i); // Perform the operation N times so // that we can get all such N numbers. for(int i = 1; i <= N; i++) { // Store the front element of queue, // in array and pop it from queue. arr[i] = q.Peek(); q.Dequeue(); // If the last digit of dequeued // integer is not 0, then enqueue // the next such number. if (arr[i] % 10 != 0) q.Enqueue(arr[i] * 10 + arr[i] % 10 - 1); // Enqueue the next such number q.Enqueue(arr[i] * 10 + arr[i] % 10); // If the last digit of dequeued // integer is not 9, then enqueue // the next such number. if (arr[i] % 10 != 9) q.Enqueue(arr[i] * 10 + arr[i] % 10 + 1); } Console.WriteLine(arr[N]);}// Driver Codepublic static void Main(String[] args){ int N = 21; findNthNumber(N);}}// This code is contributed by Rohit_ranjan |
Javascript
<script> // JavaScript program to find Nth number with // absolute difference between all // adjacent digits at most 1. // Return Nth number with // absolute difference between all // adjacent digits at most 1. function findNthNumber(N) { // To store all such numbers let arr = new Array(N + 1); let q = []; // Enqueue all integers from 1 to 9 // in increasing order. for(let i = 1; i <= 9; i++) q.push(i); // Perform the operation N times so // that we can get all such N numbers. for(let i = 1; i <= N; i++) { // Store the front element of queue, // in array and pop it from queue. arr[i] = q[0]; q.shift(); // If the last digit of dequeued // integer is not 0, then enqueue // the next such number. if (arr[i] % 10 != 0) q.push(arr[i] * 10 + arr[i] % 10 - 1); // Enqueue the next such number q.push(arr[i] * 10 + arr[i] % 10); // If the last digit of dequeued // integer is not 9, then enqueue // the next such number. if (arr[i] % 10 != 9) q.push(arr[i] * 10 + arr[i] % 10 + 1); } document.write(arr[N] + "</br>"); } let N = 21; findNthNumber(N);</script> |
Output:
45
Time Complexity: O(N)
Auxiliary Space: O(N)
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!
