Given an array arr[] consisting of N integers representing N rods each with a disk of radius arr[i], the task is to check if it is possible to move all disks to a single rod based on the condition that a disk at ith rod can be moved to jth rod only if:
- abs(i – j) = 1 and ith rod has only a single disk.
- The disk at ith rod has radii less than the top disk of the jth rod or the jth rod is empty.
Examples:
Input: arr[] = {1, 3, 5, 2}, N = 4
Output: YES
Explanation:
One of the possible ways is:
First move the disk of radii 3 at rod 2 to the top of rod 3.
The array arr[] modifies to {1, 0, (5, 3), 2}.
Move the disk of radii 2 at rod 4 to the top of rod 3. The array arr[] modifies to {1, 0, (5, 3, 2), 0}.
Move the disk of radii 1 at rod 1 to top of the rod 2. The array arr[] modifies to {0, 1, (5, 3, 2), 0}.
Now, move the disk of radii 1 at rod 2 to the top of rod 3. The array arr[] modifies to {0, 0, (5, 3, 2, 1), 0}.Input: arr[] = {4, 1, 2}, N = 3
Output: NO
Approach: The given problem can be solved based on the following observations:
- It can be observed that if there exists an index i such that arr[i] < arr[i – 1] and arr[i] < arr[i + 1], then it is impossible to move all the disks at a single rod.
- Therefore, the task is reduced to finding if there exists any index i such that arr[i] < arr[i – 1] and arr[i] < arr[i + 1].
Follow the steps below to solve the problem:
- Initialize a variable, say flag = false, to store if any such index exists in the array or not.
- Traverse the array over the indices [1, N – 2]. For every ith index, check if arr[i] < arr[i – 1] and arr[i] < arr[i + 1], then set flag = true and break.
- Print “YES” if flag is false. Otherwise, print “NO”.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to check if it is possible// to move all disks to a single rodbool check(int a[], int n){ // Stores if it is possible to move // all disks to a single rod bool flag = 0; // Traverse the array for (int i = 1; i < n - 1; i++) { // If i-th element is smaller than // both its adjacent elements if (a[i + 1] > a[i] && a[i] < a[i - 1]) flag = 1; } // If flag is true if (flag) return false; // Otherwise else return true;}// Driver Codeint main(){ int arr[] = { 1, 3, 5, 2 }; int N = sizeof(arr) / sizeof(arr[0]); if (check(arr, N)) cout << "YES"; else cout << "NO"; return 0;} |
Java
// Java program to implement// the above approachimport java.io.*;import java.util.*;class GFG{// Function to check if it is possible// to move all disks to a single rodstatic boolean check(int a[], int n){ // Stores if it is possible to move // all disks to a single rod boolean flag = false; // Traverse the array for (int i = 1; i < n - 1; i++) { // If i-th element is smaller than // both its adjacent elements if (a[i + 1] > a[i] && a[i] < a[i - 1]) flag = true; } // If flag is true if (flag) return false; // Otherwise else return true;}// Driver Codepublic static void main(String[] args){ int arr[] = { 1, 3, 5, 2 }; int N = arr.length; if (check(arr, N)) System.out.print("YES"); else System.out.print("NO");}}// This code is contributed by code_hunt. |
Python3
# Python 3 program for the above approach# Function to check if it is possible# to move all disks to a single roddef check(a, n) : # Stores if it is possible to move # all disks to a single rod flag = 0 # Traverse the array for i in range(1, n - 1): # If i-th element is smaller than # both its adjacent elements if (a[i + 1] > a[i] and a[i] < a[i - 1]) : flag = 1 # If flag is true if (flag != 0) : return False # Otherwise else : return True# Driver Codearr = [ 1, 3, 5, 2 ]N = len(arr) if (check(arr, N) != 0): print("YES")else : print("NO") # This code is contributed by splevel62. |
C#
// C# program to implement// the above approachusing System;public class GFG{ // Function to check if it is possible// to move all disks to a single rodstatic bool check(int[] a, int n){ // Stores if it is possible to move // all disks to a single rod bool flag = false; // Traverse the array for (int i = 1; i < n - 1; i++) { // If i-th element is smaller than // both its adjacent elements if (a[i + 1] > a[i] && a[i] < a[i - 1]) flag = true; } // If flag is true if (flag) return false; // Otherwise else return true;} // Driver Code public static void Main(String[] args) { int[] arr = { 1, 3, 5, 2 }; int N = arr.Length; if (check(arr, N)) Console.Write("YES"); else Console.Write("NO"); }}// This code is contributed by susmitakundugoaldanga. |
Javascript
<script>// javascript program to implement// the above approach// Function to check if it is possible// to move all disks to a single rodfunction check(a , n){ // Stores if it is possible to move // all disks to a single rod flag = false; // Traverse the array for (i = 1; i < n - 1; i++) { // If i-th element is smaller than // both its adjacent elements if (a[i + 1] > a[i] && a[i] < a[i - 1]) flag = true; } // If flag is true if (flag) return false; // Otherwise else return true;}// Driver Codevar arr = [ 1, 3, 5, 2 ];var N = arr.length;if (check(arr, N)) document.write("YES");else document.write("NO");// This code contributed by Princi Singh </script> |
Output:
YES
Time Complexity: O(N)
Auxiliary Space: O(1)
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