# Check if all disks can be placed at a single rod based on given conditions

• Last Updated : 22 Mar, 2021

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:

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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 ``using` `namespace` `std;` `// Function to check if it is possible``// to move all disks to a single rod``bool` `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 Code``int` `main()``{``    ``int` `arr[] = { 1, 3, 5, 2 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``if` `(check(arr, N))``        ``cout << ``"YES"``;``    ``else``        ``cout << ``"NO"``;` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG``{` `// Function to check if it is possible``// to move all disks to a single rod``static` `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 Code``public` `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 rod``def` `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 Code``arr ``=` `[ ``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 approach``using` `System;``public` `class` `GFG``{``  ` `// Function to check if it is possible``// to move all disks to a single rod``static` `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

 ``

Output:
`YES`

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

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