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:
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++
#include <bits/stdc++.h>
using namespace std;
bool check(int a[], int n)
{
bool flag = 0;
for (int i = 1; i < n - 1; i++) {
if (a[i + 1] > a[i]
&& a[i] < a[i - 1])
flag = 1;
}
if (flag)
return false;
else
return true;
}
int 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
import java.io.*;
import java.util.*;
class GFG
{
static boolean check(int a[], int n)
{
boolean flag = false;
for (int i = 1; i < n - 1; i++)
{
if (a[i + 1] > a[i]
&& a[i] < a[i - 1])
flag = true;
}
if (flag)
return false;
else
return true;
}
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");
}
}
|
Python3
def check(a, n) :
flag = 0
for i in range(1, n - 1):
if (a[i + 1] > a[i]
and a[i] < a[i - 1]) :
flag = 1
if (flag != 0) :
return False
else :
return True
arr = [ 1, 3, 5, 2 ]
N = len(arr)
if (check(arr, N) != 0):
print("YES")
else :
print("NO")
|
C#
using System;
public class GFG
{
static bool check(int[] a, int n)
{
bool flag = false;
for (int i = 1; i < n - 1; i++)
{
if (a[i + 1] > a[i]
&& a[i] < a[i - 1])
flag = true;
}
if (flag)
return false;
else
return true;
}
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");
}
}
|
Javascript
<script>
function check(a , n)
{
flag = false;
for (i = 1; i < n - 1; i++)
{
if (a[i + 1] > a[i]
&& a[i] < a[i - 1])
flag = true;
}
if (flag)
return false;
else
return true;
}
var arr = [ 1, 3, 5, 2 ];
var N = arr.length;
if (check(arr, N))
document.write("YES");
else
document.write("NO");
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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