Check if a given array is sorted in Spiral manner or not
Last Updated :
22 Apr, 2021
Given an array arr[] of size N, the task is to check if the array is spirally sorted or not. If found to be true, then print “YES”. Otherwise, print “NO”.
Note: An array is spirally sorted if arr[0] ? arr[N – 1] ? arr[1] ? arr[N – 2] …
Examples:
Input: arr[] = { 1, 10, 14, 20, 18, 12, 5 }
Output: YES
Explanation:
arr[0] < arr[6]
arr[1] < arr[5]
arr[2] < arr[4]
Therefore, the required output is YES.
Input: arr[] = { 1, 2, 4, 3 }
Output: NO
Approach: The idea is to traverse the array and for every array element, say arr[i], check if arr[i] is less than or equal to arr[N – i – 1] and arr[N – i – 1] less than or equal to arr[i + 1] or not. If found to be false, then print “NO”. Otherwise, if all array elements satisfy the condition, then print “YES”. Follow the steps below to solve the problem:
- Initialize two variables, say start and end, to store the start and end indices of the given array.
- Iterate a loop while start is less than end, and check if arr[start] less than or equal to arr[end] and arr[end] is less than or equal to arr[start + 1] or not. If found to be false, then print “NO”.
- Otherwise, print “YES”.
Below is the implementation of the above approach:
C++14
#include <iostream>
using namespace std;
bool isSpiralSorted( int arr[], int n)
{
int start = 0;
int end = n - 1;
while (start < end) {
if (arr[start] > arr[end]) {
return false ;
}
start++;
if (arr[end] > arr[start]) {
return false ;
}
end--;
}
return true ;
}
int main()
{
int arr[] = { 1, 10, 14, 20, 18, 12, 5 };
int N = sizeof (arr) / sizeof (arr[0]);
if (isSpiralSorted(arr, N))
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
|
Java
import java.util.*;
class GFG{
static boolean isSpiralSorted( int [] arr, int n)
{
int start = 0 ;
int end = n - 1 ;
while (start < end)
{
if (arr[start] > arr[end])
{
return false ;
}
start++;
if (arr[end] > arr[start])
{
return false ;
}
end--;
}
return true ;
}
public static void main(String[] args)
{
int [] arr = { 1 , 10 , 14 , 20 , 18 , 12 , 5 };
int N = arr.length;
if (isSpiralSorted(arr, N) != false )
System.out.print( "YES" );
else
System.out.print( "NO" );
}
}
|
Python3
def isSpiralSorted(arr, n) :
start = 0 ;
end = n - 1 ;
while (start < end) :
if (arr[start] > arr[end]) :
return False ;
start + = 1 ;
if (arr[end] > arr[start]) :
return False ;
end - = 1 ;
return True ;
if __name__ = = "__main__" :
arr = [ 1 , 10 , 14 , 20 , 18 , 12 , 5 ];
N = len (arr);
if (isSpiralSorted(arr, N)) :
print ( "YES" );
else :
print ( "NO" );
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static bool isSpiralSorted( int [] arr, int n)
{
int start = 0;
int end = n - 1;
while (start < end)
{
if (arr[start] > arr[end])
{
return false ;
}
start++;
if (arr[end] > arr[start])
{
return false ;
}
end--;
}
return true ;
}
static void Main()
{
int [] arr = { 1, 10, 14, 20, 18, 12, 5 };
int N = arr.Length;
if (isSpiralSorted(arr, N))
Console.WriteLine( "YES" );
else
Console.WriteLine( "NO" );
}
}
|
Javascript
<script>
function isSpiralSorted(arr, n)
{
let start = 0;
let end = n - 1;
while (start < end)
{
if (arr[start] > arr[end])
{
return false ;
}
start++;
if (arr[end] > arr[start])
{
return false ;
}
end--;
}
return true ;
}
let arr = [ 1, 10, 14, 20, 18, 12, 5 ];
let N = arr.length;
if (isSpiralSorted(arr, N))
document.write( "YES" );
else
document.write( "NO" );
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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