Preorder Traversal and BST

PROBLEM :
Given an array, write a program that prints 1 if given array can represent preorder traversal of a BST, else prints 0.

Input:
The first line of input contains an integer T denoting the number of test cases.
The first line of each test case is N,N is the size of array.
The second line of each test case contains N input A[i].

Output:
Should print 1 if given array can represent preorder traversal of BST. Otherwise 0.

Constraints:
1 <=T<= 55
1 <= N <= 1000
1 <= arr[i] <= 1000

Example:

Input:
3
5
40 30 35 80 100
8
40 30 32 35 80 90 100 120
8
7  9 6 1 4 2 3 40

Output:
1
1
0

--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------

#include<iostream>
using namespace std;
int Preorder_Traversal_BST(int [],int ,int ) ;
int find_Rlarge(int [],int ,int ) ;
int allR_greater(int [],int ,int ,int ) ;
int main()
 {
int t,no,arr[1000],i ;
cin>>t ;
while(t--)
{
   cin>>no ;
   for(i=0;i<no;i++)
       cin>>arr[i] ;
     
   if(Preorder_Traversal_BST(arr,0,no-1))
       cout<<1 ;
   else
       cout<<0 ;
     
   cout<<endl ;
}
return 0;
}

int Preorder_Traversal_BST(int arr[],int start,int end)
{
    int i ;
    if(start==end||start>end)
        return 1 ;
       
    int pos=find_Rlarge(arr,start,end) ;

       
    if(allR_greater(arr,pos,end,arr[start]))
    {
        if(Preorder_Traversal_BST(arr,start+1,pos-1)&&Preorder_Traversal_BST(arr,pos+1,end))
            return 1 ;
    }
    return 0 ;
}

int find_Rlarge(int arr[],int start,int end)
{
    int i ;
    for(i=start+1;i<=end;i++)
    {
        if(arr[i]>arr[start])
            return i ;
    }
    return end ;
}

int allR_greater(int arr[],int start,int end,int ele)
{
    int i ;
    for(i=start;i<=end;i++)
        if(arr[i]<ele)
            return 0 ;
           
    return 1 ;
}

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