Floor in a Sorted Array

PROBLEM :
Given a sorted array having no duplicates, arr[] and a value, x, find floor of x in given array. Floor of x is the largest element in arr[] such that the element is smaller than or equal to x. If floor exists, then return index of it, else return -1.

Examples:

Input : arr[] = {1, 2, 8, 10, 11, 12, 19}, x = 5
Output : 1
1 is index of 2. 2 is the largest element in arr[]
smaller than 5.

Input : arr[] = {1, 2, 8, 10, 11, 12, 19}, x = 20
Output : 6
19 is the largest element in arr[] smaller than 20.

Input : arr[] = {1, 2, 8, 10, 11, 12, 19}, x = 0
Output : -1
Since floor doesn't exist, output is -1.
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 and x, N is the size of array.
The second line of each test case contains N array elements.

Output:
Print index of floor of x if it exists, else print -1

Constraints:
1 = T = 500
1 = N = 1000
0 = X = 1000
1 = arr[i] = 10000

Example:
Input:
3
7 0
1 2 8 10 11 12 19
7 5
1 2 8 10 11 12 19
7 10
1 2 8 10 11 12 19

Output:
-1
1
3

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

#include<iostream>
using namespace std;
int FloorSortedArray(int [],int ,int ) ;
#define max 1000
int main()
{
    int t,no,i,k ;
    int arr[max] ;
    cin>>t ;
    while(t--)
    {
        cin>>no>>k ;
        for(i=0;i<no;i++)
            cin>>arr[i] ;
           
        cout<<FloorSortedArray(arr,no,k)<<endl ;
    }
return 0;
}

int FloorSortedArray(int arr[],int no,int ele)
{
    int start,end,ans,mid ;
    start=0 ;
    end=no-1 ;
    ans=-1 ;
   
    while(start<=end)
    {
        mid=(start+end)/2 ;
       
        if(arr[mid]==ele)
        {
            ans=mid ;
            break ;
        }
        else if(mid==no-1)
        {
            if(arr[mid]<end)
                ans=mid ;
            break ;
        }
        else if(arr[mid]<ele&&arr[mid+1]>ele)
        {
            ans=mid ;
            break ;
        }
        else if(arr[mid]>ele)
            end=mid-1 ;
        else
            start=mid+1 ;
    }
   
    if(ans==-1&&arr[no-1]<ele)
        return no-1 ;
    return ans ;
}

---------------------------------------------------------------------------------

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