Given a sorted array and a value x, the floor of x is the largest element in the array smaller than or equal to x. Write efficient functions to find the floor of x
Examples:
Input: arr[] = {1, 2, 8, 10, 10, 12, 19}, x = 5
Output: 2
Explanation: 2 is the largest element in
arr[] smaller than 5Input: arr[] = {1, 2, 8, 10, 10, 12, 19}, x = 20
Output: 19
Explanation: 19 is the largest element in
arr[] smaller than 20Input : arr[] = {1, 2, 8, 10, 10, 12, 19}, x = 0
Output : -1
Explanation: Since floor doesn’t exist, output is -1.
Naive Approach: To solve the problem follow the below idea:
The idea is simple, traverse through the array and find the first element greater than x. The element just before the found element is the floor of x
Follow the given steps to solve the problem:
- Traverse through the array from start to end.
- If the current element is greater than x print the previous number and break out of the loop
- If there is no number greater than x then print the last element
- If the first number is greater than x then print that the floor of x doesn’t exist
Below is the implementation of the above approach:
// C++ program to find floor of a given number // in a sorted array #include <iostream> using namespace std;
/* An inefficient function to get index of floor of x in arr[0..n-1] */ int floorSearch( int arr[], int n, int x)
{ // If last element is smaller than x
if (x >= arr[n - 1])
return n - 1;
// If first element is greater than x
if (x < arr[0])
return -1;
// Linearly search for the first element
// greater than x
for ( int i = 1; i < n; i++)
if (arr[i] > x)
return (i - 1);
return -1;
} /* Driver program to check above functions */ int main()
{ int arr[] = { 1, 2, 4, 6, 10, 12, 14 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 7;
int index = floorSearch(arr, n - 1, x);
if (index == -1)
cout << "Floor of " << x
<< " doesn't exist in array " ;
else
cout << "Floor of " << x << " is " << arr[index];
return 0;
} // This code is contributed by shivanisinghss2110 |
// C/C++ program to find floor of a given number // in a sorted array #include <stdio.h> /* An inefficient function to get index of floor of x in arr[0..n-1] */ int floorSearch( int arr[], int n, int x)
{ // If last element is smaller than x
if (x >= arr[n - 1])
return n - 1;
// If first element is greater than x
if (x < arr[0])
return -1;
// Linearly search for the first element
// greater than x
for ( int i = 1; i < n; i++)
if (arr[i] > x)
return (i - 1);
return -1;
} /* Driver program to check above functions */ int main()
{ int arr[] = { 1, 2, 4, 6, 10, 12, 14 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 7;
int index = floorSearch(arr, n - 1, x);
if (index == -1)
printf ( "Floor of %d doesn't exist in array " , x);
else
printf ( "Floor of %d is %d" , x, arr[index]);
return 0;
} |
// Java program to find floor of // a given number in a sorted array import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
/* An inefficient function to get index of floor
of x in arr[0..n-1] */ static int floorSearch( int arr[], int n, int x)
{
// If last element is smaller than x
if (x >= arr[n - 1 ])
return n - 1 ;
// If first element is greater than x
if (x < arr[ 0 ])
return - 1 ;
// Linearly search for the first element
// greater than x
for ( int i = 1 ; i < n; i++)
if (arr[i] > x)
return (i - 1 );
return - 1 ;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 4 , 6 , 10 , 12 , 14 };
int n = arr.length;
int x = 7 ;
int index = floorSearch(arr, n - 1 , x);
if (index == - 1 )
System.out.print( "Floor of " + x
+ " doesn't exist in array " );
else
System.out.print( "Floor of " + x + " is "
+ arr[index]);
}
} // This code is contributed // by Akanksha Rai(Abby_akku) |
# Python3 program to find floor of a # given number in a sorted array # Function to get index of floor # of x in arr[low..high] def floorSearch(arr, n, x):
# If last element is smaller than x
if (x > = arr[n - 1 ]):
return n - 1
# If first element is greater than x
if (x < arr[ 0 ]):
return - 1
# Linearly search for the first element
# greater than x
for i in range ( 1 , n):
if (arr[i] > x):
return (i - 1 )
return - 1
# Driver Code arr = [ 1 , 2 , 4 , 6 , 10 , 12 , 14 ]
n = len (arr)
x = 7
index = floorSearch(arr, n - 1 , x)
if (index = = - 1 ):
print ( "Floor of" , x, "doesn't exist \
in array ", end=" ")
else :
print ( "Floor of" , x, "is" , arr[index])
# This code is contributed by Smitha Dinesh Semwal. |
// C# program to find floor of a given number // in a sorted array using System;
class GFG {
/* An inefficient function to get index of floor
of x in arr[0..n-1] */ static int floorSearch( int [] arr, int n, int x)
{
// If last element is smaller than x
if (x >= arr[n - 1])
return n - 1;
// If first element is greater than x
if (x < arr[0])
return -1;
// Linearly search for the first element
// greater than x
for ( int i = 1; i < n; i++)
if (arr[i] > x)
return (i - 1);
return -1;
}
// Driver Code
static void Main()
{
int [] arr = { 1, 2, 4, 6, 10, 12, 14 };
int n = arr.Length;
int x = 7;
int index = floorSearch(arr, n - 1, x);
if (index == -1)
Console.WriteLine( "Floor of " + x
+ " doesn't exist in array " );
else
Console.WriteLine( "Floor of " + x + " is "
+ arr[index]);
}
} // This code is contributed // by mits |
<?php // PHP program to find floor of // a given number in a sorted array /* An inefficient function to get index of floor
of x in arr[0..n-1] */
function floorSearch( $arr , $n , $x )
{ // If last element is smaller
// than x
if ( $x >= $arr [ $n - 1])
return $n - 1;
// If first element is greater
// than x
if ( $x < $arr [0])
return -1;
// Linearly search for the
// first element greater than x
for ( $i = 1; $i < $n ; $i ++)
if ( $arr [ $i ] > $x )
return ( $i - 1);
return -1;
} // Driver Code $arr = array (1, 2, 4, 6, 10, 12, 14);
$n = sizeof( $arr );
$x = 7;
$index = floorSearch( $arr , $n - 1, $x );
if ( $index == -1)
echo "Floor of " , $x ,
"doesn't exist in array " ;
else echo "Floor of " , $x ,
" is " , $arr [ $index ];
// This code is contributed by ajit ?> |
<script> // JavaScript program to find floor of // a given number in a sorted array /* An inefficient function to get index of floor
of x in arr[0..n-1] */ function floorSearch(arr, n, x)
{
// If last element is smaller than x
if (x >= arr[n - 1])
return n - 1;
// If first element is greater than x
if (x < arr[0])
return -1;
// Linearly search for the first element
// greater than x
for (let i = 1; i < n; i++)
if (arr[i] > x)
return (i - 1);
return -1;
}
// Driver Code let arr = [ 1, 2, 4, 6, 10, 12, 14 ];
let n = arr.length;
let x = 7;
let index = floorSearch(arr, n - 1, x);
if (index == -1)
document.write(
"Floor of " + x
+ " doesn't exist in array " );
else
document.write(
"Floor of " + x + " is "
+ arr[index]);
</script> |
Floor of 7 is 6
Time Complexity: O(N). To traverse an array only one loop is needed.
Auxiliary Space: O(1). No extra space is required
Floor in a Sorted Array using binary search:
To solve the problem follow the below idea:
There is a catch in the problem, the given array is sorted. The idea is to use Binary Search to find the floor of a number x in a sorted array by comparing it to the middle element and dividing the search space into half
Follow the given steps to solve the problem:
- The algorithm can be implemented recursively or through iteration, but the basic idea remains the same.
- There are some base cases to handle
- If there is no number greater than x then print the last element
- If the first number is greater than x then print -1
- create three variables low = 0, mid and high = n-1 and another variable to store the answer
- Run a loop or recurse until and unless low is less than or equal to high.
- check if the middle ( (low + high) /2) element is less than x, if yes then update the low, i.e low = mid + 1, and update the answer with the middle element. In this step we are reducing the search space to half.
- Else update the high , i.e high = mid – 1
- Print the answer
Below is the implementation of the above approach:
// A C/C++ program to find floor // of a given number in a sorted array #include <bits/stdc++.h> using namespace std;
/* Function to get index of floor of x in arr[low..high] */
int floorSearch( int arr[], int low, int high, int x)
{ // If low and high cross each other
if (low > high)
return -1;
// If last element is smaller than x
if (x >= arr[high])
return high;
// Find the middle point
int mid = (low + high) / 2;
// If middle point is floor.
if (arr[mid] == x)
return mid;
// If x lies between mid-1 and mid
if (mid > 0 && arr[mid - 1] <= x && x < arr[mid])
return mid - 1;
// If x is smaller than mid, floor
// must be in left half.
if (x < arr[mid])
return floorSearch(arr, low, mid - 1, x);
// If mid-1 is not floor and x is
// greater than arr[mid],
return floorSearch(arr, mid + 1, high, x);
} // Driver code int main()
{ int arr[] = { 1, 2, 4, 6, 10, 12, 14 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 7;
// Function call
int index = floorSearch(arr, 0, n - 1, x);
if (index == -1)
cout << "Floor of " << x
<< " doesn't exist in array " ;
else
cout << "Floor of " << x << " is " << arr[index];
return 0;
} // this code is contributed by shivanisinghss2110 |
// A C/C++ program to find floor // of a given number in a sorted array #include <stdio.h> /* Function to get index of floor of x in arr[low..high] */
int floorSearch( int arr[], int low, int high, int x)
{ // If low and high cross each other
if (low > high)
return -1;
// If last element is smaller than x
if (x >= arr[high])
return high;
// Find the middle point
int mid = (low + high) / 2;
// If middle point is floor.
if (arr[mid] == x)
return mid;
// If x lies between mid-1 and mid
if (mid > 0 && arr[mid - 1] <= x && x < arr[mid])
return mid - 1;
// If x is smaller than mid, floor
// must be in left half.
if (x < arr[mid])
return floorSearch(arr, low, mid - 1, x);
// If mid-1 is not floor and x is
// greater than arr[mid],
return floorSearch(arr, mid + 1, high, x);
} // Driver code int main()
{ int arr[] = { 1, 2, 4, 6, 10, 12, 14 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 7;
// Function call
int index = floorSearch(arr, 0, n - 1, x);
if (index == -1)
printf ( "Floor of %d doesn't exist in array " , x);
else
printf ( "Floor of %d is %d" , x, arr[index]);
return 0;
} |
// Java program to find floor of // a given number in a sorted array import java.io.*;
class GFG {
/* Function to get index of floor of x in
arr[low..high] */
static int floorSearch( int arr[], int low, int high,
int x)
{
// If low and high cross each other
if (low > high)
return - 1 ;
// If last element is smaller than x
if (x >= arr[high])
return high;
// Find the middle point
int mid = (low + high) / 2 ;
// If middle point is floor.
if (arr[mid] == x)
return mid;
// If x lies between mid-1 and mid
if (mid > 0 && arr[mid - 1 ] <= x && x < arr[mid])
return mid - 1 ;
// If x is smaller than mid, floor
// must be in left half.
if (x < arr[mid])
return floorSearch(arr, low, mid - 1 , x);
// If mid-1 is not floor and x is
// greater than arr[mid],
return floorSearch(arr, mid + 1 , high, x);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 4 , 6 , 10 , 12 , 14 };
int n = arr.length;
int x = 7 ;
// Function call
int index = floorSearch(arr, 0 , n - 1 , x);
if (index == - 1 )
System.out.println(
"Floor of " + x
+ " doesn't exist in array " );
else
System.out.println( "Floor of " + x + " is "
+ arr[index]);
}
} // This code is contributed by Prerna Saini |
# Python3 program to find floor of a # given number in a sorted array # Function to get index of floor # of x in arr[low..high] def floorSearch(arr, low, high, x):
# If low and high cross each other
if (low > high):
return - 1
# If last element is smaller than x
if (x > = arr[high]):
return high
# Find the middle point
mid = int ((low + high) / 2 )
# If middle point is floor.
if (arr[mid] = = x):
return mid
# If x lies between mid-1 and mid
if (mid > 0 and arr[mid - 1 ] < = x
and x < arr[mid]):
return mid - 1
# If x is smaller than mid,
# floor must be in left half.
if (x < arr[mid]):
return floorSearch(arr, low, mid - 1 , x)
# If mid-1 is not floor and x is greater than
# arr[mid],
return floorSearch(arr, mid + 1 , high, x)
# Driver Code if __name__ = = "__main__" :
arr = [ 1 , 2 , 4 , 6 , 10 , 12 , 14 ]
n = len (arr)
x = 7
# Function call
index = floorSearch(arr, 0 , n - 1 , x)
if (index = = - 1 ):
print ( "Floor of" , x, "doesn't exist\
in array ", end=" ")
else :
print ( "Floor of" , x, "is" , arr[index])
# This code is contributed by Smitha Dinesh Semwal. |
// C# program to find floor of // a given number in a sorted array using System;
class GFG {
/* Function to get index of floor of x in
arr[low..high] */
static int floorSearch( int [] arr, int low, int high,
int x)
{
// If low and high cross each other
if (low > high)
return -1;
// If last element is smaller than x
if (x >= arr[high])
return high;
// Find the middle point
int mid = (low + high) / 2;
// If middle point is floor.
if (arr[mid] == x)
return mid;
// If x lies between mid-1 and mid
if (mid > 0 && arr[mid - 1] <= x && x < arr[mid])
return mid - 1;
// If x is smaller than mid, floor
// must be in left half.
if (x < arr[mid])
return floorSearch(arr, low, mid - 1, x);
// If mid-1 is not floor and x is
// greater than arr[mid],
return floorSearch(arr, mid + 1, high, x);
}
// Driver code
public static void Main()
{
int [] arr = { 1, 2, 4, 6, 10, 12, 14 };
int n = arr.Length;
int x = 7;
// Function call
int index = floorSearch(arr, 0, n - 1, x);
if (index == -1)
Console.Write( "Floor of " + x
+ " doesn't exist in array " );
else
Console.Write( "Floor of " + x + " is "
+ arr[index]);
}
} // This code is contributed by nitin mittal. |
<script> // javascript program to find floor of // a given number in a sorted array /* Function to get index of floor of x in arr[low..high] */ function floorSearch(
arr , low,
high , x)
{ // If low and high cross each other
if (low > high)
return -1;
// If last element is smaller than x
if (x >= arr[high])
return high;
// Find the middle point
var mid = (low + high) / 2;
// If middle point is floor.
if (arr[mid] == x)
return mid;
// If x lies between mid-1 and mid
if (
mid > 0 && arr[mid - 1] <= x && x < arr[mid])
return mid - 1;
// If x is smaller than mid, floor
// must be in left half.
if (x < arr[mid])
return floorSearch(
arr, low,
mid - 1, x);
// If mid-1 is not floor and x is
// greater than arr[mid],
return floorSearch(
arr, mid + 1, high,
x);
} /* Driver program to check above functions */ var arr = [ 1, 2, 4, 6, 10, 12, 14 ];
var n = arr.length;
var x = 7;
var index = floorSearch(
arr, 0, n - 1,
x);
if (index == -1)
document.write(
"Floor of " + x + " doesn't exist in array " );
else document.write(
"Floor of " + x + " is " + arr[index]);
// This code is contributed by Amit Katiyar </script> |
Floor of 7 is 6
Time Complexity: O(log N). To run a binary search.
Auxiliary Space: O(1). As no extra space is required.