Floor in a Sorted Array - GeeksforGeeks

Given a sorted array and a value x, the floor of x is the largest element in array smaller than or equal to x. Write efficient functions to find floor of x.

Examples:

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

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

Input : arr[] = {1, 2, 8, 10, 10, 12, 19}, x = 0
Output : -1
Since floor doesn't exist, output is -1.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 (Simple)
A simple solution is linearly traverse input sorted array and search for the first element greater than x. The element just before the found element is floor of x.

C++

// 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

// Java program to find floor of a given number  
// in a sorted array  
import java.io.*; 
import java.util.*; 
import java.lang.*; 
  
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

# 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  
arr = [1, 2, 4, 6, 10, 12, 14]  
n = len(arr)  
x = 7
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#

// 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 
// 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 
?> 

Output:

Floor of 7 is 6.

Time Complexity : O(n)

Method 2 (Efficient)
The idea is to use Binary Search.

C++

// 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 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, 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

// 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 program to check above functions */
    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, 0, n - 1,  
                                          x); 
        if (index == -1) 
            System.out.println("Floor of " + x + 
                     " dosen't exist in array "); 
        else
            System.out.println("Floor of " + x + 
                            " is " + arr[index]); 
    } 
} 
// This code is contributed by Prerna Saini 

Python3

# 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 
arr = [1, 2, 4, 6, 10, 12, 14] 
n = len(arr) 
x = 7
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#

// 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 program to check above functions */
    public static void Main() 
    { 
        int []arr = {1, 2, 4, 6, 10, 12, 14}; 
        int n = arr.Length; 
        int x = 7; 
        int index = floorSearch(arr, 0, n - 1,  
                                          x); 
        if (index == -1) 
            Console.Write("Floor of " + x + 
                     " dosen't exist in array "); 
        else
            Console.Write("Floor of " + x + 
                            " is " + arr[index]); 
    } 
} 
  
// This code is contributed by nitin mittal. 

Output:

Floor of 7 is 6.

Time Complexity : O(Log n)

This article is contributed by Mayank Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

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