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.

Simple Method

Approach: 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.

Algorithm:

Traverse through the array from start to end.
If the current element is greater than x print the previous number and break 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 -1

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.

Complexity Analysis:

Time Complexity : O(n).
To traverse an array only one loop is needed so the time complexity is O(n).
Space Complexity: O(1).
No extra space is required, So the space complexity is constant

Efficient Method

Approach: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.

Algorithm:

The algorithm can be implemented recursively or through iteration, but the basic idea remains the same.
There is come base cases to handle.
1. If there is no number greater than x then print the last element
2. 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.elow = mid + 1, and update answer with the middle element. In this step we are reducing the search space to half.
Else update the low, i.ehigh = mid – 1
Print the answer.

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.

Complexity Analysis:

Time Complexity : O(log n).
To run a binary search, the time complexity required is O(log n).
Space Complexity: O(1).
As no extra space is required, so the space complexity is constant.

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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