Floor in a Sorted Array

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.

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:

  1. Traverse through the array from start to end.
  2. If the current element is greater than x print the previous number and break the loop.
  3. If there is no number greater than x then print the last element
  4. If the first number is greater than x then print -1

C++

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// 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;
}

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Java

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// 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)

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Python3

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# 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] <=
                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. 

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C#

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// 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

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PHP

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<?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
?>

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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:

  1. The algorithm can be implemented recursively or through iteration, but the basic idea remains the same.
  2. 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
  3. create three variables low = 0, mid and high = n-1 and another variable to store the answer
  4. Run a loop or recurse until and unless low is less than or equal to high.
  5. 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.
  6. Else update the low, i.ehigh = mid – 1
  7. Print the answer.

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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] <=
                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.

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C#

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// 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.

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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.

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