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:
- 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 = 7index = 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.
- 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.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 = 7index = 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.
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