Value of continuous floor function : F(x) = F(floor(x/2)) + x

Given an array of positive integers. For every element x of array, we need to find the value of continuous floor function defined as F(x) = F(floor(x/2)) + x, where F(0) = 0.

Examples :-

Input : arr[] = {6, 8}
Output : 10 15

Explanation : F(6) = 6 + F(3)
                   = 6 + 3 + F(1)
                   = 6 + 3 + 1 + F(0)
                   = 10
Similarly F(8) = 15

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Basic Approach : For given value of x, we can calculate F(x) by using simple recursive function as:

int func(int x)
{
    if (x == 0) 
        return 0;
    return (x + func(floor(x/2)));
}

In this approach if we have n queries then it will take O(x) for every query element

An Efficient Approach is to use memoization We construct an array which holds the value of F(x) for each possible value of x. We compute the value if not already computed. Else we return the value.

C++

// C++ program for finding value  
// of continuous floor function 
#include <bits/stdc++.h> 
  
#define max 10000 
using namespace std; 
  
int dp[max]; 
  
void initDP() 
    { 
         for (int i = 0; i < max; i++) 
             dp[i] = -1; 
    } 
  
// function to return value of F(n) 
int func(int x) 
    { 
        if (x == 0) 
            return 0; 
        if (dp[x] == -1) 
            dp[x] = x + func(x / 2); 
  
        return dp[x]; 
    } 
  
void printFloor(int arr[], int n) 
    { 
        for (int i = 0; i < n; i++) 
            cout << func(arr[i]) << " "; 
    } 
  
// Driver code 
int main() 
    { 
        // call the initDP() to fill DP array 
        initDP(); 
  
        int arr[] = { 8, 6 }; 
        int n = sizeof(arr) / sizeof(arr[0]); 
  
        printFloor(arr, n); 
  
        return 0; 
    } 

Java

// Java program for finding value 
// of continuous floor function 
class GFG  
{ 
    static final int max = 10000; 
    static int dp[] = new int[max]; 
      
    static void initDP()  
    { 
        for (int i = 0; i < max; i++) 
            dp[i] = -1; 
    } 
      
    // function to return value of F(n) 
    static int func(int x)  
    { 
        if (x == 0) 
            return 0; 
        if (dp[x] == -1) 
            dp[x] = x + func(x / 2); 
      
        return dp[x]; 
    } 
      
    static void printFloor(int arr[], int n)  
    { 
        for (int i = 0; i < n; i++) 
            System.out.print(func(arr[i]) + " "); 
    } 
      
    // Driver code 
    public static void main(String[] args)  
    { 
          
        // call the initDP() to fill DP array 
        initDP(); 
      
        int arr[] = {8, 6}; 
        int n = arr.length; 
      
        printFloor(arr, n); 
    } 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

# Python3 program for finding value  
# of continuous floor function  
  
max = 10000
  
dp = [0] * max
  
# function to initialize the DP array 
def initDP() : 
      
        for i in range(max) :  
                dp[i] = -1
      
# function to return value of F(n)  
def func(x) : 
  
    if (x == 0) : 
        return 0
          
    if (dp[x] == -1) : 
        dp[x] = x + func(x // 2) 
  
    return dp[x]  
      
def printFloor(arr, n) : 
      
    for i in range(n) : 
                  
        print(func(arr[i]), end = " ") 
  
# Driver Code 
if __name__ == "__main__" : 
  
        # call the initDP() to  
        # fill DP array          
        initDP() 
  
        arr = [8, 6] 
        n = len(arr) 
  
        printFloor(arr, n) 
  
# This code is contributed by Ryuga 

C#

// C# program for finding value  
// of continuous floor function 
using System; 
  
class GFG  
{ 
    static int max = 10000; 
    static int []dp = new int[max]; 
      
    static void initDP()  
    { 
        for (int i = 0; i < max; i++) 
            dp[i] = -1; 
    } 
      
    // function to return value of F(n) 
    static int func(int x)  
    { 
        if (x == 0) 
            return 0; 
        if (dp[x] == -1) 
            dp[x] = x + func(x / 2); 
      
        return dp[x]; 
    } 
      
    static void printFloor(int []arr, int n)  
    { 
        for (int i = 0; i < n; i++) 
            Console.Write(func(arr[i]) + " "); 
    } 
      
    // Driver code 
    public static void Main()  
    { 
          
        // call the initDP() to fill DP array 
        initDP(); 
      
        int []arr = {8, 6}; 
        int n = arr.Length; 
      
        printFloor(arr, n); 
    } 
} 
  
// This code is contributed by nitin mittal 

PHP

<?php 
// PHP program for finding value  
// of continuous floor function 
$max = 10000; 
  
$dp = array_fill(0, $max, NULL); 
  
function initDP() 
{ 
    global $max,$dp; 
    for ($i = 0; $i < $max; $i++) 
        $dp[$i] = -1; 
} 
  
// function to return value of F(n) 
function func($x) 
{ 
    global $dp; 
    if ($x == 0) 
        return 0; 
    if ($dp[$x] == -1) 
        $dp[$x] = $x + func(intval($x / 2)); 
  
    return $dp[$x]; 
} 
  
function printFloor(&$arr, $n) 
{ 
    for ($i = 0; $i < $n; $i++) 
        echo func($arr[$i]) . " "; 
} 
  
// Driver code 
  
// call the initDP() to fill DP array 
initDP(); 
  
$arr = array(8, 6); 
$n = sizeof($arr); 
  
printFloor($arr, $n); 
  
// This code is contributed by ita_c 
?> 

Output:

15 10

This article is contributed by Shivam Pradhan (anuj_charm). 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 try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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