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
Basic Approach: For a 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.
Below is the implementation of the above approach:
C++
#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;
}
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]) << " " ;
}
int main()
{
initDP();
int arr[] = { 8, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
printFloor(arr, n);
return 0;
}
|
Java
import java.io.*;
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 ;
}
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]) + " " );
}
public static void main(String[] args)
{
initDP();
int arr[] = { 8 , 6 };
int n = arr.length;
printFloor(arr, n);
}
}
|
Python3
max = 10000
dp = [ 0 ] * max
def initDP() :
for i in range ( max ) :
dp[i] = - 1
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 = " " )
if __name__ = = "__main__" :
initDP()
arr = [ 8 , 6 ]
n = len (arr)
printFloor(arr, n)
|
C#
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;
}
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]) + " " );
}
public static void Main()
{
initDP();
int []arr = {8, 6};
int n = arr.Length;
printFloor(arr, n);
}
}
|
PHP
<?php
$max = 10000;
$dp = array_fill (0, $max , NULL);
function initDP()
{
global $max , $dp ;
for ( $i = 0; $i < $max ; $i ++)
$dp [ $i ] = -1;
}
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 ]) . " " ;
}
initDP();
$arr = array (8, 6);
$n = sizeof( $arr );
printFloor( $arr , $n );
?>
|
Javascript
<script>
let max = 10000;
let dp = Array.from({length: max}, (_, i) => 0);
function initDP()
{
for (let i = 0; i < max; i++)
dp[i] = -1;
}
function func(x)
{
if (x == 0)
return 0;
if (dp[x] == -1)
dp[x] = x + func(Math.floor(x / 2));
return dp[x];
}
function prletFloor(arr, n)
{
for (let i = 0; i < n; i++)
document.write(func(arr[i]) + " " );
}
initDP();
let arr = [8, 6];
let n = arr.length;
prletFloor(arr, n);
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(n)
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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