Count of divisors having more set bits than quotient on dividing N

Given a positive integer N. The task is to find the number of divisor (say d) which give quotient (say q) on integer division (i.e q = N/d) such that quotient have less or equal set bit than divisor. In other words, find number of possible value of ‘d’ which will produce ‘q’ on integer dividing ‘n’ (q = N/d) such that ‘q’ have less or equal set bits (atleast 1) than ‘d’.

Examples :

Input : N = 5
Output : 4
for d = 1 (set bit = 1), 
    q = 5/1 = 5 (set bit = 2), count = 0.
for d = 2 (set bit = 1), 
    q = 5/2 = 2 (set bit = 1), count = 1.
for d = 3 (set bit = 2), 
    q = 5/3 = 1 (set bit = 1), count = 2.
for d = 4 (set bit = 1), 
    q = 5/4 = 1 (set bit = 1), count = 3.
for d = 5 (set bit = 2), 
    q = 5/5 = 1 (set bit = 1), count = 4.

Input : N = 3
Output : 2

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

Observe, for all d > n, q have 0 set bit so there is no point to go above n.
Also, for n/2 < d <= n, it will return q = 1, which have 1 set bit which will always be less
than or equal to d. And, minimum possible value of d is 1. So, possible value of d is from 1 to n.
Now, observe on dividing n by d, where 1 <= d <= n, it will give q in sorted order (decreasing).
So, with increasing d we will get decreasing q. So, we get the two sorted sequence, one for increasing d
and other for decreasing q. Now, observe initially the number of set bit of d is greater than q but after
a point number of set bit of d is less than or equal to q.
So, our problem reduce to find that point i.e. ‘x’ for the value of d, from 1 <= d <= n, such that q have less than or equal set bit to d.
So, count of possible d such that q have less than or equal set bit to d will be equal to n – x + 1.

Below is the implementation of this approach :

C++

// C++ Program to find number of Divisors  
// which on integer division produce quotient  
// having less set bit than divisor 
#include <bits/stdc++.h> 
using namespace std; 
  
// Return the count of set bit. 
int bit(int x) 
{ 
    int ans = 0; 
  
    while (x) { 
        x /= 2; 
        ans++; 
    } 
  
    return ans; 
} 
  
// check if q and d have same number of set bit. 
bool check(int d, int x) 
{ 
    if (bit(x / d) <= bit(d)) 
        return true; 
  
    return false; 
} 
  
// Binary Search to find the point at which 
// number of set in q is less than or equal to d. 
int bs(int n) 
{ 
    int l = 1, r = sqrt(n); 
  
    // while left index is less than right index 
    while (l < r) { 
        // finding the middle. 
        int m = (l + r) / 2; 
  
        // check if q and d have same number of 
       // set it or not. 
        if (check(m, n)) 
            r = m; 
        else
            l = m + 1; 
    } 
  
    if (!check(l, n)) 
        return l + 1; 
  
    else
        return l; 
} 
  
int countDivisor(int n) 
{ 
    return n - bs(n) + 1; 
} 
// Driven Program 
int main() 
{ 
    int n = 5; 
    cout << countDivisor(n) << endl; 
  
    return 0; 
} 

Java

// Java Program to find number  
// of Divisors which on integer  
// division produce quotient  
// having less set bit than divisor 
import java .io.*; 
  
class GFG  
{ 
  
// Return the count of set bit. 
static int bit(int x) 
{ 
    int ans = 0; 
    while (x > 0) 
    { 
        x /= 2; 
        ans++; 
    } 
  
    return ans; 
} 
  
// check if q and d have  
// same number of set bit. 
static boolean check(int d, int x) 
{ 
    if (bit(x / d) <= bit(d)) 
        return true; 
  
    return false; 
} 
  
// Binary Search to find  
// the point at which 
// number of set in q is  
// less than or equal to d. 
static int bs(int n) 
{ 
    int l = 1, r = (int)Math.sqrt(n); 
  
    // while left index is  
    // less than right index 
    while (l < r)  
    { 
        // finding the middle. 
        int m = (l + r) / 2; 
  
    // check if q and d have 
    // same number of 
    // set it or not. 
        if (check(m, n)) 
            r = m; 
        else
            l = m + 1; 
    } 
  
    if (!check(l, n)) 
        return l + 1; 
  
    else
        return l; 
} 
  
static int countDivisor(int n) 
{ 
    return n - bs(n) + 1; 
} 
  
    // Driver Code 
    static public void main (String[] args) 
    { 
        int n = 5; 
        System.out.println(countDivisor(n)); 
    } 
} 
  
// This code is contributed by anuj_67. 

Python3

# Python3 Program to find number 
# of Divisors which on integer 
# division produce quotient  
# having less set bit than divisor 
import math 
# Return the count of set bit. 
def bit(x) : 
    ans = 0
    while (x) : 
        x /= 2
        ans = ans + 1
  
    return ans 
  
# check if q and d have 
# same number of set bit. 
def check(d, x) : 
    if (bit(x /d) <= bit(d)) : 
        return True
  
    return False; 
      
# Binary Search to find 
# the point at which 
# number of set in q is  
# less than or equal to d. 
def bs(n) : 
    l = 1
    r = int(math.sqrt(n)) 
  
    # while left index is less  
    # than right index 
    while (l < r) :          
        # finding the middle. 
        m = int((l + r) / 2) 
  
        # check if q and d have  
        # same number of 
        # set it or not. 
        if (check(m, n)) : 
            r = m 
        else : 
            l = m + 1
  
    if (check(l, n) == False) : 
        return math.floor(l + 1) 
    else : 
        return math.floor(l) 
  
def countDivisor(n) : 
    return n - bs(n) + 1
  
# Driver Code 
n = 5
print (countDivisor(n)) 
  
# This code is contributed by  
# Manish Shaw (manishshaw1) 

C#

// C# Program to find number of Divisors  
// which on integer division produce quotient  
// having less set bit than divisor 
using System; 
class GFG { 
  
// Return the count of set bit. 
static int bit(int x) 
{ 
    int ans = 0; 
    while (x > 0) { 
        x /= 2; 
        ans++; 
    } 
  
    return ans; 
} 
  
// check if q and d have  
// same number of set bit. 
static bool check(int d, int x) 
{ 
    if (bit(x / d) <= bit(d)) 
        return true; 
  
    return false; 
} 
  
// Binary Search to find  
// the point at which 
// number of set in q is  
// less than or equal to d. 
static int bs(int n) 
{ 
    int l = 1, r = (int)Math.Sqrt(n); 
  
    // while left index is  
    // less than right index 
    while (l < r) { 
        // finding the middle. 
        int m = (l + r) / 2; 
  
    // check if q and d have 
    // same number of 
    // set it or not. 
        if (check(m, n)) 
            r = m; 
        else
            l = m + 1; 
    } 
  
    if (!check(l, n)) 
        return l + 1; 
  
    else
        return l; 
} 
  
static int countDivisor(int n) 
{ 
    return n - bs(n) + 1; 
} 
  
    // Driver Code 
    static public void Main () 
    { 
        int n = 5; 
        Console.WriteLine(countDivisor(n)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP Program to find number of Divisors  
// which on integer division produce quotient  
// having less set bit than divisor 
  
// Return the count of set bit. 
function bit( $x) 
{ 
    $ans = 0; 
  
    while ($x)  
    { 
        $x /= 2; 
        $ans++; 
    } 
  
    return $ans; 
} 
  
// check if q and d have 
// same number of set bit. 
function check( $d, $x) 
{ 
    if (bit($x /$d) <= bit($d)) 
        return true; 
  
    return false; 
} 
  
// Binary Search to find 
// the point at which 
// number of set in q is  
// less than or equal to d. 
function bs(int $n) 
{ 
    $l = 1; 
    $r = sqrt($n); 
  
    // while left index is less  
    // than right index 
    while ($l < $r)  
    { 
          
        // finding the middle. 
        $m = ($l + $r) / 2; 
  
        // check if q and d have  
        // same number of 
        // set it or not. 
        if (check($m, $n)) 
            $r = $m; 
        else
            $l = $m + 1; 
    } 
  
    if (!check($l, $n)) 
        return floor($l + 1); 
  
    else
        return floor($l); 
} 
  
function countDivisor( $n) 
{ 
    return $n - bs($n) + 1; 
} 
  
    // Driver Code 
    $n = 5; 
    echo countDivisor($n) ; 
  
// This code is contributed by anuj_67. 
?> 

Output:

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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