Sum of n numbers with exactly 2 active bits set
In this article, we will explore a program to find the sum of n numbers that have exactly two active bits set in their binary representation. An "active bit" refers to a set bit (1) in the binary representation of a number. For example, the number 3 has two active bits (11 in binary), while the number 5 also has two active bits (101 in binary).
Problem Statement
The task is to find the sum of all numbers with exactly two active bits among the first n positive integers.
Example with n = 10: To better understand the problem, let's find the sum of all numbers with exactly two active bits among the first 10 positive integers.
-
Binary representation of numbers with exactly two active bits:
3 (11) 5 (101) 6 (110) 9 (1001) 10 (1010) 12 (1100) 17 (10001) 18 (10010) 20 (10100) 24 (11000)
-
Sum of the above numbers: 3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 = 124
Another example, if n = 6, the numbers with exactly 2 active bits set in their binary representation are:
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100Result 45 = [3+5+6+9+10+12].
Pseudocode
Before diving into the algorithm, let's define the pseudocode for a function named twoActiveBits:
twoActiveBits(n):
sum = 0
num = 1
count = n
temp = 1
while count > 0:
num = num << 1
while temp < num and count > 0:
sum = sum + (num | temp)
temp = temp << 1
count = count - 1
temp = 1
print "Given n : ", n
print "Result : ", sum
Algorithm Explanation
- Initialize the variables
sum,num,count, andtemp. - Loop while
count > 0. - Left-shift
numby 1 (equivalent to multiplyingnumby 2) to get the next number with two active bits. - Inner loop: While
tempis less thannumandcount > 0, calculate the sum by performing a bitwise OR operation betweennumandtempand add it to thesum. - Left-shift
tempby 1 (equivalent to multiplyingtempby 2) to get the next number with a single active bit. - Decrement
countby 1 to keep track of the remaining numbers to be processed. - Repeat steps 3-6 until all required numbers have been added to the
sum. - Print the value of
nand the final calculatedsum.
Code Solution
// C Program for
// Sum of n numbers with exactly 2 active bits set
#include <stdio.h>
void twoActiveBits(int n)
{
int count = n;
int temp = 1;
int num = 1;
int sum = 0;
while (count > 0)
{
num = (num << 1);
while (temp < num && count > 0)
{
sum += (num | temp);
temp = temp << 1;
count--;
}
temp = 1;
}
// Display number of element
printf(" Given n : %d\n", n);
// Display calculate sum
printf(" Result : %d\n", sum);
}
int main(int argc, char
const *argv[])
{
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
twoActiveBits(10);
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
twoActiveBits(30);
return 0;
}Output
Given n : 10
Result : 124
Given n : 30
Result : 2300/*
Java Program
Sum of n numbers with exactly 2 active bits set
*/
public class BinaryBits
{
public void twoActiveBits(int n)
{
int count = n;
int temp = 1;
int num = 1;
int sum = 0;
while (count > 0)
{
num = (num << 1);
while (temp < num && count > 0)
{
sum += (num | temp);
temp = temp << 1;
count--;
}
temp = 1;
}
// Display number of element
System.out.println(" Given n : " + n);
// Display calculate sum
System.out.println(" Result : " + sum);
}
public static void main(String[] args)
{
BinaryBits task = new BinaryBits();
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
task.twoActiveBits(10);
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
task.twoActiveBits(30);
}
}Output
Given n : 10
Result : 124
Given n : 30
Result : 2300// Include header file
#include <iostream>
using namespace std;
/*
C++ Program
Sum of n numbers with exactly 2 active bits set
*/
class BinaryBits
{
public: void twoActiveBits(int n)
{
int count = n;
int temp = 1;
int num = 1;
int sum = 0;
while (count > 0)
{
num = (num << 1);
while (temp < num && count > 0)
{
sum += (num | temp);
temp = temp << 1;
count--;
}
temp = 1;
}
// Display number of element
cout << " Given n : " << n << endl;
// Display calculate sum
cout << " Result : " << sum << endl;
}
};
int main()
{
BinaryBits *task = new BinaryBits();
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
task->twoActiveBits(10);
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
task->twoActiveBits(30);
return 0;
}Output
Given n : 10
Result : 124
Given n : 30
Result : 2300// Include namespace system
using System;
/*
Csharp Program
Sum of n numbers with exactly 2 active bits set
*/
public class BinaryBits
{
public void twoActiveBits(int n)
{
int count = n;
int temp = 1;
int num = 1;
int sum = 0;
while (count > 0)
{
num = (num << 1);
while (temp < num && count > 0)
{
sum += (num | temp);
temp = temp << 1;
count--;
}
temp = 1;
}
// Display number of element
Console.WriteLine(" Given n : " + n);
// Display calculate sum
Console.WriteLine(" Result : " + sum);
}
public static void Main(String[] args)
{
BinaryBits task = new BinaryBits();
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
task.twoActiveBits(10);
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
task.twoActiveBits(30);
}
}Output
Given n : 10
Result : 124
Given n : 30
Result : 2300package main
import "fmt"
/*
Go Program
Sum of n numbers with exactly 2 active bits set
*/
type BinaryBits struct {}
func getBinaryBits() * BinaryBits {
var me *BinaryBits = &BinaryBits {}
return me
}
func(this BinaryBits) twoActiveBits(n int) {
var count int = n
var temp int = 1
var num int = 1
var sum int = 0
for (count > 0) {
num = (num << 1)
for (temp < num && count > 0) {
sum += (num | temp)
temp = temp << 1
count--
}
temp = 1
}
// Display number of element
fmt.Println(" Given n : ", n)
// Display calculate sum
fmt.Println(" Result : ", sum)
}
func main() {
var task * BinaryBits = getBinaryBits()
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
task.twoActiveBits(10)
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
task.twoActiveBits(30)
}Output
Given n : 10
Result : 124
Given n : 30
Result : 2300<?php
/*
Php Program
Sum of n numbers with exactly 2 active bits set
*/
class BinaryBits
{
public function twoActiveBits($n)
{
$count = $n;
$temp = 1;
$num = 1;
$sum = 0;
while ($count > 0)
{
$num = ($num << 1);
while ($temp < $num && $count > 0)
{
$sum += ($num | $temp);
$temp = $temp << 1;
$count--;
}
$temp = 1;
}
// Display number of element
echo(" Given n : ".$n.
"\n");
// Display calculate sum
echo(" Result : ".$sum.
"\n");
}
}
function main()
{
$task = new BinaryBits();
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
$task->twoActiveBits(10);
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
$task->twoActiveBits(30);
}
main();Output
Given n : 10
Result : 124
Given n : 30
Result : 2300/*
Node JS Program
Sum of n numbers with exactly 2 active bits set
*/
class BinaryBits
{
twoActiveBits(n)
{
var count = n;
var temp = 1;
var num = 1;
var sum = 0;
while (count > 0)
{
num = (num << 1);
while (temp < num && count > 0)
{
sum += (num | temp);
temp = temp << 1;
count--;
}
temp = 1;
}
// Display number of element
console.log(" Given n : " + n);
// Display calculate sum
console.log(" Result : " + sum);
}
}
function main()
{
var task = new BinaryBits();
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
task.twoActiveBits(10);
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
task.twoActiveBits(30);
}
main();Output
Given n : 10
Result : 124
Given n : 30
Result : 2300# Python 3 Program
# Sum of n numbers with exactly 2 active bits set
class BinaryBits :
def twoActiveBits(self, n) :
count = n
temp = 1
num = 1
sum = 0
while (count > 0) :
num = (num << 1)
while (temp < num and count > 0) :
sum += (num | temp)
temp = temp << 1
count -= 1
temp = 1
# Display number of element
print(" Given n : ", n)
# Display calculate sum
print(" Result : ", sum)
def main() :
task = BinaryBits()
# Test A
# n = 10
# --------------
# Binary
# 3 11
# 5 101
# 6 110
# 9 1001
# 10 1010
# 12 1100
# 17 10001
# 18 10010
# 20 10100
# 24 11000
# [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
# ------------------------------------------------
# Sum : 124
task.twoActiveBits(10)
# Test B
# num = 30
# --------------
# Binary
# 3 11
# 5 101
# 6 110
# 9 1001
# 10 1010
# 12 1100
# 17 10001
# 18 10010
# 20 10100
# 24 11000
# 33 100001
# 34 100010
# 36 100100
# 40 101000
# 48 110000
# 65 1000001
# 66 1000010
# 68 1000100
# 72 1001000
# 80 1010000
# 96 1100000
# 129 10000001
# 130 10000010
# 132 10000100
# 136 10001000
# 144 10010000
# 160 10100000
# 192 11000000
# 257 100000001
# 258 100000010
# [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
# 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
# 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
# ---------------------------------------------------
# Sum : 2300
task.twoActiveBits(30)
if __name__ == "__main__": main()Output
Given n : 10
Result : 124
Given n : 30
Result : 2300# Ruby Program
# Sum of n numbers with exactly 2 active bits set
class BinaryBits
def twoActiveBits(n)
count = n
temp = 1
num = 1
sum = 0
while (count > 0)
num = (num << 1)
while (temp < num && count > 0)
sum += (num | temp)
temp = temp << 1
count -= 1
end
temp = 1
end
# Display number of element
print(" Given n : ", n, "\n")
# Display calculate sum
print(" Result : ", sum, "\n")
end
end
def main()
task = BinaryBits.new()
# Test A
# n = 10
# --------------
# Binary
# 3 11
# 5 101
# 6 110
# 9 1001
# 10 1010
# 12 1100
# 17 10001
# 18 10010
# 20 10100
# 24 11000
# [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
# ------------------------------------------------
# Sum : 124
task.twoActiveBits(10)
# Test B
# num = 30
# --------------
# Binary
# 3 11
# 5 101
# 6 110
# 9 1001
# 10 1010
# 12 1100
# 17 10001
# 18 10010
# 20 10100
# 24 11000
# 33 100001
# 34 100010
# 36 100100
# 40 101000
# 48 110000
# 65 1000001
# 66 1000010
# 68 1000100
# 72 1001000
# 80 1010000
# 96 1100000
# 129 10000001
# 130 10000010
# 132 10000100
# 136 10001000
# 144 10010000
# 160 10100000
# 192 11000000
# 257 100000001
# 258 100000010
# [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
# 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
# 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
# ---------------------------------------------------
# Sum : 2300
task.twoActiveBits(30)
end
main()Output
Given n : 10
Result : 124
Given n : 30
Result : 2300
/*
Scala Program
Sum of n numbers with exactly 2 active bits set
*/
class BinaryBits()
{
def twoActiveBits(n: Int): Unit = {
var count: Int = n;
var temp: Int = 1;
var num: Int = 1;
var sum: Int = 0;
while (count > 0)
{
num = (num << 1);
while (temp < num && count > 0)
{
sum += (num | temp);
temp = temp << 1;
count -= 1;
}
temp = 1;
}
// Display number of element
println(" Given n : " + n);
// Display calculate sum
println(" Result : " + sum);
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: BinaryBits = new BinaryBits();
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
task.twoActiveBits(10);
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
task.twoActiveBits(30);
}
}Output
Given n : 10
Result : 124
Given n : 30
Result : 2300/*
Swift 4 Program
Sum of n numbers with exactly 2 active bits set
*/
class BinaryBits
{
func twoActiveBits(_ n: Int)
{
var count: Int = n;
var temp: Int = 1;
var num: Int = 1;
var sum: Int = 0;
while (count > 0)
{
num = (num << 1);
while (temp < num && count > 0)
{
sum += (num | temp);
temp = temp << 1;
count -= 1;
}
temp = 1;
}
// Display number of element
print(" Given n : ", n);
// Display calculate sum
print(" Result : ", sum);
}
}
func main()
{
let task: BinaryBits = BinaryBits();
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
task.twoActiveBits(10);
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
task.twoActiveBits(30);
}
main();Output
Given n : 10
Result : 124
Given n : 30
Result : 2300/*
Kotlin Program
Sum of n numbers with exactly 2 active bits set
*/
class BinaryBits
{
fun twoActiveBits(n: Int): Unit
{
var count: Int = n;
var temp: Int = 1;
var num: Int = 1;
var sum: Int = 0;
while (count > 0)
{
num = (num shl 1);
while (temp < num && count > 0)
{
sum += (num or temp);
temp = temp shl 1;
count -= 1;
}
temp = 1;
}
// Display number of element
println(" Given n : " + n);
// Display calculate sum
println(" Result : " + sum);
}
}
fun main(args: Array < String > ): Unit
{
val task: BinaryBits = BinaryBits();
// Test A
// n = 10
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24]
// ------------------------------------------------
// Sum : 124
task.twoActiveBits(10);
// Test B
// num = 30
// --------------
// Binary
// 3 11
// 5 101
// 6 110
// 9 1001
// 10 1010
// 12 1100
// 17 10001
// 18 10010
// 20 10100
// 24 11000
// 33 100001
// 34 100010
// 36 100100
// 40 101000
// 48 110000
// 65 1000001
// 66 1000010
// 68 1000100
// 72 1001000
// 80 1010000
// 96 1100000
// 129 10000001
// 130 10000010
// 132 10000100
// 136 10001000
// 144 10010000
// 160 10100000
// 192 11000000
// 257 100000001
// 258 100000010
// [3 + 5 + 6 + 9 + 10 + 12 + 17 + 18 + 20 + 24 + 33 +
// 34 + 36 + 40 + 48 + 65 + 66 + 68 + 72 + 80 + 96 +
// 129 + 130 + 132 + 136 + 144 + 160 + 192 + 257 + 258]
// ---------------------------------------------------
// Sum : 2300
task.twoActiveBits(30);
}Output
Given n : 10
Result : 124
Given n : 30
Result : 2300Resultant Output Explanation
For the test cases with n = 10 and n = 30, we use the twoActiveBits function
to find the sums of numbers with exactly two active bits among the first 10 and 30 positive integers, respectively.
-
twoActiveBits(10)produces the output: Given n : 10 Result : 124 -
twoActiveBits(30)produces the output: Given n : 30 Result : 2300
Time Complexity
The time complexity of the algorithm depends on the value of n. Let k be the number of
bits needed to represent the largest number among the first n numbers. In the worst case, the outer
loop runs for n iterations, and the inner loop runs for k iterations in each iteration of
the outer loop. Therefore, the overall time complexity is O(n * k).
Finally
In this article, we have discussed a program to find the sum of numbers with exactly two active bits among the first n positive integers. We explained the problem, provided the pseudocode and algorithm, and demonstrated the outputs for two test cases. The algorithm's time complexity is also analyzed, allowing us to understand the efficiency of the solution for different input values.
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