Lucky alive person in a circle | Set – 2

0 0 17 minutes read

Given that N person (numbered 1 to N) standing as to form a circle. They all have the gun in their hand which is pointed to their leftmost Partner.

Every one shoots such that 1 shoot 2, 3 shoots 4, 5 shoots 6 …. (N-1)the shoot N (if N is even otherwise N shoots 1).
Again on the second iteration, they shoot the rest of remains as above mentioned logic (now for n as even, 1 will shoot to 3, 5 will shoot to 7 and so on).

The task is to find which person is the luckiest(didn’t die)?

Examples:

Input: N = 3
Output: 3
As N = 3 then 1 will shoot 2, 3 will shoot 1 hence 3 is the luckiest person.

Input: N = 8
Output: 1
Here as N = 8, 1 will shoot 1, 3 will shoot 4, 5 will shoot 6, 7 will shoot 8, Again 1 will shoot 3, 5 will shoot 7, Again 1 will shoot 5 and hence 1 is the luckiest person.

This problem has already been discussed in Lucky alive person in a circle | Code Solution to sword puzzle. In this post, a different approach is discussed.

Approach:

Take the Binary Equivalent of N.
Find its 1’s compliment and convert its equal decimal number N`.
find |N – N`|.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to convert string to number
int stringToNum(string s)
{
 
    // object from the class stringstream
    stringstream geek(s);
 
    // The object has the value 12345 and stream
    // it to the integer x
    int x = 0;
    geek >> x;
 
    return x;
}
 
// Function to convert binary to decimal
int binaryToDecimal(string n)
{
    int num = stringToNum(n);
    int dec_value = 0;
 
    // Initializing base value to 1, i.e 2^0
    int base = 1;
 
    int temp = num;
    while (temp) {
        int last_digit = temp % 10;
        temp = temp / 10;
 
        dec_value += last_digit * base;
 
        base = base * 2;
    }
 
    return dec_value;
}
 
string itoa(int num, string str, int base)
{
    int i = 0;
    bool isNegative = false;
 
    /* Handle 0 explicitly, otherwise 
    empty string is printed for 0 */
    if (num == 0) {
        str[i++] = '0';
        return str;
    }
 
    // In standard itoa(), negative numbers
    // are handled only with base 10.
    // Otherwise numbers are considered unsigned.
    if (num < 0 && base == 10) {
        isNegative = true;
        num = -num;
    }
 
    // Process individual digits
    while (num != 0) {
        int rem = num % base;
        str[i++] = (rem > 9) ? (rem - 10) + 'a' : rem + '0';
        num = num / base;
    }
 
    // If the number is negative, append '-'
    if (isNegative)
        str[i++] = '-';
 
    // Reverse the string
    reverse(str.begin(), str.end());
 
    return str;
}
 
char flip(char c)
{
    return (c == '0') ? '1' : '0';
}
 
// Function to find the ones complement
string onesComplement(string bin)
{
    int n = bin.length(), i;
 
    string ones = "";
 
    // for ones complement flip every bit
    for (i = 0; i < n; i++)
        ones += flip(bin[i]);
 
    return ones;
}
 
// Driver code
int main()
{
    // Taking the number of a person
    // standing in a circle.
    int N = 3;
    string arr = "";
 
    // Storing the binary equivalent in a string.
    string ans(itoa(N, arr, 2));
 
    // taking one's compelement and
    // convert it to decimal value
    int N_dash = binaryToDecimal(onesComplement(ans));
 
    int luckiest_person = N - N_dash;
 
    cout << luckiest_person;
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
 
public class Main {
 
  // Function to convert string to number
  public static int stringToNum(String s)
  {
    // The object has the value 12345 and stream
    // it to the integer x
    int x = Integer.parseInt(s);
    return x;
  }
 
  // Function to convert binary to decimal
  public static int binaryToDecimal(String n)
  {
    int num = stringToNum(n);
    int dec_value = 0;
 
    // Initializing base value to 1, i.e 2^0
    int base = 1;
    int temp = num;
    while (temp != 0) {
      int last_digit = temp % 10;
      temp = temp / 10;
      dec_value += last_digit * base;
      base = base * 2;
    }
    return dec_value;
  }
 
  public static String itoa(int num, String str, int base)
  {
    int i = 0;
    boolean isNegative = false;
 
    /* Handle 0 explicitly, otherwise 
        empty string is printed for 0 */
    if (num == 0) {
      str += '0';
      return str;
    }
 
    // In standard itoa(), negative numbers
    // are handled only with base 10.
    // Otherwise numbers are considered unsigned.
    if (num < 0 && base == 10) {
      isNegative = true;
      num = -num;
    }
 
    // Process individual digits
    while (num != 0) {
      int rem = num % base;
      str += (rem > 9) ? (char)(rem - 10 + 'a')
        : (char)(rem + '0');
      num = num / base;
    }
 
    // If the number is negative, append '-'
    if (isNegative)
      str += '-';
    StringBuilder sb = new StringBuilder(str);
 
    // Reverse the string
    sb.reverse();
    return sb.toString();
  }
 
  public static char flip(char c)
  {
    return (c == '0') ? '1' : '0';
  }
 
  // Function to find the ones complement
  public static String onesComplement(String bin)
  {
    int n = bin.length(), i;
    String ones = "";
 
    // for ones complement flip every bit
    for (i = 0; i < n; i++)
      ones += flip(bin.charAt(i));
    return ones;
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    // Taking the number of a person
    // standing in a circle.
    int N = 3;
    String arr = "";
 
    // Storing the binary equivalent in a string.
    String ans = itoa(N, arr, 2);
 
    // taking one's compelement and
    // convert it to decimal value
    int N_dash = binaryToDecimal(onesComplement(ans));
    int luckiest_person = N - N_dash;
    System.out.println(luckiest_person);
  }
}
// This code is contributed by Prajwal Kandekar

Python3

# Function to convert string to number
def string_to_num(s):
    return int(s)
 
# Function to convert binary to decimal integer
def binary_to_decimal(n):
    num = string_to_num(n)
    dec_value = 0
     
    # Initializing base value to 1, i.e 2^0
    base = 1
    temp = num
    while temp:
        last_digit = temp % 10
        temp = temp // 10
        dec_value += last_digit * base
        base = base * 2
    return dec_value 
 
def itoa(num, base):
    i = 0
    is_negative = False
    str_list = []
 
    # Handle 0 explicitly, otherwise empty string is printed for 0
    if num == 0:
        str_list.append('0')
        return ''.join(str_list)
 
    # In standard itoa(), negative numbers are handled only with base 10. Otherwise numbers are considered unsigned.
    if num < 0 and base == 10:
        is_negative = True
        num = -num
 
    # Process individual digits
    while num != 0:
        rem = num % base
        str_list.append(str(rem) if rem <= 9 else chr(ord('A') + rem - 10))
        num //= base
 
    # If the number is negative, append '-'
    if is_negative:
        str_list.append('-')
     
    # Reverse the string
    str_list.reverse()
     
    return ''.join(str_list)
 
def flip(c):
    return '1' if c == '0' else '0'
 
# Function to find the ones complement
def ones_complement(bin):
    n = len(bin)
    ones = ""
    # for ones complement flip every bit
    for i in range(n):
        ones += flip(bin[i])
    return ones
 
# Driver code
 
# Taking the number of a person standing in a circle.
N = 3
 
# Storing the binary equivalent in a string.
ans = itoa(N, 2)
 
# Taking one's complement and convert it to decimal value
N_dash = binary_to_decimal(ones_complement(ans))
 
luckiest_person = N - N_dash
 
print(luckiest_person)

C#

using System;
 
class MainClass {
 
  // Function to convert string to number
  public static int stringToNum(string s)
  {
    // The object has the value 12345 and stream
    // it to the integer x
    int x = int.Parse(s);
    return x;
  }
 
  // Function to convert binary to decimal
  public static int binaryToDecimal(string n)
  {
    int num = stringToNum(n);
    int dec_value = 0;
 
    // Initializing base value to 1, i.e 2^0
    int base_value = 1;
    int temp = num;
    while (temp != 0) {
      int last_digit = temp % 10;
      temp = temp / 10;
      dec_value += last_digit * base_value;
      base_value = base_value * 2;
    }
    return dec_value;
  }
 
  public static string itoa(int num, string str, int base_value)
  {
    int i = 0;
    bool isNegative = false;
 
    /* Handle 0 explicitly, otherwise 
        empty string is printed for 0 */
    if (num == 0) {
      str += '0';
      return str;
    }
 
    // In standard itoa(), negative numbers
    // are handled only with base 10.
    // Otherwise numbers are considered unsigned.
    if (num < 0 && base_value == 10) {
      isNegative = true;
      num = -num;
    }
 
    // Process individual digits
    while (num != 0) {
      int rem = num % base_value;
      str += (rem > 9) ? (char)(rem - 10 + 'a')
        : (char)(rem + '0');
      num = num / base_value;
    }
 
    // If the number is negative, append '-'
    if (isNegative)
      str += '-';
    char[] charArray = str.ToCharArray();
 
    // Reverse the string
    Array.Reverse(charArray);
    return new string(charArray);
  }
 
  public static char flip(char c)
  {
    return (c == '0') ? '1' : '0';
  }
 
  // Function to find the ones complement
  public static string onesComplement(string bin)
  {
    int n = bin.Length, i;
    string ones = "";
 
    // for ones complement flip every bit
    for (i = 0; i < n; i++)
      ones += flip(bin[i]);
    return ones;
  }
 
  // Driver code
  public static void Main()
  {
 
    // Taking the number of a person
    // standing in a circle.
    int N = 3;
    string arr = "";
 
    // Storing the binary equivalent in a string.
    string ans = itoa(N, arr, 2);
 
    // taking one's compelement and
    // convert it to decimal value
    int N_dash = binaryToDecimal(onesComplement(ans));
    int luckiest_person = N - N_dash;
    Console.WriteLine(luckiest_person);
  }
}
// This code is contributed by Prajwal Kandekar

Javascript

// JavaScript implementation of the above approach
 
// Function to convert string to number
function stringToNum(s)
{
    return parseInt(s);
}
 
// Function to convert binary to decimal integer
function binaryToDecimal(n)
{
     
    let num = stringToNum(n);
    let dec_value = 0;
 
    // Initializing base value to 1, i.e 2^0
    let base = 1;
 
    let temp = num;
    while (temp) {
        let last_digit = temp % 10;
        temp = Math.floor(temp / 10);
 
        dec_value += last_digit * base;
 
        base = base * 2;
    }
 
    return dec_value; 
}
 
function itoa(num, str, base)
{
    let i = 0;
    let isNegative = false;
    str = str.split("");
 
    /* Handle 0 explicitly, otherwise 
    empty string is printed for 0 */
    if (num == 0) {
        str[i++] = '0';
        return str.join("");
    }
 
    // In standard itoa(), negative numbers
    // are handled only with base 10.
    // Otherwise numbers are considered unsigned.
    if (num < 0 && base == 10) {
        isNegative = true;
        num = -num;
    }
 
    // Process individual digits
    while (num != 0) {
        let rem = num % base;
        str[i++] = (rem > 9) ? (rem - 10): rem;
        num = Math.floor(num / base);
    }
 
    // If the number is negative, append '-'
    if (isNegative)
        str[i++] = '-';
     
    // Reverse the string
    str.reverse();
     
 
    return str.join("");
}
 
function flip(c)
{
    return (c == '0') ? '1' : '0';
}
 
// Function to find the ones complement
function onesComplement(bin)
{
    let n = bin.length;
    let i;
 
    let ones = "";
 
    // for ones complement flip every bit
    for (i = 0; i < n; i++)
        ones += flip(bin[i]);
 
    return ones;
}
 
// Driver code
 
// Taking the number of a person
// standing in a circle.
let N = 3;
let arr = "";
 
// Storing the binary equivalent in a string.
let ans = itoa(N, arr, 2);
 
// taking one's compelement and
// convert it to decimal value
let N_dash = binaryToDecimal(onesComplement(ans));
 
let luckiest_person = N - N_dash;
 
console.log(luckiest_person);
 
// This code is contributed by phasing17

Output

Alternate Shorter Implementation :
The approach used here is same.

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
int luckiest_person(int n)
{
    // to calculate the number of bits in
    // the binary equivalent of n
    int len = log2(n) + 1;
 
    // Finding complement by inverting the
    // bits one by one from last
    int n2 = n;
    for (int i = 0; i < len; i++) {
 
        // XOR of n2 with (1<<i) to flip
        // the last (i+1)th bit of binary equivalent of n2
        n2 = n2 ^ (1 << i);
    }
 
    // n2 will be the one's complement of n
    return abs(n - n2);
}
int main()
{
    int N = 3;
    int lucky_p = luckiest_person(N);
    cout << lucky_p;
 
    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;
 
class GFG {
 
    static int luckiest_person(int n)
    {
 
        // To calculate the number of bits in
        // the binary equivalent of n
        int len = (int)(Math.log(n) / Math.log(2)) + 1;
 
        // Finding complement by inverting the
        // bits one by one from last
        int n2 = n;
        for (int i = 0; i < len; i++) {
 
            // XOR of n2 with (1<<i) to flip the last
            // (i+1)th bit of binary equivalent of n2
            n2 = n2 ^ (1 << i);
        }
 
        // n2 will be the one's complement of n
        return Math.abs(n - n2);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 3;
        int lucky_p = luckiest_person(N);
 
        System.out.println(lucky_p);
    }
}
 
// This code is contributed by rishavmahato348

Python3

# Python3 implementation of the above approach
import math
 
 
def luckiest_person(n):
 
    # to calculate the number of bits in
    # the binary equivalent of n
    len_ = int(math.log(n, 2)) + 1
 
    # Finding complement by inverting the
    # bits one by one from last
    n2 = n
    for i in range(len_):
 
        # XOR of n2 with (1<<i) to flip
        # the last (i+1)th bit of binary equivalent of n2
        n2 = n2 ^ (1 << i)
 
    # n2 will be the one's complement of n
    return abs(n - n2)
 
 
# Driver Code
N = 3
lucky_p = luckiest_person(N)
print(lucky_p)
 
# this code is contributed by phasing17

C#

// C# implementation of the above approach
using System;
 
class GFG {
 
    static int luckiest_person(int n)
    {
 
        // To calculate the number of bits in
        // the binary equivalent of n
        int len = (int)(Math.Log(n) / Math.Log(2)) + 1;
 
        // Finding complement by inverting the
        // bits one by one from last
        int n2 = n;
        for (int i = 0; i < len; i++) {
 
            // XOR of n2 with (1<<i) to flip the last
            // (i+1)th bit of binary equivalent of n2
            n2 = n2 ^ (1 << i);
        }
 
        // n2 will be the one's complement of n
        return Math.Abs(n - n2);
    }
 
    // Driver code
    public static void Main()
    {
        int N = 3;
        int lucky_p = luckiest_person(N);
 
        Console.Write(lucky_p);
    }
}
 
// This code is contributed by subhammahato348

Javascript

<script>
 
// JavaScript implementation of the above approach
 
function luckiest_person(n)
{
    // to calculate the number of bits in
    // the binary equivalent of n
    let len = parseInt(Math.log(n) / Math.log(2)) + 1; 
 
    // Finding complement by inverting the
    // bits one by one from last
    let n2 = n;
    for (let i = 0; i < len; i++) {
 
        // XOR of n2 with (1<<i) to flip 
        // the last (i+1)th bit of binary equivalent of n2
        n2 = n2 ^ (1 << i); 
    } 
 
    // n2 will be the one's complement of n
    return Math.abs(n - n2);
}
 
// Driver Code
    let N = 3;
    let lucky_p = luckiest_person(N);
    document.write(lucky_p);
 
</script>

Output

Alternate Implementation in O(1) : The approach used here is same, but the operations used are of constant time.

C++

// Here is a O(1) solution for this problem
// it will work for 32 bit integers]
#include <bits/stdc++.h>
using namespace std;
 
// function to find the highest power of 2
// which is less than n
int setBitNumber(int n)
{
    // Below steps set bits after
    // MSB (including MSB)
 
    // Suppose n is 273 (binary
    // is 100010001). It does following
    // 100010001 | 010001000 = 110011001
    n |= n >> 1;
 
    // This makes sure 4 bits
    // (From MSB and including MSB)
    // are set. It does following
    // 110011001 | 001100110 = 111111111
    n |= n >> 2;
 
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
 
    // Increment n by 1 so that
    // there is only one set bit
    // which is just before original
    // MSB. n now becomes 1000000000
    n = n + 1;
 
    // Return original MSB after shifting.
    // n now becomes 100000000
    return (n >> 1);
}
 
int luckiest_person(int n)
{
    // to calculate the highest number which
    // is power of 2 and less than n
    int nearestPower = setBitNumber(n);
 
    // return the correct answer as per the
    // approach in above article
    return 2 * (n - nearestPower) + 1;
}
int main()
{
    int N = 8;
    int lucky_p = luckiest_person(N);
    cout << lucky_p;
    return 0;
}
 
// This code is contributed by Ujesh Maurya

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
// Here is a O(1) solution for this problem
// it will work for 32 bit integers]
class GFG {
    static int setBitNumber(int n)
    {
 
        // Below steps set bits after
        // MSB (including MSB)
 
        // Suppose n is 273 (binary
        // is 100010001). It does following
        // 100010001 | 010001000 = 110011001
        n |= n >> 1;
 
        // This makes sure 4 bits
        // (From MSB and including MSB)
        // are set. It does following
        // 110011001 | 001100110 = 111111111
        n |= n >> 2;
 
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
 
        // Increment n by 1 so that
        // there is only one set bit
        // which is just before original
        // MSB. n now becomes 1000000000
        n = n + 1;
 
        // Return original MSB after shifting.
        // n now becomes 100000000
        return (n >> 1);
    }
 
    static int luckiest_person(int n)
    {
        // to calculate the highest number which
        // is power of 2 and less than n
        int nearestPower = setBitNumber(n);
 
        // return the correct answer as per the
        // approach in above article
        return 2 * (n - nearestPower) + 1;
    }
   
    // Driver Code
    public static void main(String[] args)
    {
        int N = 8;
        int lucky_p = luckiest_person(N);
 
        System.out.print(lucky_p);
    }
}
 
// This code is contributed by Ujesh Maurya

C#

// Here is a O(1) solution for this problem
// it will work for 32 bit integers]
using System;
 
class GFG {
 
    // function to find the highest power of 2
    // which is less than n
    static int setBitNumber(int n)
    {
        // Below steps set bits after
        // MSB (including MSB)
 
        // Suppose n is 273 (binary
        // is 100010001). It does following
        // 100010001 | 010001000 = 110011001
        n |= n >> 1;
 
        // This makes sure 4 bits
        // (From MSB and including MSB)
        // are set. It does following
        // 110011001 | 001100110 = 111111111
        n |= n >> 2;
 
        n |= n >> 4;
        n |= n >> 8;
        n |= n >> 16;
 
        // Increment n by 1 so that
        // there is only one set bit
        // which is just before original
        // MSB. n now becomes 1000000000
        n = n + 1;
 
        // Return original MSB after shifting.
        // n now becomes 100000000
        return (n >> 1);
    }
 
    static int luckiest_person(int n)
    {
 
        // to calculate the highest number which
        // is power of 2 and less than n
        int nearestPower = setBitNumber(n);
 
        // return the correct answer as per the
        // approach in above article
        return 2 * (n - nearestPower) + 1;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 8;
        int lucky_p = luckiest_person(N);
 
        Console.Write(lucky_p);
    }
}
 
// This code is contributed by Ujesh Maurya

Javascript

<script>
// Javascript program for the above approach
 
// Here is a O(1) solution for this problem
// it will work for 32 bit integers]
function setBitNumber(n) {
 
    // Below steps set bits after
    // MSB (including MSB)
 
    // Suppose n is 273 (binary
    // is 100010001). It does following
    // 100010001 | 010001000 = 110011001
    n |= n >> 1;
 
    // This makes sure 4 bits
    // (From MSB and including MSB)
    // are set. It does following
    // 110011001 | 001100110 = 111111111
    n |= n >> 2;
 
    n |= n >> 4;
    n |= n >> 8;
    n |= n >> 16;
 
    // Increment n by 1 so that
    // there is only one set bit
    // which is just before original
    // MSB. n now becomes 1000000000
    n = n + 1;
 
    // Return original MSB after shifting.
    // n now becomes 100000000
    return (n >> 1);
}
 
function luckiest_person(n) {
    // to calculate the highest number which
    // is power of 2 and less than n
    let nearestPower = setBitNumber(n);
 
    // return the correct answer as per the
    // approach in above article
    return 2 * (n - nearestPower) + 1;
}
 
// Driver Code
let N = 8;
let lucky_p = luckiest_person(N);
 
document.write(lucky_p);
 
// This code is contributed by Saurabh Jaiswal
</script>

Python3

def set_bit_number(n):
    # Below steps set bits after
    # MSB (including MSB)
 
    # Suppose n is 273 (binary
    # is 100010001). It does following
    # 100010001 | 010001000 = 110011001
    n |= n >> 1
 
    # This makes sure 4 bits
    # (From MSB and including MSB)
    # are set. It does following
    # 110011001 | 001100110 = 111111111
    n |= n >> 2
 
    n |= n >> 4
    n |= n >> 8
    n |= n >> 16
 
    # Increment n by 1 so that
    # there is only one set bit
    # which is just before original
    # MSB. n now becomes 1000000000
    n = n + 1
 
    # Return original MSB after shifting.
    # n now becomes 100000000
    return (n >> 1)
 
 
def luckiest_person(n):
    # to calculate the highest number which
    # is power of 2 and less than n
    nearest_power = set_bit_number(n)
 
    # return the correct answer as per the
    # approach in above article
    return 2 * (n - nearest_power) + 1
 
 
# Driver Code
if __name__ == "__main__":
    N = 8
    lucky_p = luckiest_person(N)
    print(lucky_p)

Output

Another approach in O(1) : On the basis of the pattern that forms in given question, which is displayed in following table.

n		1		2	3		4	5	6	7		8	9	10	11	12	13	14	15		16
		1		1	3		1	3	5	7		1	3	5	7	9	11	13	15		1

C++

#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
// Driven code
int find(int n)
{
    // Obtain number less n in 2's power
    int twospower = pow(2, (int)log2(n));
 
    // Find p-position of odd number, in odd series
    int diff = n - twospower + 1;
 
    // Value of pth odd number
    int diffthodd = (2 * diff) - 1;
 
    return diffthodd;
}
// Driver code
int main()
{
    int n = 5;
    cout << find(n);
    return 0;
}
 
// This code is contributed by Dharmik Parmar

Java

import java.util.*;
 
public class GFG {
  public static void main(String[] args) {
    int n = 5;
    System.out.println(find(n));
  }
 
  public static int find(int n) {
 
    // Obtain number less n in 2's power
    int twospower = (int) Math.pow(2, (int) (Math.log(n) / Math.log(2)));
 
    // Find p-position of odd number, in odd series
    int diff = n - twospower + 1;
 
    // Value of pth odd number
    int diffthodd = (2 * diff) - 1;
 
    return diffthodd;
  }
}

Python3

import math
 
def find(n):
    # Obtain number less n in 2's power
    twospower = int(math.pow(2, int(math.log2(n))))
 
    # Find p-position of odd number, in odd series
    diff = n - twospower + 1
 
    # Value of pth odd number
    diffthodd = (2 * diff) - 1
 
    return diffthodd
 
# Driver code
n = 5
print(find(n))

C#

using System;
 
public class GFG {
    static public void Main()
    {
 
        int n = 5;
        Console.WriteLine(find(n));
    }
 
    public static int find(int n)
    {
 
        // Obtain number less n in 2's power
        int twospower = (int)Math.Pow(
            2, (int)(Math.Log(n) / Math.Log(2)));
 
        // Find p-position of odd number, in odd series
        int diff = n - twospower + 1;
 
        // Value of pth odd number
        int diffthodd = (2 * diff) - 1;
 
        return diffthodd;
    }
}

Javascript

// JS program to implement the approach
 
function find(n) {
  // Obtain number less n in 2's power
  let twospower = Math.pow(2, Math.floor(Math.log(n) / Math.log(2)));
 
  // Find p-position of odd number, in odd series
  let diff = n - twospower + 1;
 
  // Value of pth odd number
  let diffthodd = 2 * diff - 1;
 
  return diffthodd;
}
 
let n = 5;
console.log(find(n));

Output

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!