Count numbers have all 1s together in binary representation

• Last Updated : 31 Mar, 2021

Given an integer n, the task is to count the total lucky numbers smaller than or equal to n. A number is said to be lucky if it has all contagious number of 1’s in binary representation from the beginning. For example 1, 3, 7, 15 are lucky numbers, and 2, 5 and 9 are not lucky numbers.
Examples:

Input :n = 7
Output :3
1, 3 and 7 are lucky numbers

Input :n = 17
Output :4

Approach:one approach is that first we find out the binary representation of each number and than check for contagious number of 1’s for each number, but this approach is time consuming and can give tle if the constraints are two large, Efficient approach can be find out by observing the numbers, we can say that every ith lucky number can be found by the formula 2i-1, and by iterating a loop upto number less than equal to n we can find out the total lucky numbers.
Below is the implementation of above approach

CPP

 #include using namespace std; int countLuckyNum(int n){    int count = 0, i = 1;     while (1) {        if (n >= ((1 << i) - 1))            count++;        else            break;        i++;    }     return count;} // Driver codeint main(){    int n = 7;    cout << countLuckyNum(n);    return 0;}

Java

 import java.util.*;import java.lang.*;import java.io.*; public class GFG {     // Function to return the count of lucky number    static int countLuckyNum(int n)    {         int count = 0, i = 1;        while (true) {            if (n >= ((1 << i) - 1))                count++;            else                break;            i++;        }        return count;    }     // Driver code    public static void main(String[] args)    {        int n = 7;        System.out.println(countLuckyNum(n));    }}

Python

 # python3 code of above problem # function to count the lucky number def countLuckyNum(n):         count, i = 0, 1         while True:        if n>= 2**i-1:            count+= 1        else:            break        i+= 1;    return count     # driver coden = 7 print(countLuckyNum(n))

C#

 // C# implementation of the approachusing System; public class GFG {     // Function to return the count of lucky number    static int countLuckyNum(int n)    {         int count = 0, i = 1;        while (true) {            if (n >= ((1 << i) - 1))                count++;            else                break;            i++;        }        return count;    }     // Driver code    public static void Main()    {        int n = 7;        Console.WriteLine(countLuckyNum(n));    }}

PHP

 = ((1 << \$i) - 1))        \$count += 1;      else        break;      \$i += 1;  }  return \$count;                    }       // Driver code  \$n = 7;echo countLuckyNum(\$n) ;  ?>

Javascript


output:3

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