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 <bits/stdc++.h>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 numberdef countLuckyNum(n): count, i = 0, 1 while True: if n>= 2**i-1: count+= 1 else: break i+= 1; return count # driver coden = 7print(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
<?php// PHP implementation of the approach // Function to count the lucky numberfunction countLuckyNum($n) { $count = 0; $i = 1; while(1) { if ($n >= ((1 << $i) - 1)) $count += 1; else break; $i += 1; } return $count; } // Driver code $n = 7; echo countLuckyNum($n) ; ?> |
Javascript
<script> // Function to return the count of lucky number function countLuckyNum(n) { var count = 0, i = 1; while (true) { if (n >= ((1 << i) - 1)) count++; else break; i++; } return count; } // Driver code var n = 7; document.write(countLuckyNum(n));// This code is contributed by aashish1995</script> |
output:3
Time Complexity: O(logn)
Auxiliary Space: O(1)
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