Given a positive integer N, the task is to find a number which contains (N – 1) set bits in its binary form at every even index (1-based) from the right.
Examples:
Input: N = 2
Output: 2
Binary representation of 2 is 10 which has
1 set bit at even position from the right.
Input: N = 4
Output: 42
Binary representation of 42 is 101010
Observation: If we check out the numbers in binary form then the result is something like this:
| n | Decimal Equivalent | Binary Equivalent |
|---|---|---|
| 1 | 0 | 0 |
| 2 | 2 | 10 |
| 3 | 10 | 1010 |
| 4 | 42 | 101010 |
| 5 | 170 | 10101010 |
Naive Approach: As we can see in the table our binary equivalent is always adding a “10” in last of the previous string. So, we can generate a binary string which is made up of sub-string “10” concatenated N-1 times and then print its decimal equivalent.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define ll long long int// Function to return the string generated// by appending "10" n-1 timesstring constructString(ll n){ // Initialising string as empty string s = ""; for (ll i = 0; i < n; i++) { s += "10"; } return s;}// Function to return the decimal equivalent// of the given binary stringll binaryToDecimal(string n){ string num = n; ll dec_value = 0; // Initializing base value to 1 // i.e 2^0 ll base = 1; ll len = num.length(); for (ll i = len - 1; i >= 0; i--) { if (num[i] == '1') dec_value += base; base = base * 2; } return dec_value;}// Function that calls the constructString// and binarytodecimal and returns the answerll findNumber(ll n){ string s = constructString(n - 1); ll num = binaryToDecimal(s); return num;}// Driver codeint main(){ ll n = 4; cout << findNumber(n); return 0;} |
Java
// Java implementation of above approachimport java.util.*;class GFG {// Function to return the String generated// by appending "10" n-1 timesstatic String constructString(int n){ // Initialising String as empty String s = ""; for (int i = 0; i < n; i++) { s += "10"; } return s;}// Function to return the decimal equivalent// of the given binary Stringstatic int binaryToDecimal(String n){ String num = n; int dec_value = 0; // Initializing base value to 1 // i.e 2^0 int base = 1; int len = num.length(); for (int i = len - 1; i >= 0; i--) { if (num.charAt(i) == '1') dec_value += base; base = base * 2; } return dec_value;}// Function that calls the constructString// and binarytodecimal and returns the answerstatic int findNumber(int n){ String s = constructString(n - 1); int num = binaryToDecimal(s); return num;}// Driver codepublic static void main(String[] args) { int n = 4; System.out.println(findNumber(n));}}/* This code is contributed by PrinciRaj1992 */ |
Python
# Python3 implementation of the approach# Function to return the generated# by appending "10" n-1 timesdef constructString(n): # Initialising as empty s = "" for i in range(n): s += "10" return s# Function to return the decimal equivaLent# of the given binary stringdef binaryToDecimal(n): num = n dec_value = 0 # Initializing base value to 1 # i.e 2^0 base = 1 Len = len(num) for i in range(Len - 1,-1,-1): if (num[i] == '1'): dec_value += base base = base * 2 return dec_value# Function that calls the constructString# and binarytodecimal and returns the answerdef findNumber(n): s = constructString(n - 1) num = binaryToDecimal(s) return num# Driver coden = 4print(findNumber(n))# This code is contributed by mohit kumar 29 |
C#
// C# implementation of above approachusing System;class GFG{ // Function to return the String generated// by appending "10" n-1 timesstatic String constructString(int n){ // Initialising String as empty String s = ""; for (int i = 0; i < n; i++) { s += "10"; } return s;}// Function to return the decimal equivalent// of the given binary Stringstatic int binaryToDecimal(String n){ String num = n; int dec_value = 0; // Initializing base value to 1 // i.e 2^0 int base_t = 1; int len = num.Length; for (int i = len - 1; i >= 0; i--) { if (num[i] == '1') dec_value = dec_value + base_t; base_t = base_t * 2; } return dec_value;}// Function that calls the constructString// and binarytodecimal and returns the answerstatic int findNumber(int n){ String s = constructString(n - 1); int num = binaryToDecimal(s); return num;}// Driver codestatic public void Main (){ int n = 4; Console.Write(findNumber(n));}}// This code is contributed by ajit |
Javascript
<script>// JavaScript implementation of above approach// Function to return the String generated// by appending "10" n-1 timesfunction constructString(n){ // Initialising String as empty var s = ""; for (var i = 0; i < n; i++) { s += "10"; } return s;}// Function to return the decimal equivalent// of the given binary Stringfunction binaryToDecimal(n){ var num = n; var dec_value = 0; // Initializing base value to 1 // i.e 2^0 var base = 1; var len = num.length; for (var i = len - 1; i >= 0; i--) { if (num.charAt(i) == '1') dec_value += base; base = base * 2; } return dec_value;}// Function that calls the constructString// and binarytodecimal and returns the answerfunction findNumber(n){ var s = constructString(n - 1); var num = binaryToDecimal(s); return num;}// Driver codevar n = 4;document.write(findNumber(n));// This code is contributed by Amit Katiyar </script> |
Output:
42
Efficient Approach: If we take the numbers and convert them to base 4 we can see an interesting pattern as follows:
| n | Decimal Equivalent | Binary Equivalent | Base_4 |
|---|---|---|---|
| 1 | 0 | 0 | 0 |
| 2 | 2 | 10 | 2 |
| 3 | 10 | 1010 | 22 |
| 4 | 42 | 101010 | 222 |
| 5 | 170 | 10101010 | 2222 |
We are actually appending “2” for every nth term in base4 i.e. for n = 7 our number in base4 would have (n – 1) i.e. 6 consecutive 2’s.
Now we have to take a point in mind as we know that if we convert from any base m to base 10 i.e. decimal than the solution is (n0 * m0 + n1 * m1 + n2 * m2 + …. + n * mn). So as our base is 4 by further calculation we can found that our required number n can be found by using the deduced formula in O(1) time complexity.
Formula:
A(n) = floor((2 / 3) * (4n – 1))
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define ll long long int// Function to compute number// using our deduced formulall findNumber(int n){ // Initialize num to n-1 ll num = n - 1; num = 2 * (ll)pow(4, num); num = floor(num / 3.0); return num;}// Driver codeint main(){ int n = 5; cout << findNumber(n); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG { // Function to compute number// using our deduced formulastatic int findNumber(int n){ // Initialize num to n-1 int num = n - 1; num = 2 * (int)Math.pow(4, num); num = (int)Math.floor(num / 3.0); return num;}// Driver codepublic static void main (String[] args) { int n = 5; System.out.println (findNumber(n));}}// The code is contributed by ajit. |
Python3
# Python3 implementation of the approach # Function to compute number # using our deduced formula def findNumber(n) : # Initialize num to n-1 num = n - 1; num = 2 * (4 ** num); num = num // 3; return num; # Driver code if __name__ == "__main__" : n = 5; print(findNumber(n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;class GFG{// Function to compute number// using our deduced formulastatic int findNumber(int n){ // Initialize num to n-1 int num = n - 1; num = 2 * (int)Math.Pow(4, num); num = (int)Math.Floor(num / 3.0); return num;}// Driver codestatic public void Main (){ int n = 5; Console.Write(findNumber(n));}}// The code is contributed by Tushil. |
Javascript
<script> // Javascript implementation of the approach // Function to compute number // using our deduced formula function findNumber(n) { // Initialize num to n-1 let num = n - 1; num = 2 * Math.pow(4, num); num = Math.floor(num / 3.0); return num; } let n = 5; document.write(findNumber(n));</script> |
Output:
170
Time Complexity: O(1)
Auxiliary Space: O(1)
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