Given L and R. The task is to find the number in whose binary representation all bits between the L-th and R-th index are set and the rest of the bits are unset. The binary representation is 32 bits.
Examples:
Input: L = 2, R = 5
Output: 60
Explanation: The binary representation is
0..0111100 => 60Input: L = 1, R = 3
Output: 14
Explanation: The binary representation is
0..01110 => 14
Naive Approach: The naive approach to finding the number is to iterate from i = L to i = R and calculate the addition of all the powers of 2i.
The program below illustrates the naive approach:
C++
// CPP program to print the integer// with all the bits set in range L-R// Naive Approach#include <bits/stdc++.h>using namespace std;// Function to return the integer// with all the bits set in range L-Rint getInteger(int L, int R){ int number = 0; // iterate from L to R // and add all powers of 2 for (int i = L; i <= R; i++) number += pow(2, i); return number;}// Driver Codeint main(){ int L = 2, R = 5; cout << getInteger(L, R); return 0;} |
Java
// Java program to print the // integer with all the bits // set in range L-R Naive Approachimport java.io.*;class GFG {// Function to return the // integer with all the // bits set in range L-Rstatic int getInteger(int L, int R){ int number = 0; // iterate from L to R // and add all powers of 2 for (int i = L; i <= R; i++) number += Math.pow(2, i); return number;}// Driver Codepublic static void main (String[] args) { int L = 2, R = 5; System.out.println(getInteger(L, R));}}// This code is contributed by anuj_67.. |
Python3
# Python 3 program to print the integer# with all the bits set in range L-R# Naive Approachfrom math import pow# Function to return the integer# with all the bits set in range L-Rdef getInteger(L, R): number = 0 # iterate from L to R # and add all powers of 2 for i in range(L, R + 1, 1): number += pow(2, i) return number# Driver Codeif __name__ == '__main__': L = 2 R = 5 print(int(getInteger(L, R)))# This code is contributed by# Surendra_Gangwar |
C#
// C# program to print the // integer with all the bits // set in range L-R Naive Approachusing System;class GFG{// Function to return the // integer with all the // bits set in range L-Rstatic int getInteger(int L, int R){ int number = 0; // iterate from L to R // and add all powers of 2 for (int i = L; i <= R; i++) number += (int)Math.Pow(2, i); return number;}// Driver Codepublic static void Main () { int L = 2, R = 5; Console.Write(getInteger(L, R));}}// This code is contributed// by shiv_bhakt. |
PHP
<?php// PHP program to print // the integer with all // the bits set in range // L-R Naive Approach// Function to return the // integer with all the // bits set in range L-Rfunction getInteger($L, $R){ $number = 0; // iterate from L to R // and add all powers of 2 for ($i = $L; $i <= $R; $i++) $number += pow(2, $i); return $number;}// Driver Code$L = 2;$R = 5;echo getInteger($L, $R);// This code is contributed// by shiv_bhakt.?> |
Javascript
<script>// Javascript program to print the integer// with all the bits set in range L-R// Naive Approach// Function to return the integer// with all the bits set in range L-Rfunction getInteger(L, R){ var number = 0; // iterate from L to R // and add all powers of 2 for (var i = L; i <= R; i++) number += Math.pow(2, i); return number;}// Driver Codevar L = 2, R = 5;document.write( getInteger(L, R));</script> |
Output
60
Time Complexity: O(R2), considering the time complexity of the pow() method.
Auxiliary Space: O(1)
An efficient approach is to compute the number with all (R) bits set from the right and subtract the number with all (L-1) bits set from the right to get the required number.
1. Compute the number which has all R set bits from the right using the below formula.
(1 << (R+1)) - 1.
2. Subtract the number which has all (L-1) set bits from the right.
(1<<L) - 1
Hence, computing ((1<<(R+1))-1)-((1<<L)-1), we get the final formula as:
(1<<(R+1))-(1<<L)
The program below illustrates the efficient approach:
C++
// CPP program to print the integer// with all the bits set in range L-R// Efficient Approach#include <bits/stdc++.h>using namespace std;// Function to return the integer// with all the bits set in range L-Rint setbitsfromLtoR(int L, int R){ return (1 << (R + 1)) - (1 << L);}// Driver Codeint main(){ int L = 2, R = 5; cout << setbitsfromLtoR(L, R); return 0;} |
Java
// Java program to print // the integer with all // the bits set in range// L-R Efficient Approachimport java.io.*;class GFG { // Function to return the // integer with all the // bits set in range L-Rstatic int setbitsfromLtoR(int L, int R){ return (1 << (R + 1)) - (1 << L);}// Driver Codepublic static void main (String[] args){ int L = 2, R = 5; System.out.println(setbitsfromLtoR(L, R));}}// This code is contributed// by shiv_bhakt. |
Python3
# Python3 program to print # the integer with all the # bits set in range L-R# Efficient Approach# Function to return the# integer with all the# bits set in range L-Rdef setbitsfromLtoR(L, R): return ((1 << (R + 1)) - (1 << L))# Driver CodeL = 2R = 5print(setbitsfromLtoR(L, R))# This code is contributed# by Smita |
C#
// C# program to print // the integer with all // the bits set in range// L-R Efficient Approachusing System;class GFG{// Function to return the // integer with all the // bits set in range L-Rstatic int setbitsfromLtoR(int L, int R){ return (1 << (R + 1)) - (1 << L);}// Driver Codepublic static void Main (){ int L = 2, R = 5; Console.WriteLine(setbitsfromLtoR(L, R));}}// This code is contributed// by shiv_bhakt. |
PHP
<?php// PHP program to print // the integer with all // the bits set in range // L-R Efficient Approach// Function to return the // integer with all the// bits set in range L-Rfunction setbitsfromLtoR($L, $R){ return (1 << ($R + 1)) - (1 << $L);}// Driver Code$L = 2;$R = 5;echo setbitsfromLtoR($L, $R);// This code is contributed // by shiv_bhakt.?> |
Javascript
<script>// Javascript program to print the integer// with all the bits set in range L-R// Efficient Approach// Function to return the integer// with all the bits set in range L-Rfunction setbitsfromLtoR(L, R){ return (1 << (R + 1)) - (1 << L);}// Driver Codevar L = 2, R = 5;document.write( setbitsfromLtoR(L, R));// This code is contributed by famously.</script> |
Output
60
Time Complexity: O(1)
Auxiliary Space: O(1)
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