Given two integers L and R, the task is to count the numbers having only one unset bit in the range [L, R].
Examples:
Input: L = 4, R = 9
Output: 2
Explanation:
The binary representation of all numbers in the range [4, 9] are
4 = (100)2
5 = (101)2
6 = (110)2
7 = (111)2
8 = (1000)2
9 = (1001)2
Out of all the above numbers, only 5 and 6 have exactly one unset bit.
Therefore, the required count is 2.Input: L = 10, R = 13
Output: 2
Explanation:
The binary representations of all numbers in the range [10, 13]
10 = (1010)2
11 = (1011)2
12 = (1100)2
13 = (1101)2
Out of all the above numbers, only 11 and 13 have exactly one unset bit.
Therefore, the required count is 2.
Naive Approach: The simplest approach is to check every number in the range [L, R] whether it has exactly one unset bit or not. For every such number, increment the count. After traversing the entire range, print the value of count.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count numbers in the range// [l, r] having exactly one unset bitint count_numbers(int L, int R){ // Stores the required count int ans = 0; // Iterate over the range for (int n = L; n <= R; n++) { // Calculate number of bits int no_of_bits = log2(n) + 1; // Calculate number of set bits int no_of_set_bits = __builtin_popcount(n); // If count of unset bits is 1 if (no_of_bits - no_of_set_bits == 1) { // Increment answer ans++; } } // Return the answer return ans;}// Driver Codeint main(){ int L = 4, R = 9; // Function Call cout << count_numbers(L, R); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to count numbers in the range// [l, r] having exactly one unset bitstatic int count_numbers(int L, int R){ // Stores the required count int ans = 0; // Iterate over the range for (int n = L; n <= R; n++) { // Calculate number of bits int no_of_bits = (int) (Math.log(n) + 1); // Calculate number of set bits int no_of_set_bits = Integer.bitCount(n); // If count of unset bits is 1 if (no_of_bits - no_of_set_bits == 1) { // Increment answer ans++; } } // Return the answer return ans;}// Driver Codepublic static void main(String[] args){ int L = 4, R = 9; // Function Call System.out.print(count_numbers(L, R));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program for the above approachfrom math import log2# Function to count numbers in the range# [l, r] having exactly one unset bitdef count_numbers(L, R): # Stores the required count ans = 0 # Iterate over the range for n in range(L, R + 1): # Calculate number of bits no_of_bits = int(log2(n) + 1) # Calculate number of set bits no_of_set_bits = bin(n).count('1') # If count of unset bits is 1 if (no_of_bits - no_of_set_bits == 1): # Increment answer ans += 1 # Return the answer return ans# Driver Codeif __name__ == '__main__': L = 4 R = 9 # Function call print(count_numbers(L, R))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Function to count numbers in the range// [l, r] having exactly one unset bitstatic int count_numbers(int L, int R){ // Stores the required count int ans = 0; // Iterate over the range for(int n = L; n <= R; n++) { // Calculate number of bits int no_of_bits = (int)(Math.Log(n) + 1); // Calculate number of set bits int no_of_set_bits = bitCount(n); // If count of unset bits is 1 if (no_of_bits - no_of_set_bits == 1) { // Increment answer ans++; } } // Return the answer return ans;}static int bitCount(long x){ int setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits;}// Driver Codepublic static void Main(String[] args){ int L = 4, R = 9; // Function call Console.Write(count_numbers(L, R));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// javascript program for the// above approach// Function to count numbers in the range// [l, r] having exactly one unset bitfunction count_numbers(L, R){ // Stores the required count let ans = 0; // Iterate over the range for (let n = L; n <= R; n++) { // Calculate number of bits let no_of_bits = Math.floor(Math.log(n) + 1); // Calculate number of set bits let no_of_set_bits = bitCount(n); // If count of unset bits is 1 if (no_of_bits - no_of_set_bits == 1) { // Increment answer ans++; } } // Return the answer return ans;} function bitCount(x){ let setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits;} // Driver Code let L = 4, R = 9; // Function Call document.write(count_numbers(L, R)); // This code is contributed by avijitmondal1998.</script> |
Output:
2
Time Complexity: O(N*Log R)
Auxiliary Space: O(1)
Efficient Approach: Since the maximum number of bits is at most log R and the number should contain exactly one zero in its binary representation, fix one position in [0, log R] as the unset bit and set all other bits and increment count if the generated number is within the given range.
Follow the steps below to solve the problem:
- Loop over all possible positions of unset bit, say zero_bit, in the range [0, log R]
- Set all bits before zero_bit.
- Iterate over the range j = zero_bit + 1 to log R and set the bit at position j and increment count if the number formed is within the range [L, R].
- Print the count after the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count numbers in the range// [L, R] having exactly one unset bitint count_numbers(int L, int R){ // Stores the count elements // having one zero in binary int ans = 0; // Stores the maximum number of // bits needed to represent number int LogR = log2(R) + 1; // Loop over for zero bit position for (int zero_bit = 0; zero_bit < LogR; zero_bit++) { // Number having zero_bit as unset // and remaining bits set int cur = 0; // Sets all bits before zero_bit for (int j = 0; j < zero_bit; j++) { // Set the bit at position j cur |= (1LL << j); } for (int j = zero_bit + 1; j < LogR; j++) { // Set the bit position at j cur |= (1LL << j); // If cur is in the range [L, R], // then increment ans if (cur >= L && cur <= R) { ans++; } } } // Return ans return ans;}// Driver Codeint main(){ long long L = 4, R = 9; // Function Call cout << count_numbers(L, R); return 0;} |
Java
// Java program for // the above approachimport java.util.*;class GFG{// Function to count numbers in the range// [L, R] having exactly one unset bitstatic int count_numbers(int L, int R){ // Stores the count elements // having one zero in binary int ans = 0; // Stores the maximum number of // bits needed to represent number int LogR = (int) (Math.log(R) + 1); // Loop over for zero bit position for (int zero_bit = 0; zero_bit < LogR; zero_bit++) { // Number having zero_bit as unset // and remaining bits set int cur = 0; // Sets all bits before zero_bit for (int j = 0; j < zero_bit; j++) { // Set the bit at position j cur |= (1L << j); } for (int j = zero_bit + 1; j < LogR; j++) { // Set the bit position at j cur |= (1L << j); // If cur is in the range [L, R], // then increment ans if (cur >= L && cur <= R) { ans++; } } } // Return ans return ans;}// Driver Codepublic static void main(String[] args){ int L = 4, R = 9; // Function Call System.out.print(count_numbers(L, R));}}// This code is contributed by shikhasingrajput |
Python3
# Python3 program for# the above approachimport math# Function to count numbers in the range# [L, R] having exactly one unset bitdef count_numbers(L, R): # Stores the count elements # having one zero in binary ans = 0; # Stores the maximum number of # bits needed to represent number LogR = (int)(math.log(R) + 1); # Loop over for zero bit position for zero_bit in range(LogR): # Number having zero_bit as unset # and remaining bits set cur = 0; # Sets all bits before zero_bit for j in range(zero_bit): # Set the bit at position j cur |= (1 << j); for j in range(zero_bit + 1, LogR): # Set the bit position at j cur |= (1 << j); # If cur is in the range [L, R], # then increment ans if (cur >= L and cur <= R): ans += 1; # Return ans return ans;# Driver Codeif __name__ == '__main__': L = 4; R = 9; # Function Call print(count_numbers(L, R));# This code is contributed by Rajput-Ji |
C#
// C# program for // the above approachusing System;public class GFG{// Function to count numbers in the range// [L, R] having exactly one unset bitstatic int count_numbers(int L, int R){ // Stores the count elements // having one zero in binary int ans = 0; // Stores the maximum number of // bits needed to represent number int LogR = (int) (Math.Log(R) + 1); // Loop over for zero bit position for (int zero_bit = 0; zero_bit < LogR; zero_bit++) { // Number having zero_bit as unset // and remaining bits set int cur = 0; // Sets all bits before zero_bit for (int j = 0; j < zero_bit; j++) { // Set the bit at position j cur |= (1 << j); } for (int j = zero_bit + 1; j < LogR; j++) { // Set the bit position at j cur |= (1 << j); // If cur is in the range [L, R], // then increment ans if (cur >= L && cur <= R) { ans++; } } } // Return ans return ans;}// Driver Codepublic static void Main(String[] args){ int L = 4, R = 9; // Function Call Console.Write(count_numbers(L, R));}}// This code contributed by shikhasingrajput |
Javascript
<script>// JavaScript program for//the above approach// Function to count numbers in the range// [L, R] having exactly one unset bitfunction count_numbers(L, R){ // Stores the count elements // having one zero in binary let ans = 0; // Stores the maximum number of // bits needed to represent number let LogR = (Math.log(R) + 1); // Loop over for zero bit position for (let zero_bit = 0; zero_bit < LogR; zero_bit++) { // Number having zero_bit as unset // and remaining bits set let cur = 0; // Sets all bits before zero_bit for (let j = 0; j < zero_bit; j++) { // Set the bit at position j cur |= (1 << j); } for (let j = zero_bit + 1; j < LogR; j++) { // Set the bit position at j cur |= (1 << j); // If cur is in the range [L, R], // then increment ans if (cur >= L && cur <= R) { ans++; } } } // Return ans return ans;} // Driver Code let L = 4, R = 9; // Function Call document.write(count_numbers(L, R)); // This code is contributed by souravghosh0416.</script> |
Output:
2
Time Complexity: O((log R)2)
Auxiliary Space: O(1)
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