Set the leftmost unset bit

This article aims to find a way to set the leftmost unset bit of a given number. The first unset bit after the most significant set bit is considered the leftmost unset bit.

Problem Statement

Given a number n, the task is to set the unset leftmost bit in the binary expansion of the number to 1. All other bits should remain unchanged. If all bits of the original number are set, then the number is returned.

Examples

Input: 46

Output: 62

The Chinese translation of

Explanation

is:

Explanation

Binary Expansion of 46 = 101110.

The leftmost unset bit is 101110.

Upon setting the underlined bit we get, 111110. This is the binary expansion of 62.

Hence answer is 62.

Input: 11

Output: 15

The Chinese translation of

Explanation

is:

Explanation

Binary Expansion of 11 = 1011.

The leftmost unset bit is 1011.

Upon changing the underlined bit, we get 1111 which is the binary expansion of 15.

Input: 30

Output: 31

The Chinese translation of

Explanation

is:

Explanation

Binary Expansion of 30 = 11110.

The leftmost unset bit is 11110.

When setting the leftmost unset bit, we get 11111, which is the binary expansion of 31.

Input: 7

Output: 7

The Chinese translation of

Explanation

is:

Explanation

Binary Expansion of 7 = 111.

Since all bits are set, there is no leftmost unset bit. So the answer remains the same as the original number.

Solution method

• Check if all bits are set. If so, return the original number as the answer.

• Find the position of the latest unset bit using bitwise AND operator, and update the counter.

• Set the bits to correspond to the counter.

• Show answer.

Find if all bits are set

The idea here is that by adding one bit, the input number will become a perfect square of 2 if all of its bits are set. Hence, the following expression will determine whether or not all the bits of the number are set: n & (n 1) == 0;

Let us understand this through an example.

Let the number be 5. We need to check if all the bits of 5 are set or not.

The Chinese translation of is:

n = 3	n 1 = 4	n & (n 1)
011
011	100	000

Thus it is safe to conclude that all the bits of n are already set and we return the number as it is.

找到最左边未设置的位

如果AND操作的输出不等于零，则继续查找最左边未设置的位。首先生成一个位数与给定整数相等的数字。新数字的最左边位最初设置为1。

然后，我们运行一个循环，从最左边的1位开始，向右搜索第一个0，通过对给定数字和新数字进行位与运算来实现。当位与运算的结果为0时，我们返回第一个未设置的最左边位的位置pos。

To Set the Leftmost Unset Bit

Generate a new number in which only the bit corresponding to pos is set. Perform bitwise OR operation between this new number and the original number.

Algorithm

的中文翻译为：

算法

Function all_bits_set()

• 计算 n & (n + 1)。

• If result == 0, return true.

• 否则返回 false。

Function find_leftmost_unset_bit()

• Initialize m = 1, pos = 0.

• while (n > m)

• 左移 m 1 位

• 将 m 右移 1 位，以使其对应于 n 的最高有效位。

• while ((n & m) != 0)

• 将 m 右移 1 位

• pos++

• 一旦循环中断，我们就可以得到最高有效位（MSB）中最左边未设置的位的位置。

• 返回 log2(n) - pos，即从最低有效位开始的位位置。

函数 set_leftmost_unset_bit()

• 初始化 k = 1

• Function Call find_leftmost_unset_bit().

• k = k

• Compute n | k.

• Update n.

Function main()

• 初始化 n

• Function Call all_bits_set()

• 函数调用 find_leftmost_unset_bit()

• 调用函数set_leftmost_unset_bit()

• 显示 n

示例：C++程序

这个程序通过将输入数字 n 的二进制展开中最左边未设置的位设置为 1 来修改它。它使用位运算符 OR，左移和右移运算符以及位与运算符来实现其目标。

// A C++ program to set the left most unset bit of a number. If all the bits of the given number are already set, it returns the number as it is.
#include <iostream>
#include <cmath>
using namespace std;
// function to check if all bits of the given number are already set
// if all bits of n are set, n + 1 will be a power of 2.
bool all_bits_set(int n){
   if ((n & (n + 1)) == 0)    {
      return true;
   }
   return false;
}

// function to find the position of the leftmost unset bit from the LSB.
int find_leftmost_unset_bit(int n){
   int m = 1, pos = 0;
   while (n > m){
      m = m << 1;
   }
   m = m >> 1; // to make the number of digits in m equal to number of digits in n
   
   // the following loop executes till the first zero is encountered, starting from the msb
   while ((n & m) != 0){
      m = m >> 1;
      pos++;
   }
   
   // since pos is the position of the unset bit from the MSB we return log2(n) - pos which is the location of the leftmost unset bit from the LSB.
   return log2(n) - pos;
}

// function to set the leftmost unset bit from the LSB.
void set_leftmost_unset_bit(int &n){
   int k = 1;
   int pos = find_leftmost_unset_bit(n);
   k = k << (pos); // left shift k by pos
   n = n | k; // to set the leftmost unset bit
}

// main function
int main(){
   int n = 46;
   cout << "Input Number: "<< n << endl;
   if (all_bits_set(n))    {
      cout << n << endl;
      return 0;
   }
   set_leftmost_unset_bit(n);
   cout << "Number after setting the Leftmost Unset Bit: " << n << endl; // display the updated number
   return 0;
}

输出

Input Number: 46
Number after setting the Leftmost Unset Bit: 62

时间和空间分析

时间复杂度：O(log2(n))，因为在函数find_leftmost_unset_bit()中，我们可能需要遍历二进制展开式的所有log2(n)位数来找到最左边的未设置位。

Space Complexity: O(1), as constant space is always used in the implementation.

Conclusion

本文讨论了一种寻找并设置给定数字最左边未设置位的方法。如果数字的所有位已经设置，我们将返回该数字。否则，为了设置该位，我们使用位左移和右移运算符生成一个新的位模式，并使用位或运算符计算结果。解决方案的概念、多个示例、使用的算法、C++程序解决方案以及时间和空间复杂度分析都被详细解释，以便更深入地理解。

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