生成二进制字符串回文所需的最小位翻转次数

给定一个整数N ，任务是找到将N的二进制表示转换为回文所需翻转的最小位数。

例子：

Input: N = 12 
Output: 2 
Explanation: 
Binary String representing 12 = “1100”. 
To make “1100” a palindrome, convert the string to “0110”. 
Therefore, minimum bits required to be flipped is 2.
Input: N = 7 
Output: 0 
Explanation: 
Binary String representing 7 = 111, which is already a palindrome. 
 

朴素的方法：最简单的方法是检查每个可能的具有相同位数的回文子集。
时间复杂度： O(N)
辅助空间： O(1)

高效方法：上述方法可以通过以下步骤进行优化：

1. 首先，检查给定数字的二进制形式的长度。
2. 取两个指针，一个指向LSB ，另一个指向MSB 。
3.  现在继续递减第一个指针并递增第二个指针。
4. 检查第一个和第二个指针位置的位是否相同。如果不是，请增加要更改的位数。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate the
// length of the binary string
int check_length(int n)
{
    // Length
    int ans = 0;
    while (n) {
 
        // Right shift of n
        n = n >> 1;
 
        // Increment the length
        ans++;
    }
 
    // Return the length
    return ans;
}
 
// Function to check if the bit present
// at i-th position is a set bit or not
int check_ith_bit(int n, int i)
{
    // Returns true if the bit is set
    return (n & (1 << (i - 1)))
               ? true
               : false;
}
 
// Function to count the minimum
// number of bit flips required
int no_of_flips(int n)
{
    // Length of the binary form
    int len = check_length(n);
 
    // Number of flips
    int ans = 0;
 
    // Pointer to the LSB
    int right = 1;
 
    // Pointer to the MSB
    int left = len;
 
    while (right < left) {
 
        // Check if the bits are equal
        if (check_ith_bit(n, right)
            != check_ith_bit(n, left))
            ans++;
 
        // Decrementing the
        // left pointer
        left--;
 
        // Incrementing the
        // right pointer
        right++;
    }
 
    // Returns the number of
    // bits to flip.
    return ans;
}
 
// Driver Code
int main()
{
 
    int n = 12;
 
    cout << no_of_flips(n);
 
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to calculate the
// length of the binary string
static int check_length(int n)
{
     
    // Length
    int ans = 0;
     
    while (n != 0)
    {
 
        // Right shift of n
        n = n >> 1;
 
        // Increment the length
        ans++;
    }
 
    // Return the length
    return ans;
}
 
// Function to check if the bit present
// at i-th position is a set bit or not
static boolean check_ith_bit(int n, int i)
{
     
    // Returns true if the bit is set
    return (n & (1<< (i - 1))) != 0 ? true : false;
}
 
// Function to count the minimum
// number of bit flips required
static int no_of_flips(int n)
{
     
    // Length of the binary form
    int len = check_length(n);
 
    // Number of flips
    int ans = 0;
 
    // Pointer to the LSB
    int right = 1;
 
    // Pointer to the MSB
    int left = len;
 
    while (right < left)
    {
         
        // Check if the bits are equal
        if (check_ith_bit(n, right) !=
            check_ith_bit(n, left))
            ans++;
 
        // Decrementing the
        // left pointer
        left--;
 
        // Incrementing the
        // right pointer
        right++;
    }
 
    // Returns the number of
    // bits to flip.
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 12;
 
    System.out.println(no_of_flips(n));
}
}
 
// This code is contributed by rutvik_56

Python3

# Python3 program to implement
# the above approach
 
# Function to calculate the
# length of the binary string
def check_length(n):
 
    # Length
    ans = 0
     
    while (n):
 
        # Right shift of n
        n = n >> 1
 
        # Increment the length
        ans += 1
 
    # Return the length
    return ans
 
# Function to check if the bit present
# at i-th position is a set bit or not
def check_ith_bit(n, i):
 
    # Returns true if the bit is set
    if (n & (1 << (i - 1))):
        return True
    else:
        return False
 
# Function to count the minimum
# number of bit flips required
def no_of_flips(n):
 
    # Length of the binary form
    ln = check_length(n)
 
    # Number of flips
    ans = 0
 
    # Pointer to the LSB
    right = 1
 
    # Pointer to the MSB
    left = ln
 
    while (right < left):
 
        # Check if the bits are equal
        if (check_ith_bit(n, right) !=
            check_ith_bit(n, left)):
            ans += 1
 
        # Decrementing the
        # left pointer
        left -= 1
 
        # Incrementing the
        # right pointer
        right += 1
 
    # Returns the number of
    # bits to flip.
    return ans
 
# Driver Code
n = 12
 
print(no_of_flips(n))
 
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to calculate the
// length of the binary string
static int check_length(int n)
{
     
    // Length
    int ans = 0;
     
    while (n != 0)
    {
 
        // Right shift of n
        n = n >> 1;
 
        // Increment the length
        ans++;
    }
 
    // Return the length
    return ans;
}
 
// Function to check if the bit present
// at i-th position is a set bit or not
static bool check_ith_bit(int n, int i)
{
     
    // Returns true if the bit is set
    return (n & (1 << (i - 1))) != 0 ?
                                true : false;
}
 
// Function to count the minimum
// number of bit flips required
static int no_of_flips(int n)
{
     
    // Length of the binary form
    int len = check_length(n);
 
    // Number of flips
    int ans = 0;
 
    // Pointer to the LSB
    int right = 1;
 
    // Pointer to the MSB
    int left = len;
 
    while (right < left)
    {
         
        // Check if the bits are equal
        if (check_ith_bit(n, right) !=
            check_ith_bit(n, left))
            ans++;
 
        // Decrementing the
        // left pointer
        left--;
 
        // Incrementing the
        // right pointer
        right++;
    }
 
    // Returns the number of
    // bits to flip.
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 12;
 
    Console.WriteLine(no_of_flips(n));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// JavaScript program for the above approach
 
// Function to calculate the
// length of the binary string
function check_length(n)
{
       
    // Length
    let ans = 0;
       
    while (n != 0)
    {
   
        // Right shift of n
        n = n >> 1;
   
        // Increment the length
        ans++;
    }
   
    // Return the length
    return ans;
}
   
// Function to check if the bit present
// at i-th position is a set bit or not
function check_ith_bit(n, i)
{
       
    // Returns true if the bit is set
    return (n & (1<< (i - 1))) != 0 ? true : false;
}
   
// Function to count the minimum
// number of bit flips required
function no_of_flips(n)
{
       
    // Length of the binary form
    let len = check_length(n);
   
    // Number of flips
    let ans = 0;
   
    // Poleter to the LSB
    let right = 1;
   
    // Poleter to the MSB
    let left = len;
   
    while (right < left)
    {
           
        // Check if the bits are equal
        if (check_ith_bit(n, right) !=
            check_ith_bit(n, left))
            ans++;
   
        // Decrementing the
        // left poleter
        left--;
   
        // Incrementing the
        // right poleter
        right++;
    }
   
    // Returns the number of
    // bits to flip.
    return ans;
}
     
// Driver Code
     
       let n = 12;
   
    document.write(no_of_flips(n));
              

输出：

2

时间复杂度： O(log N)
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live