最小翻转以使所有1都在左侧，而0则在右侧|设置1(使用位掩码)

给定一个二进制数组，我们可以将所有1翻转到左侧，将所有0翻转到右侧。计算使左边的全1和右边的全0所需的最小翻转。

例子：

Input: 1011000  
Output: 1
1 flip is required to make it 1111000.

Input : 00001 
Output : 2
2 flips required to make it 10000.

为了解决这个问题，我们使用位屏蔽。首先，我们将此数组转换为字符串，然后找到该二进制字符串的等效十进制数。我们尝试所有可能的掩码，其中左侧为1，右侧为0。我们迭代一个循环，直到十进制数变为零。每次我们将对数字与掩码进行按位XOR运算时，XOR值中的1数即为所需的翻转次数。我们将n减1并更新掩码。
1-以二进制数组作为输入
2-将数组转换为字符串，然后转换为等效的十进制数字(num)
3-获取初始掩码值并迭代直到num <= 0
4-使用(数字XOR掩码)查找所需的翻转
5-查找最小翻转并减少数量并更新遮罩
6-返回最小计数

C++

// C++ program to find
// of flips till that
// all 1s in lef
#include 
using namespace std;
 
int countones(long n);
  
// Function to count minimum
// number of flips
int findMiniFlip(int nums[], int n)
{
  string s = "";
  for (int i = 0; i < n; i++)
    s += nums[i];
 
  char *end;
  char tmp[s.length()];
  strcpy(tmp, s.c_str());
 
  // This is converting string
  // s into integer of base 2
  // (if s = '100' then num = 4)
  long num = strtol(tmp, &end, 2);
 
  // Initialize minXor
  // with n that can be
  // maximum number of flips
  int minXor = n;
 
  // right shift 1 by (n-1) bits
  long mask = (1 << (n - 1));
  while (n - 1 > 0)
  {
    // Calculate bitwise
    // XOR of num and mask
    long temp = (num ^ mask);
 
    // Math.min(a, b) returns
    // minimum of a and b
    // return minimum number
    // of flips till that digit
    minXor = min(minXor, countones(temp));
    n--;
    mask = (mask | (1 << (n - 1)));
  }
  return minXor;
}
 
// Function to count number of 1s
int countones(long n)
{
  int c = 0;
  while (n > 0)
  {
    n = n & (n - 1);
    c++;
  }
  return c;
}
  
// Driver code
int main()
{
  int nums[] = {1, 0, 1,
                1, 0, 0, 0};
  int size = sizeof(nums) /
             sizeof(nums[0]);
  int n = findMiniFlip(nums, size);
  cout << n;
}
 
// This code is contributed by Rutvik_56

Java

// Java program to find minimum flips to make
// all 1s in left
import java.io.*;
 
class GFG {
 
    // function to count minimum number of flips
    public static int findMiniFlip(int[] nums)
    {
        int n = nums.length;
        String s = "";
        for (int i = 0; i < n; i++)
            s += nums[i];
 
        // This is converting string s into integer
        // of base 2 (if s = '100' then num = 4)
        long num = Integer.parseInt(s, 2);
 
        // initialize minXor with n that can be maximum
        // number of flips
        int minXor = n;
 
        // right shift 1 by (n-1) bits
        long mask = (1 << (n-1));
        while (n-1 > 0) {
 
            // calculate bitwise XOR of num and mask
            long temp = (num ^ mask);
 
            // Math.min(a, b) returns minimum of a and b
            // return minimum number of flips till that
            // digit
            minXor = Math.min(minXor, countones(temp));
            n--;
 
            mask = (mask | (1 << n -1));
        }
        return minXor;
    }
 
    // function to count number of 1s
    public static int countones(long n)
    {
        int c = 0;
        while (n > 0) {
            n = n & (n-1);
            c++;
        }
        return c;
    }
 
    public static void main(String[] args)
    {
        int[] nums = { 1, 0, 1, 1, 0, 0, 0 };
        int n = findMiniFlip(nums);
        System.out.println(n);
    }
}

Python3

# Python3 program to find minimum flips to make
# all 1s in left
 
# Function to count minimum number of flips
def findMiniFlip(nums):
 
    n = len(nums)
    s = ''
     
    for i in range(n):
        s += str(nums[i])
 
    # This is converting string s into integer
    # of base 2 (if s='100' then num=4)
    num = int(s, 2)
 
    # Initialize minXor with n that can be maximum
    # number of flips
    minXor = n;
 
    # Right shift 1 by(n-1) bits
    mask = (1 << (n - 1))
    while (n - 1 > 0):
 
        # Calculate bitwise XOR of num and mask
        temp = (num ^ mask)
 
        # Math.min(a, b) returns minimum of a and b
        # return minimum number of flips till that
        # digit
        minXor = min(minXor, countones(temp))
        n -= 1
        mask = (mask | (1 << n - 1))
         
    return minXor
     
# Function to count number of 1s
def countones(n):
     
    c = 0
    while (n > 0):
        n = n & (n - 1)
        c += 1
         
    return c
 
# Driver code
if __name__ == "__main__":
 
    nums = [ 1, 0, 1, 1, 0, 0, 0 ]
    n = findMiniFlip(nums)
     
    print(n)
 
# This code is contributed by chitranayal

C#

// C# program to find
// minimum flips to make
// all 1s in left
using System;
class GFG{
 
// Function to count minimum
// number of flips
public static int findMiniFlip(int[] nums)
{
  int n = nums.Length;
  String s = "";
  for (int i = 0; i < n; i++)
    s += nums[i];
 
  // This is converting string
  // s into integer of base 2
  // (if s = '100' then num = 4)
  long num = Convert.ToInt32(s, 2);
 
  // initialize minXor with n
  // that can be maximum
  // number of flips
  int minXor = n;
 
  // right shift 1 by (n-1) bits
  long mask = (1 << (n - 1));
  while (n - 1 > 0)
  {
    // calculate bitwise XOR
    // of num and mask
    long temp = (num ^ mask);
 
    // Math.Min(a, b) returns
    // minimum of a and b
    // return minimum number
    // of flips till that
    // digit
    minXor = Math.Min(minXor,
                      countones(temp));
    n--;
    mask = (mask | (1 << n - 1));
  }
  return minXor;
}
 
// Function to count number of 1s
public static int countones(long n)
{
  int c = 0;
  while (n > 0)
  {
    n = n & (n - 1);
    c++;
  }
  return c;
}
 
// Driver code
public static void Main(String[] args)
{
  int[] nums = {1, 0, 1, 1,
                0, 0, 0};
  int n = findMiniFlip(nums);
  Console.WriteLine(n);
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// JavaScript program to find minimum
// flips to make all 1s in left
 
// Function to count minimum number of flips
function findMiniFlip(nums)
{
    let n = nums.length;
    let s = "";
    for(let i = 0; i < n; i++)
        s += nums[i];
 
    // This is converting string s into integer
    // of base 2 (if s = '100' then num = 4)
    let num = parseInt(s, 2);
 
    // Initialize minXor with n that
    // can be maximum number of flips
    let minXor = n;
 
    // Right shift 1 by (n-1) bits
    let mask = (1 << (n - 1));
    while (n - 1 > 0)
    {
         
        // Calculate bitwise XOR of num and mask
        let temp = (num ^ mask);
 
        // Math.min(a, b) returns minimum of a and b
        // return minimum number of flips till that
        // digit
        minXor = Math.min(minXor, countones(temp));
        n--;
 
        mask = (mask | (1 << n -1));
    }
    return minXor;
}
 
// Function to count number of 1s
function countones(n)
{
    let c = 0;
    while (n > 0)
    {
        n = n & (n - 1);
        c++;
    }
    return c;
}
 
// Driver Code
let nums = [ 1, 0, 1, 1, 0, 0, 0 ];
let n = findMiniFlip(nums);
 
document.write(n);
 
// This code is contributed by code_hunt
 

输出：

1