使二进制字符串可被2 ^ k整除所需的最小交换

给定长度为N且整数K的二进制字符串S ，任务是找到使二进制字符串可被2 ^K整除所需的最小相邻交换数。如果不可能，则打印-1 。

例子：

Input: S = “100111”, K = 2
Output: 6
Swapping the right-most zero 3 times
to the right, we get “101110”.
Swapping the second right-most zero
3 times to the right, we get “111100”.

Input: S = “1011”, K = 2
Output: -1

方法1：一种方法将从最右边的零交换。因此，让我们将问题改写为更简单的方法。需要最小数量的交换，以便在右端至少提供K个连续的零。
一种方法是模拟交换。从最右边的零开始，交换它直到它的右边有1而不是字符串的末尾。将对K个最右边的零执行此操作。这种方法的时间复杂度将是O(N * K) 。

方法2：在此处取得更好性能的关键是避免进行仿真。
观察：

Among the K right-most zeros, each zero will need to be swapped X number of times, where X is the number of 1s to the right of that zero.

因此，对于K个最右边的零，任务是找到每个右边的1个数的总和。

算法：

• 初始化变量c_zero = 0 ， c_one = 0和ans = 0 。
• 从i = N – 1到i = 0运行循环。
  • 如果S [i] = 0，则更新c_zero = c_zero + 1和ans = ans + c_one 。
  • 如果S [i] = 1，则更新c_one = c_one + 1 。
  • 如果c_zero = K，则断开。
• 如果c_zero 则返回-1 。
• 最后，返回ans 。

因此，该方法的时间复杂度将为O(N) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum swaps required
int findMinSwaps(string s, int k)
{
    // To store the final answer
    int ans = 0;
  
    // To store the count of one and zero
    int c_one = 0, c_zero = 0;
  
    // Loop from end of the string
    for (int i = s.size() - 1; i >= 0; i--) {
  
        // If s[i] = 1
        if (s[i] == '1')
            c_one++;
  
        // If s[i] = 0
        if (s[i] == '0')
            c_zero++, ans += c_one;
  
        // If c_zero = k
        if (c_zero == k)
            break;
    }
  
    // If the result can't
    // be achieved
    if (c_zero < k)
        return -1;
  
    // Return the final answer
    return ans;
}
  
// Driver code
int main()
{
    string s = "100111";
    int k = 2;
  
    cout << findMinSwaps(s, k);
  
    return 0;
}

Java

// Java implementation of the approach
  
class GFG 
{
      
    // Function to return the minimum swaps required 
    static int findMinSwaps(String s, int k) 
    { 
        // To store the final answer 
        int ans = 0; 
      
        // To store the count of one and zero 
        int c_one = 0, c_zero = 0; 
      
        // Loop from end of the string 
        for (int i = s.length() - 1; i >= 0; i--)
        { 
      
            // If s[i] = 1 
            if (s.charAt(i) == '1') 
                c_one++; 
      
            // If s[i] = 0 
            if (s.charAt(i) == '0') 
            {
                c_zero++;
                ans += c_one; 
            }
      
            // If c_zero = k 
            if (c_zero == k) 
                break; 
        } 
      
        // If the result can't 
        // be achieved 
        if (c_zero < k) 
            return -1; 
      
        // Return the final answer 
        return ans; 
    } 
      
    // Driver code 
    public static void main (String[] args)
    { 
        String s = "100111"; 
        int k = 2; 
      
        System.out.println(findMinSwaps(s, k)); 
      
    } 
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach 
  
# Function to return the minimum swaps required 
def findMinSwaps(s, k) : 
  
    # To store the final answer 
    ans = 0; 
  
    # To store the count of one and zero 
    c_one = 0; c_zero = 0; 
  
    # Loop from end of the string 
    for i in range(len(s)-1, -1, -1) :
  
        # If s[i] = 1 
        if (s[i] == '1') :
            c_one += 1; 
  
        # If s[i] = 0 
        if (s[i] == '0') :
            c_zero += 1;
            ans += c_one; 
  
        # If c_zero = k 
        if (c_zero == k) :
            break; 
  
    # If the result can't 
    # be achieved 
    if (c_zero < k) :
        return -1; 
  
    # Return the final answer 
    return ans; 
  
# Driver code 
if __name__ == "__main__" : 
  
    s = "100111"; 
    k = 2; 
  
    print(findMinSwaps(s, k)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach 
using System;
  
class GFG 
{ 
      
    // Function to return the minimum swaps required 
    static int findMinSwaps(string s, int k) 
    { 
        // To store the final answer 
        int ans = 0; 
      
        // To store the count of one and zero 
        int c_one = 0, c_zero = 0; 
      
        // Loop from end of the string 
        for (int i = s.Length - 1; i >= 0; i--) 
        { 
      
            // If s[i] = 1 
            if (s[i] == '1') 
                c_one++; 
      
            // If s[i] = 0 
            if (s[i] == '0') 
            { 
                c_zero++; 
                ans += c_one; 
            } 
      
            // If c_zero = k 
            if (c_zero == k) 
                break; 
        } 
      
        // If the result can't 
        // be achieved 
        if (c_zero < k) 
            return -1; 
      
        // Return the final answer 
        return ans; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        string s = "100111"; 
        int k = 2; 
      
        Console.WriteLine(findMinSwaps(s, k)); 
    } 
} 
  
// This code is contributed by AnkitRai01

输出：

6