最小化将二进制字符串的所有字符转换为 0 的成本

给定一个二进制字符串str 、两个整数数组R[]和大小为N 的C[] 。将所有字符从索引i翻转到R[i]需要C[i]成本。任务是最小化将给定二进制字符串转换为仅0所需的成本。

例子：

Input: str = “1010”, R[] = {1, 2, 2, 3}, C[] = {3, 1, 2, 3}
Output: 4
Explanation:
Flipping all the characters from index 1 to 2, modifies str to “1100”. Therefore, cost = 1
Flipping all the characters from index 1 to 2, modifies str to “0000”. Therefore, cost = cost + 3 = 4
Therefore, minimum cost required is 4.

Input: str = “01100”, R[] = {1, 2, 3, 4, 5}, C[] = {1, 5, 5, 2, 3}
Output: 10

方法：该问题可以使用贪心技术解决。这个想法是从左到右遍历给定的字符串并检查当前字符是否为非零字符。如果发现为真，则将所有字符从当前索引 (= i)翻转到第R[i]^个索引。请按照以下步骤解决问题：

• 初始化一个变量，比如flip ，以存储当前字符可以翻转的次数。
• 创建一个优先级队列，比如pq来存储当前翻转值大于 0 的字符右侧的字符索引范围。
• 初始化一个变量，比如cost来存储获得所需字符串的最小成本。
• 遍历给定的字符串并检查当前字符是否为“ 1 ”。如果发现为真，则将i到R[i]范围内的所有字符翻转，并将 cost 的值增加C[i] 。
• 最后，打印cost的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to get the minimum
// Cost to convert all characters
// of given string to 0s
int minCost(string str, int N, int R[], int C[])
{
    // Stores the range of indexes
    // of characters that need
    // to be flipped
    priority_queue, greater > pq;
 
    // Stores the number of times
    // current character is flipped
    int flip = 0;
 
    // Stores minimum cost to get
    // the required string
    int cost = 0;
 
    // Traverse the given string
    for (int i = 0; i < N; i++)
    {
        // Remove all value from pq
        // whose value is less than i
        while (pq.size() > 0 and pq.top() < i)
        {
            pq.pop();
 
            // Update flip
            flip--;
        }
 
        // If current character
        // is flipped odd times
        if (flip % 2 == 1)
        {
            str[i] = '1' - str[i] + '0';
        }
 
        // If current character contains
        // non-zero value
        if (str[i] == '1')
        {
            // Update flip
            flip++;
 
            // Update cost
            cost += C[i];
 
            // Append R[i] into pq
            pq.push(R[i]);
        }
    }
    return cost;
}
 
// Driver Code
int main()
{
    string str = "1010";
    int R[] = { 1, 2, 2, 3 };
    int C[] = { 3, 1, 2, 3 };
    int N = str.length();
   
    // Function call
    cout << minCost(str, N, R, C);
}

Java

// Java program to implement
// the above approach
 
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG {
 
    // Function to get the minimum
    // Cost to convert all characters
    // of given string to 0s
    public static int minCost(String s,
                              int R[],
                              int C[],
                              int N)
    {
        char ch[] = s.toCharArray();
 
        // Stores the range of indexes
        // of characters that need
        // to be flipped
        PriorityQueue pq = new PriorityQueue<>();
 
        // Stores the number of times
        // current character is flipped
        int flip = 0;
 
        // Stores minimum cost to get
        // the required string
        int cost = 0;
 
        // Traverse the given string
        for (int i = 0; i < N; i++) {
 
            // Remove all value from pq
            // whose value is less than i
 
            while (pq.size() > 0 && pq.peek() < i)
            {
                pq.poll();
 
                // Update flip
                flip--;
            }
 
            // Get the current number
            int cn = ch[i] - '0';
 
            // If current character
            // is flipped odd times
            if (flip % 2 == 1)
                cn = 1 - cn;
 
            // If current character contains
            // non-zero value
            if (cn == 1)
            {
                // Update flip
                flip++;
 
                // Update cost
                cost += C[i];
 
                // Append R[i] into pq
                pq.add(R[i]);
            }
        }
        return cost;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 4;
        String s = "1010";
        int R[] = { 1, 2, 2, 3 };
        int C[] = { 3, 1, 2, 3 };
 
        // Function call
        System.out.println(minCost(s, R, C, N));
    }
}

Python3

# Python3 program to implement
# the above approach
 
# Function to get the minimum
# Cost to convert all characters
# of given string to 0s
def minCost(s, R, C, N) :
    ch = list(s)
  
    # Stores the range of indexes
    # of characters that need
    # to be flipped
    pq = []
  
    # Stores the number of times
    # current character is flipped
    flip = 0
  
    # Stores minimum cost to get
    # the required string
    cost = 0
  
    # Traverse the given string
    for i in range(N) :
          
        # Remove all value from pq
        # whose value is less than i
        while (len(pq) > 0 and pq[0] < i) :
     
            pq.pop(0);
  
            # Update flip
            flip -= 1
  
        # Get the current number
        cn = ord(ch[i]) - ord('0')
  
        # If current character
        # is flipped odd times
        if (flip % 2 == 1) :
            cn = 1 - cn
  
        # If current character contains
        # non-zero value
        if (cn == 1) :
              
            # Update flip
            flip += 1
  
            # Update cost
            cost += C[i]
  
            # Append R[i] into pq
            pq.append(R[i])
              
    return cost
   
  # Driver code
N = 4
s = "1010"
R = [ 1, 2, 2, 3 ]
C = [ 3, 1, 2, 3 ]
 
# Function call
print(minCost(s, R, C, N))
 
# This code is contributed by divyeshrabadiya07.

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to get the minimum
// Cost to convert all characters
// of given string to 0s
public static int minCost(String s, int []R,
                           int []C, int N)
{
    char []ch = s.ToCharArray();
 
    // Stores the range of indexes
    // of characters that need
    // to be flipped
    Queue pq = new Queue();
 
    // Stores the number of times
    // current character is flipped
    int flip = 0;
 
    // Stores minimum cost to get
    // the required string
    int cost = 0;
 
    // Traverse the given string
    for(int i = 0; i < N; i++)
    {
         
        // Remove all value from pq
        // whose value is less than i
        while (pq.Count > 0 && pq.Peek() < i)
        {
            pq.Dequeue();
 
            // Update flip
            flip--;
        }
 
        // Get the current number
        int cn = ch[i] - '0';
 
        // If current character
        // is flipped odd times
        if (flip % 2 == 1)
            cn = 1 - cn;
 
        // If current character contains
        // non-zero value
        if (cn == 1)
        {
             
            // Update flip
            flip++;
 
            // Update cost
            cost += C[i];
 
            // Append R[i] into pq
            pq.Enqueue(R[i]);
             
        }
    }
    return cost;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 4;
    String s = "1010";
    int []R = { 1, 2, 2, 3 };
    int []C = { 3, 1, 2, 3 };
 
    // Function call
    Console.WriteLine(minCost(s, R, C, N));
}
}
 
// This code is contributed by Amit Katiyar

输出

4

时间复杂度： O(N * log N)
辅助空间： O(N)

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