给定一个二进制字符串str ，两个整数数组R []和C []的大小为N。将所有字符从索引i翻转到R [i]需要C [i]成本。任务是使将给定的二进制字符串转换为仅0所需的成本最小化。
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：可以使用贪婪技术解决问题。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。如果发现是正确的，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，例如flip ，以存储当前字符可以翻转的次数。
- 创建一个优先级队列，例如说pq，以将当前字符右侧的字符索引范围存储在翻转值大于0的当前字符的右侧。
- 初始化一个变量，例如cost，以存储获得所需字符串的最低成本。
- 遍历给定的字符串，并检查当前字符是否为“ 1 ”。如果确定为真，则将范围i中的所有字符翻转到R [i]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)