将字符串的所有字符转换为给定字符所需的最少操作

给定字符串str 、字符ch和整数K ，任务是找到将字符串str 的所有字符转换为ch所需的最少操作次数。每个操作都涉及从索引i 的每一侧转换K 个最接近的字符，即索引[i – K, i + K]中的字符可以转换为ch 。
注意：每个索引只能是单个操作的一部分。
例子：

Input: str = “abcsdx”, K = 2, ch = ‘#’ 
Output: 2 
Explanation: 
Operation 1: Select i = 1, therefore, str = “abcsdx” modifies to “###sdx”. 
Operation 2: Select i = 6, therefore, str = “###sdx” modifies to “######”.
 

Input: str = “Hellomypkfsg”, k = 3, ch = ‘$’ 
Output:2 
Explanation: 
Operation 1: Select i = 2, therefore, str = “Hellomypkfsg” modifies to “$$$$$mypkfsg”. 
Operation 2: Select i = 9, therefore, str = “$$$$$mypkfsg” modifies to “$$$$$$$$$$$$”.

方法：
请按照以下步骤解决问题：

1. 对于任何索引i ，可以转换的最大字符数为2 * K + 1 。因此，如果字符串中的字符总数不超过2 * K ，则只需一次操作即可将整个字符串转换为ch 。
2. 否则，所需的操作次数将为ceil(n / (2*k+1)) 。
3. 迭代字符串，从可用于操作的最左边的可能索引开始，并在它之后打印每个(2*k+1) ^th索引。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to find the minimum
// number of operations required
void countOperations(int n, int k)
{
    // Maximum number of characters that
    // can be changed in one operation
    int div = 2 * k + 1;
 
    // If length of the string less than
    // maximum number of characters that
    // can be changed in an operation
    if (n / 2 <= k) {
        cout << 1 << "\n";
 
        // Set the last index as the
        // index for the operation
        if (n > k)
            cout << k + 1;
 
        else
            cout << n;
    }
 
    // Otherwise
    else {
 
        // If size of the string is
        // equal to the maximum number
        // of characters in an operation
        if (n % div == 0) {
 
            // Find the number of
            // operations required
            int oprn = n / div;
 
            cout << oprn << "\n";
 
            // Find the starting postion
            int pos = k + 1;
 
            cout << pos << " ";
            for (int i = 1; i <= oprn; i++) {
                // Print i-th index
                cout << pos << " ";
 
                // Shift to next index
                pos += div;
            }
        }
 
        // Otherwise
        else {
 
            // Find the number of
            // operations required
            int oprn = n / div + 1;
            cout << oprn << "\n";
 
            int pos = n % div;
 
            // If n % div exceeds k
            if (n % div > k)
                pos -= k;
 
            for (int i = 1; i <= oprn; i++) {
 
                // Print i-th index
                cout << pos << " ";
 
                // Shift to next index
                pos += div;
            }
        }
    }
}
 
// Driver Code
int main()
{
    string str = "edfreqwsazxet";
    char ch = '$';
    int n = str.size();
    int k = 4;
    countOperations(n, k);
 
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to find the minimum
// number of operations required
static void countOperations(int n, int k)
{
     
    // Maximum number of characters that
    // can be changed in one operation
    int div = 2 * k + 1;
 
    // If length of the String less than
    // maximum number of characters that
    // can be changed in an operation
    if (n / 2 <= k)
    {
        System.out.print(1 + "\n");
 
        // Set the last index as the
        // index for the operation
        if (n > k)
            System.out.print(k + 1);
 
        else
            System.out.print(n);
    }
 
    // Otherwise
    else
    {
         
        // If size of the String is
        // equal to the maximum number
        // of characters in an operation
        if (n % div == 0)
        {
             
            // Find the number of
            // operations required
            int oprn = n / div;
 
            System.out.print(oprn + "\n");
 
            // Find the starting postion
            int pos = k + 1;
 
            System.out.print(pos + " ");
            for(int i = 1; i <= oprn; i++)
            {
                 
                // Print i-th index
                System.out.print(pos + " ");
 
                // Shift to next index
                pos += div;
            }
        }
 
        // Otherwise
        else
        {
             
            // Find the number of
            // operations required
            int oprn = n / div + 1;
            System.out.print(oprn + "\n");
 
            int pos = n % div;
 
            // If n % div exceeds k
            if (n % div > k)
                pos -= k;
 
            for(int i = 1; i <= oprn; i++)
            {
                 
                // Print i-th index
                System.out.print(pos + " ");
 
                // Shift to next index
                pos += div;
            }
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "edfreqwsazxet";
    char ch = '$';
    int n = str.length();
    int k = 4;
     
    countOperations(n, k);
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 program to implement
# the above approach
 
# Function to find the minimum
# number of operations required
def countOperations(n, k):
     
    # Maximum number of characters that
    # can be changed in one operation
    div = 2 * k + 1
 
    # If length of the less than
    # maximum number of characters that
    # can be changed in an operation
    if (n // 2 <= k):
        print(1)
 
        # Set the last index as the
        # index for the operation
        if (n > k):
            print(k + 1)
        else:
            print(n)
 
    # Otherwise
    else:
 
        # If size of the is
        # equal to the maximum number
        # of characters in an operation
        if (n % div == 0):
 
            # Find the number of
            # operations required
            oprn = n // div
 
            print(oprn)
 
            # Find the starting postion
            pos = k + 1
 
            print(pos, end = " ")
            for i in range(1, oprn + 1):
                 
                # Print i-th index
                print(pos, end = " ")
 
                # Shift to next index
                pos += div
 
        # Otherwise
        else:
 
            # Find the number of
            # operations required
            oprn = n // div + 1
            print(oprn)
 
            pos = n % div
 
            # If n % div exceeds k
            if (n % div > k):
                pos -= k
 
            for i in range(1, oprn + 1):
 
                # Print i-th index
                print(pos, end = " ")
 
                # Shift to next index
                pos += div
 
# Driver Code
if __name__ == '__main__':
     
    str = "edfreqwsazxet"
    ch = '$'
    n = len(str)
    k = 4
     
    countOperations(n, k)
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to find the minimum
// number of operations required
static void countOperations(int n, int k)
{
     
    // Maximum number of characters that
    // can be changed in one operation
    int div = 2 * k + 1;
 
    // If length of the String less than
    // maximum number of characters that
    // can be changed in an operation
    if (n / 2 <= k)
    {
        Console.Write(1 + "\n");
 
        // Set the last index as the
        // index for the operation
        if (n > k)
            Console.Write(k + 1);
 
        else
            Console.Write(n);
    }
 
    // Otherwise
    else
    {
         
        // If size of the String is
        // equal to the maximum number
        // of characters in an operation
        if (n % div == 0)
        {
             
            // Find the number of
            // operations required
            int oprn = n / div;
 
            Console.Write(oprn + "\n");
 
            // Find the starting postion
            int pos = k + 1;
 
            Console.Write(pos + " ");
            for(int i = 1; i <= oprn; i++)
            {
                 
                // Print i-th index
                Console.Write(pos + " ");
 
                // Shift to next index
                pos += div;
            }
        }
 
        // Otherwise
        else
        {
             
            // Find the number of
            // operations required
            int oprn = n / div + 1;
            Console.Write(oprn + "\n");
 
            int pos = n % div;
 
            // If n % div exceeds k
            if (n % div > k)
                pos -= k;
 
            for(int i = 1; i <= oprn; i++)
            {
                 
                // Print i-th index
                Console.Write(pos + " ");
 
                // Shift to next index
                pos += div;
            }
        }
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "edfreqwsazxet";
    int n = str.Length;
    int k = 4;
     
    countOperations(n, k);
}
}
  
// This code is contributed by Rohit_ranjan

Javascript

// javascript program to implement
// the above approach   
// Function to find the minimum
    // number of operations required
    function countOperations(n , k) {
 
        // Maximum number of characters that
        // can be changed in one operation
        var div = 2 * k + 1;
 
        // If length of the String less than
        // maximum number of characters that
        // can be changed in an operation
        if (n / 2 <= k) {
            document.write(1 + "<br/>");
 
            // Set the last index as the
            // index for the operation
            if (n > k)
                document.write(k + 1);
 
            else
                document.write(n);
        }
 
        // Otherwise
        else {
 
            // If size of the String is
            // equal to the maximum number
            // of characters in an operation
            if (n % div == 0) {
 
                // Find the number of
                // operations required
                var oprn = parseInt(n / div);
 
                document.write(oprn + "<br/>");
 
                // Find the starting postion
                var pos = k + 1;
 
                document.write(pos + " ");
                for (i = 1; i <= oprn; i++) {
 
                    // Prvar i-th index
                    document.write(pos + " ");
 
                    // Shift to next index
                    pos += div;
                }
            }
 
            // Otherwise
            else {
 
                // Find the number of
                // operations required
                var oprn = parseInt(n / div + 1);
                document.write(oprn + "<br/>");
 
                var pos = n % div;
 
                // If n % div exceeds k
                if (n % div > k)
                    pos -= k;
 
                for (i = 1; i <= oprn; i++) {
 
                    // Prvar i-th index
                    document.write(pos + " ");
 
                    // Shift to next index
                    pos += div;
                }
            }
        }
    }
 
    // Driver Code
     
        var str = "edfreqwsazxet";
        var ch = '$';
        var n = str.length;
        var k = 4;
 
        countOperations(n, k);
 
// This code contributed by Rajput-Ji

输出：

2
4 13

时间复杂度： O(N)
辅助空间： O(1)

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