通过重复将 K 个连续字符转换为 1 来最小化将给定的二进制字符串作为全 1 的操作

给定一个包含 N 个字符的二进制字符串str一个整数K ，任务是找到将字符串的所有字符转换为1所需的最小移动，其中在每次移动时， K个连续字符可以转换为1 。

例子：

Input: str=”0010″, K=3
Output: 2
Explanation:
Move 1: Select the substring from 0 to 2 and replace it with all 1.
Move 2: Select the substring from 1 to 3 and replace it with all 1.

Input: str=”0000010″, K=1
Output: 6

方法：这个问题可以使用贪心方法来解决。要解决此问题，请执行以下步骤：

1. 创建一个变量cnt ，以存储所需的最小移动次数。用0初始化它。
2. 使用变量i和str[i] = '0'遍历字符串，然后将 i 更新为i + K并将cnt增加1 ，因为无论这些字符是什么，总是需要移动来转换字符str[i ]到'1' 。
3. 循环结束后，返回cnt这将是所需的答案。

下面是上述方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to find the minimum
// operations required to convert
// the binary string str to all 1s
int minMoves(string str, int K)
{
    int N = str.size();
 
    // Variable to store number
    // of minimum moves required
    int cnt = 0;
 
    int i = 0;
 
    // Loop to traverse str
    while (i < N) {
 
        // If element is '0'
        if (str[i] == '0') {
            i += K;
            cnt += 1;
        }
 
        // If element is '1'
        else {
            i++;
        }
    }
 
    return cnt;
}
 
// Driver Code
int main()
{
    string str = "0010";
    int K = 3;
    cout << minMoves(str, K);
}

Java

// Java code for the above approach
class GFG {
 
    // Function to find the minimum
    // operations required to convert
    // the binary String str to all 1s
    static int minMoves(String str, int K) {
        int N = str.length();
 
        // Variable to store number
        // of minimum moves required
        int cnt = 0;
 
        int i = 0;
 
        // Loop to traverse str
        while (i < N) {
 
            // If element is '0'
            if (str.charAt(i) == '0') {
                i += K;
                cnt += 1;
            }
 
            // If element is '1'
            else {
                i++;
            }
        }
 
        return cnt;
    }
 
    // Driver Code
    public static void main(String args[]) {
        String str = "0010";
        int K = 3;
        System.out.println(minMoves(str, K));
    }
}
 
// This code is contributed by Saurabh Jaiswal

Python3

# Python code for the above approach
 
# Function to find the minimum
# operations required to convert
# the binary string str to all 1s
def minMoves(str, K):
    N = len(str)
 
    # Variable to store number
    # of minimum moves required
    cnt = 0
 
    i = 0
 
    # Loop to traverse str
    while (i < N):
 
        # If element is '0'
        if (str[i] == '0'):
            i += K
            cnt += 1
        # If element is '1'
        else:
            i += 1
 
    return cnt
 
# Driver Code
str = "0010"
K = 3
print(minMoves(str, K))
 
# This code is contributed by Saurabh Jaiswal

C#

// C# code for the above approach
using System;
 
class GFG
{
   
// Function to find the minimum
// operations required to convert
// the binary string str to all 1s
static int minMoves(string str, int K)
{
    int N = str.Length;
 
    // Variable to store number
    // of minimum moves required
    int cnt = 0;
 
    int i = 0;
 
    // Loop to traverse str
    while (i < N) {
 
        // If element is '0'
        if (str[i] == '0') {
            i += K;
            cnt += 1;
        }
 
        // If element is '1'
        else {
            i++;
        }
    }
 
    return cnt;
}
 
// Driver Code
public static void Main()
{
    string str = "0010";
    int K = 3;
    Console.Write(minMoves(str, K));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

       // JavaScript code for the above approach
 
 
       // Function to find the minimum
       // operations required to convert
       // the binary string str to all 1s
       function minMoves(str, K) {
           let N = str.length;
 
           // Variable to store number
           // of minimum moves required
           let cnt = 0;
 
           let i = 0;
 
           // Loop to traverse str
           while (i < N) {
 
               // If element is '0'
               if (str[i] == '0') {
                   i += K;
                   cnt += 1;
               }
 
               // If element is '1'
               else {
                   i++;
               }
           }
 
           return cnt;
       }
 
       // Driver Code
 
       let str = "0010";
       let K = 3;
       document.write(minMoves(str, K));
 
 // This code is contributed by Potta Lokesh
   

输出

2

时间复杂度： O(N)
辅助空间： O(1)