删除给定二进制字符串中任何连续 3 个 0 或 1 的最小翻转次数

给定一个由N个字符组成的二进制字符串S ，任务是找到所需的最小翻转次数，使得不存在三个连续的相同字符。

例子：

Input: S = “1100011”
Output: 1
Explanation:
Flip the character at index 3 modifies the string S “1101011” that have no three consecutive same characters. Therefore, the minimum number of flips required is 1.

Input: S = “0001111101”
Output: 2

方法：给定的问题可以通过考虑每三个连续字符来解决，如果它们相同，则增加所需的翻转次数，因为需要翻转三个字符之一。请按照以下步骤解决问题：

• 初始化变量，例如count为0 ，存储所需的最小翻转次数。
• 如果字符串的大小小于等于2，则返回0 ，因为不需要任何翻转。
• 使用变量i迭代范围[0, N – 2)并执行以下步骤：
  • 如果索引i 、 (i + 1)和(i + 2)字符处的字符相同，则将count的值增加1并将i的值增加3 。
  • 否则，将i的值增加1 。
• 执行上述步骤后，打印count的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum number
// of flips to make all three pairs of
// consecutive characters different
int minFlips(string str)
{
    // Stores resultant count of pairs
    int count = 0;
 
    // Base Case
    if (str.size() <= 2) {
        return 0;
    }
 
    // Iterate over the range [0, N - 2]
    for (int i = 0; i < str.size() - 2;) {
 
        // If the consecutive 3 numbers
        // are the same then increment
        // the count and the counter
        if (str[i] == str[i + 1]
            && str[i + 2] == str[i + 1]) {
            i = i + 3;
            count++;
        }
        else {
            i++;
        }
    }
 
    // Return the answer
    return count;
}
 
// Driver Code
int main()
{
    string S = "0011101";
    cout << minFlips(S);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG
{
 
// Function to find the minimum number
// of flips to make all three pairs of
// consecutive characters different
static int minFlips(String str)
{
    // Stores resultant count of pairs
    int count = 0;
   
    // Base Case
    if (str.length() <= 2) {
        return 0;
    }
   
    // Iterate over the range [0, N - 2]
    for (int i = 0; i < str.length() - 2😉 {
   
        // If the consecutive 3 numbers
        // are the same then increment
        // the count and the counter
        if (str.charAt(i) == str.charAt(i+1)
            && str.charAt(i+2) == str.charAt(i+1)) {
            i = i + 3;
            count++;
        }
        else {
            i++;
        }
    }
   
    // Return the answer
    return count;
}
   
// Driver Code
public static void main(String[] args)
{
     String S = "0011101";
   System.out.println(minFlips(S));
}
}
 
// This code is contributed by dwivediyash

Python3

# python 3 program for the above approach
#
# Function to find the minimum number
# of flips to make all three pairs of
# consecutive characters different
def minFlips(st):
 
    # Stores resultant count of pairs
    count = 0
 
    # Base Case
    if (len(st) <= 2):
        return 0
 
    # Iterate over the range [0, N - 2]
    for i in range(len(st) - 2):
 
        # If the consecutive 3 numbers
        # are the same then increment
        # the count and the counter
        if (st[i] == st[i + 1]
                and st[i + 2] == st[i + 1]):
            i = i + 3
            count += 1
 
        else:
            i += 1
 
    # Return the answer
    return count
 
# Driver Code
if __name__ == "__main__":
 
    S = "0011101"
    print(minFlips(S))
 
    # This code is contributed by ukasp.

Javascript

        // JavaScript Program to implement
        // the above approach
 
 
        // Function to find the minimum number
        // of flips to make all three pairs of
        // consecutive characters different
        function minFlips(str) {
            // Stores resultant count of pairs
            let count = 0;
 
            // Base Case
            if (str.length <= 2) {
                return 0;
            }
 
            // Iterate over the range [0, N - 2]
            for (let i = 0; i < str.length - 2;) {
 
                // If the consecutive 3 numbers
                // are the same then increment
                // the count and the counter
                if (str[i] == str[i + 1]
                    && str[i + 2] == str[i + 1]) {
                    i = i + 3;
                    count++;
                }
                else {
                    i++;
                }
            }
 
            // Return the answer
            return count;
        }
 
        // Driver Code
 
        let S = "0011101";
        document.write(minFlips(S));
 
// This code is contributed by Potta Lokesh
    

C#

// C# program for the above approach
using System;
 
 
public class GFG
{
 
// Function to find the minimum number
// of flips to make all three pairs of
// consecutive characters different
static int minFlips(string str)
{
    // Stores resultant count of pairs
    int count = 0;
   
    // Base Case
    if (str.Length <= 2) {
        return 0;
    }
   
    // Iterate over the range [0, N - 2]
    for (int i = 0; i < str.Length - 2;) {
   
        // If the consecutive 3 numbers
        // are the same then increment
        // the count and the counter
        if (str[i] == str[i+1]
            && str[i+2] == str[i+1]) {
            i = i + 3;
            count++;
        }
        else {
            i++;
        }
    }
   
    // Return the answer
    return count;
}
   
// Driver Code
public static void Main(string[] args)
{
     string S = "0011101";
     Console.WriteLine(minFlips(S));
}
}
 
// This code is contributed by AnkThon

输出：

1

时间复杂度： O(N)
辅助空间： O(1)