给定二进制字符串中不重叠的子字符串“101”和“010”的计数

给定二进制字符串str ，任务是找到“010”或“101”形式的非重叠子字符串的计数。

例子：

Input: str = “10101010101” 
Output: 3 
str[0..2] = “101” 
str[3..5] = “010” 
str[6..8] = “101”
Input: str = “111111111111110” 
Output: 0 

方法：初始化count = 0并为给定字符串中的每个索引i检查从当前索引i开始的大小为3的子字符串是否与“010”或“101”匹配。如果匹配，则更新count = count + 1和i = i + 3 （以避免子字符串重叠）否则将i增加1 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count of
// required non-overlapping sub-strings
int countSubStr(string s, int n)
{
 
    // To store the required count
    int count = 0;
    for (int i = 0; i < n - 2;) {
 
        // If "010" matches the sub-string
        // starting at current index i
        if (s[i] == '0' && s[i + 1] == '1'
            && s[i + 2] == '0') {
            count++;
            i += 3;
        }
 
        // If "101" matches the sub-string
        // starting at current index i
        else if (s[i] == '1' && s[i + 1] == '0'
                 && s[i + 2] == '1') {
            count++;
            i += 3;
        }
        else {
            i++;
        }
    }
 
    return count;
}
 
// Driver code
int main()
{
    string s = "10101010101";
    int n = s.length();
 
    cout << countSubStr(s, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    // Function to return the count of
    // required non-overlapping sub-strings
    static int countSubStr(char[] s, int n)
    {
 
        // To store the required count
        int count = 0;
        for (int i = 0; i < n - 2😉
        {
 
            // If "010" matches the sub-string
            // starting at current index i
            if (s[i] == '0' && s[i + 1] == '1'
                    && s[i + 2] == '0')
            {
                count++;
                i += 3;
            }
            // If "101" matches the sub-string
            // starting at current index i
            else if (s[i] == '1' && s[i + 1] == '0'
                    && s[i + 2] == '1')
            {
                count++;
                i += 3;
            }
            else
             
            {
                i++;
            }
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        char[] s = "10101010101".toCharArray();
        int n = s.length;
 
        System.out.println(countSubStr(s, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the count of
# required non-overlapping sub-strings
def countSubStr(s, n) :
 
    # To store the required count
    count = 0;
    i = 0
     
    while i < (n-2) :
 
        # If "010" matches the sub-string
        # starting at current index i
        if (s[i] == '0' and s[i + 1] == '1'and s[i + 2] == '0') :
            count += 1;
            i += 3;
 
        # If "101" matches the sub-string
        # starting at current index i
        elif (s[i] == '1' and s[i + 1] == '0'and s[i + 2] == '1') :
            count += 1;
            i += 3;
         
        else :
            i += 1;
 
    return count;
 
 
# Driver code
if __name__ == "__main__" :
 
    s = "10101010101";
    n = len(s);
 
    print(countSubStr(s, n));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
     
class GFG
{
    // Function to return the count of
    // required non-overlapping sub-strings
    static int countSubStr(char[] s, int n)
    {
 
        // To store the required count
        int count = 0;
        for (int i = 0; i < n - 2;)
        {
 
            // If "010" matches the sub-string
            // starting at current index i
            if (s[i] == '0' &&
                s[i + 1] == '1' &&
                s[i + 2] == '0')
            {
                count++;
                i += 3;
            }
             
            // If "101" matches the sub-string
            // starting at current index i
            else if (s[i] == '1' &&
                     s[i + 1] == '0' &&
                     s[i + 2] == '1')
            {
                count++;
                i += 3;
            }
            else
            {
                i++;
            }
        }
 
        return count;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        char[] s = "10101010101".ToCharArray();
        int n = s.Length;
 
        Console.WriteLine(countSubStr(s, n));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

// javascript implementation of the approach
 
    // Function to return the count of
    // required non-overlapping sub-strings
    function countSubStr( s , n) {
 
        // To store the required count
        var count = 0;
        for (i = 0; i < n - 2;) {
 
            // If "010" matches the sub-string
            // starting at current index i
            if (s[i] == '0' && s[i + 1] == '1' && s[i + 2] == '0') {
                count++;
                i += 3;
            }
            // If "101" matches the sub-string
            // starting at current index i
            else if (s[i] == '1' && s[i + 1] == '0' && s[i + 2] == '1') {
                count++;
                i += 3;
            } else
 
            {
                i++;
            }
        }
 
        return count;
    }
 
    // Driver code
     
        var s = "10101010101";
        var n = s.length;
 
        document.write(countSubStr(s, n));
 
// This code contributed by Rajput-Ji

输出：

3

时间复杂度： O(n)，其中 n 是字符串的长度。

辅助空间： O(n)