通过给定的操作检查 str1 是否可以转换为 str2

给定两个二进制字符串str1和str2 。任务是检查是否可以通过在 str1 上应用以下操作任意次数来将str1转换为str2 。
在每个操作中，可以将两个连续的 0 组合成一个1 。
例子：

Input: str1 = “00100”, str2 = “111”
Output: Yes
Combine first two zeros to one and combine last two zeros to one.
Input: str1 = “00”, str2 = “000”
Output: No
It is not possible to convert str1 to str2.

编程需要懂一点英语

方法：让我们从左到右逐个字符字符并行处理 str1 和 str2。让我们使用两个索引 i 和 j：索引 i 用于 str1，索引 j 用于 str2。现在，有两种情况：

如果str1[i] = str2[j]然后增加i和j 。
如果str1[i] != str2[j]那么，
- 如果 str1 中有两个连续的 0，即str1[i] = 0和str1[i + 1] = 0和str2[j] = 1 ，这意味着可以将两个 0 组合以匹配str2中的1 。因此，将i增加2并将j增加1 。
- 否则str1无法转换为str2 。
如果最后i和j都位于它们各自字符串的末尾，即str1和str2 ，那么答案是Yes否则答案是No 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true if str1 can be
// converted to str2 with the given operations
bool canConvert(string str1, string str2)
{
    int i = 0, j = 0;
 
    // Traverse from left to right
    while (i < str1.size() && j < str2.size()) {
 
        // If the two characters do not match
        if (str1[i] != str2[j]) {
 
            // If possible to combine
            if (str1[i] == '0' && str2[j] == '1'
                && i + 1 < str1.size()
                && str1[i + 1] == '0') {
                i += 2;
                j++;
            }
 
            // If not possible to combine
            else {
                return false;
            }
        }
 
        // If the two characters match
        else {
            i++;
            j++;
        }
    }
 
    // If possible to convert one string to another
    if (i == str1.size() && j == str2.size())
        return true;
    return false;
}
 
// Driver code
int main()
{
    string str1 = "00100", str2 = "111";
 
    if (canConvert(str1, str2))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
    // Function that returns true if str1 can be
    // converted to str2 with the given operations
    static boolean canConvert(String str1, String str2)
    {
        int i = 0, j = 0;
 
        // Traverse from left to right
        while (i < str1.length() && j < str2.length())
        {
 
            // If the two characters do not match
            if (str1.charAt(i) != str2.charAt(j))
            {
 
                // If possible to combine
                if (str1.charAt(i) == '0' && str2.charAt(j) == '1'
                    && i + 1 < str1.length() && str1.charAt(i+1) == '0')
                {
                    i += 2;
                    j++;
                }
 
                // If not possible to combine
                else
                {
                    return false;
                }
            }
 
            // If the two characters match
            else
            {
                i++;
                j++;
            }
        }
 
        // If possible to convert one string to another
        if (i == str1.length() && j == str2.length())
            return true;
        return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        String str1 = "00100", str2 = "111";
 
        if (canConvert(str1, str2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python implementation of the approach
 
# Function that returns true if str1 can be
# converted to str2 with the given operations
def canConvert(str1, str2):
    i, j = 0, 0;
 
    # Traverse from left to right
    while (i < len(str1) and j < len(str2)):
 
        # If the two characters do not match
        if (str1[i] != str2[j]):
 
            # If possible to combine
            if (str1[i] == '0' and str2[j] == '1'
                and i + 1 < len(str1)
                and str1[i + 1] == '0'):
                i += 2;
                j+=1;
 
            # If not possible to combine
            else:
                return False;
        # If the two characters match
        else:
            i += 1;
            j += 1;
             
    # If possible to convert one string to another
    if (i == len(str1) and j == len(str2)):
        return True;
    return False;
 
# Driver code
str1 = "00100";
str2 = "111";
 
if (canConvert(str1, str2)):
    print("Yes");
else:
    print("No");
 
# This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function that returns true if str1 can be
    // converted to str2 with the given operations
    static bool canConvert(string str1, string str2)
    {
        int i = 0, j = 0;
 
        // Traverse from left to right
        while (i < str1.Length && j < str2.Length)
        {
 
            // If the two characters do not match
            if (str1[i] != str2[j])
            {
 
                // If possible to combine
                if (str1[i] == '0' && str2[j] == '1'
                    && i + 1 < str1.Length && str1[i+1] == '0')
                {
                    i += 2;
                    j++;
                }
 
                // If not possible to combine
                else
                {
                    return false;
                }
            }
 
            // If the two characters match
            else
            {
                i++;
                j++;
            }
        }
 
        // If possible to convert one string to another
        if (i == str1.Length && j == str2.Length)
            return true;
        return false;
    }
 
    // Driver code
    public static void Main()
    {
 
        string str1 = "00100", str2 = "111";
 
        if (canConvert(str1, str2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by AnkitRai01

Javascript

输出：

Yes