通过将 B 向右移动和 A 向左移动而不交叉将字符串S1 转换为 S2

给定由'A'、'B' 和 '#'组成的字符串s1和s2 ，其中：

• '#' 代表一个空单元格
• “A”代表可以向左移动的机器人
• “B”代表可以朝正确方向移动的机器人

任务是检查是否可以通过机器人的移动将 s1 转换为 s2。

例子：

Input: s1 = “#B#A#”, s2 = “##BA#”
Output: Yes
Explanation: In one step ‘B’ moves one position to the right.
So the string becomes “##BA#” which is same as the final s2 string

Input: s1 = “#B#A#”, s2 = “#A#B#”  
Output: No
Explanation: As the robots cannot cross each other.

方法：解决问题的思路是基于 以下观察：

The robots can reach the final position when the strings satisfy the condition:

• If the empty spaces ‘#’ are removed from string then both the strings should be identical as no crossing of A and B are allowed.
• After removing the ‘#’ from both the strings, if positions of A is lesser in s2 than in s1 and positions of ‘B’ is greater in s2 than in s1.

按照下面提到的步骤来实现上述观察：

• 从头开始迭代数组：
  • 从两个字符串中删除空格 ( '#' )。
  • 将A和B的位置存储在两个字符串中
• 如果修改后的字符串s1 和 s2 完全相同，那么它们可以在以下情况下达到最终状态：
  • 'A'在s1中的位置大于或等于'A'在s2中的位置， 'B'在s1中的位置小于或等于'B'在s2中的位置。
• 在所有其他情况下，机器人无法到达s2中提到的最终位置。

以下是上述方法的实现。

C++

// C++ code for the above approach:
 
#include 
using namespace std;
 
// Function to check if robots can move
string moveRobots(string s1, string s2)
{
 
    // Strings to save s1 and s2 without '#'
    string a, b;
    for (int i = 0; i < s1.size(); i++)
        if (s1[i] != '#')
            a += s1[i];
    for (int i = 0; i < s2.size(); i++)
        if (s2[i] != '#')
            b += s2[i];
 
 
    // Condition 1: strings s1 and s2
    // without empty spaces should be
    // exactly same
    if (a == b) {
        int n = a.size();
 
        // v1 and v2 will store the
        // positions of 'A' or 'B'
        // in s1 and s2 respectively
        vector v1;
        vector v2;
        for (int i = 0; i < s1.size(); i++) {
            if (s1[i] != '#')
                v1.push_back(i);
        }
        for (int i = 0; i < s2.size(); i++)
            if (s2[i] != '#')
                v2.push_back(i);
 
        // Condition 2:
        // Position of 'A' in s1 should be
        // greater than or equal to
        // Position of 'A' in s2 and
        // Position of 'B' in s1 should be
        // less than or equal to
        // Position of 'B' in s2
        if (a[0] == 'A' and v1[0] < v2[0])
            return "No";
        if (a[0] == 'B' and v1[0] > v2[0])
            return "No";
        for (int i = 1; i < n; i++) {
            if (a[i] == 'A') {
                if (v1[i] < v2[i])
                    return "No";
            }
            else {
                if (v1[i] > v2[i])
                    return "No";
            }
        }
        return "Yes";
    }
    return "No";
}
 
// Driver code
int main()
{
    string s1 = "#B#A#";
    string s2 = "##BA#";
    cout << moveRobots(s1, s2) << endl;
    return 0;
}

Java

// Java code for the above approach:
import java.util.*;
class GFG {
 
  // Function to check if robots can move
  static String moveRobots(String s1, String s2)
  {
 
    // Strings to save s1 and s2 without '#'
    String a = "", b = "";
    for (int i = 0; i < s1.length(); i++)
      if (s1.charAt(i) != '#')
        a += s1.charAt(i);
    for (int i = 0; i < s2.length(); i++)
      if (s2.charAt(i) != '#')
        b += s2.charAt(i);
 
    // Condition 1: strings s1 and s2
    // without empty spaces should be
    // exactly same
    if (a.equals(b)) {
      int n = a.length();
 
      // v1 and v2 will store the
      // positions of 'A' or 'B'
      // in s1 and s2 respectively
      ArrayList v1
        = new ArrayList();
      ArrayList v2
        = new ArrayList();
      for (int i = 0; i < s1.length(); i++) {
        if (s1.charAt(i) != '#')
          v1.add(i);
      }
      for (int i = 0; i < s2.length(); i++)
        if (s2.charAt(i) != '#')
          v2.add(i);
 
 
      // Condition 2:
      // Position of 'A' in s1 should be
      // greater than or equal to
      // Position of 'A' in s2 and
      // Position of 'B' in s1 should be
      // less than or equal to
      // Position of 'B' in s2
      if (a.charAt(0) == 'A'
          && (int)v1.get(0) < (int)v2.get(0))
        return "No";
      if (a.charAt(0) == 'B'
          && (int)v1.get(0) > (int)v2.get(0))
        return "No";
      for (int i = 1; i < n; i++) {
        if (a.charAt(i) == 'A') {
          if ((int)v1.get(i) < (int)v2.get(i))
            return "No";
        }
        else {
          if ((int)v1.get(i) > (int)v2.get(i))
            return "No";
        }
      }
      return "Yes";
    }
    return "No";
  }
 
  // Driver code
  public static void main(String[] args)
  {
    String s1 = "#B#A#";
    String s2 = "##BA#";
    System.out.println(moveRobots(s1, s2));
  }
}
 
// This code is contributed by ukasp.

Python3

# python3 code for the above approach:
 
# Function to check if robots can move
def moveRobots(s1, s2):
 
    # Strings to save s1 and s2 without '#'
    a, b = "", ""
    for i in range(0, len(s1)):
        if (s1[i] != '#'):
            a += s1[i]
    for i in range(0, len(s2)):
        if (s2[i] != '#'):
            b += s2[i]
 
    # Condition 1: strings s1 and s2
    # without empty spaces should be
    # exactly same
    if (a == b):
        n = len(a)
 
        # v1 and v2 will store the
        # positions of 'A' or 'B'
        # in s1 and s2 respectively
        v1 = []
        v2 = []
        for i in range(0, len(s1)):
            if (s1[i] != '#'):
                v1.append(i)
 
        for i in range(0, len(s2)):
            if (s2[i] != '#'):
                v2.append(i)
 
        # Condition 2:
        # Position of 'A' in s1 should be
        # greater than or equal to
        # Position of 'A' in s2 and
        # Position of 'B' in s1 should be
        # less than or equal to
        # Position of 'B' in s2
        if (a[0] == 'A' and v1[0] < v2[0]):
            return "No"
        if (a[0] == 'B' and v1[0] > v2[0]):
            return "No"
        for i in range(1, n):
            if (a[i] == 'A'):
                if (v1[i] < v2[i]):
                    return "No"
 
            else:
                if (v1[i] > v2[i]):
                    return "No"
 
        return "Yes"
 
    return "No"
 
# Driver code
if __name__ == "__main__":
 
    s1 = "#B#A#"
    s2 = "##BA#"
    print(moveRobots(s1, s2))
 
# This code is contributed by rakeshsahni

C#

// C# code for the above approach:
using System;
using System.Collections;
 
class GFG
{
 
  // Function to check if robots can move
  static string moveRobots(string s1, string s2)
  {
 
    // Strings to save s1 and s2 without '#'
    string a = "", b = "";
    for (int i = 0; i < s1.Length; i++)
      if (s1[i] != '#')
        a += s1[i];
    for (int i = 0; i < s2.Length; i++)
      if (s2[i] != '#')
        b += s2[i];
 
    // Condition 1: strings s1 and s2
    // without empty spaces should be
    // exactly same
    if (a == b) {
      int n = a.Length;
 
      // v1 and v2 will store the
      // positions of 'A' or 'B'
      // in s1 and s2 respectively
      ArrayList v1 = new ArrayList();
      ArrayList v2 = new ArrayList();
      for (int i = 0; i < s1.Length; i++) {
        if (s1[i] != '#')
          v1.Add(i);
      }
      for (int i = 0; i < s2.Length; i++)
        if (s2[i] != '#')
          v2.Add(i);
 
      // Condition 2:
      // Position of 'A' in s1 should be
      // greater than or equal to
      // Position of 'A' in s2 and
      // Position of 'B' in s1 should be
      // less than or equal to
      // Position of 'B' in s2
      if (a[0] == 'A' && (int)v1[0] < (int)v2[0])
        return "No";
      if (a[0] == 'B' && (int)v1[0] > (int)v2[0])
        return "No";
      for (int i = 1; i < n; i++) {
        if (a[i] == 'A') {
          if ((int)v1[i] < (int)v2[i])
            return "No";
        }
        else {
          if ((int)v1[i] > (int)v2[i])
            return "No";
        }
      }
      return "Yes";
    }
    return "No";
  }
 
  // Driver code
  public static void Main()
  {
    string s1 = "#B#A#";
    string s2 = "##BA#";
    Console.Write(moveRobots(s1, s2));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code for the above approach:
 
    // Function to check if robots can move
    const moveRobots = (s1, s2) => {
 
        // Strings to save s1 and s2 without '#'
        let a, b;
        for (let i = 0; i < s1.length; i++)
            if (s1[i] != '#')
                a += s1[i];
        for (let i = 0; i < s2.length; i++)
            if (s2[i] != '#')
                b += s2[i];
 
        // Condition 1: strings s1 and s2
        // without empty spaces should be
        // exactly same
        if (a == b) {
            let n = a.length;
 
            // v1 and v2 will store the
            // positions of 'A' or 'B'
            // in s1 and s2 respectively
            let v1 = [];
            let v2 = [];
            for (let i = 0; i < s1.length; i++) {
                if (s1[i] != '#')
                    v1.push(i);
            }
            for (let i = 0; i < s2.length; i++)
                if (s2[i] != '#')
                    v2.push(i);
 
            // Condition 2:
            // Position of 'A' in s1 should be
            // greater than or equal to
            // Position of 'A' in s2 and
            // Position of 'B' in s1 should be
            // less than or equal to
            // Position of 'B' in s2
            if (a[0] == 'A' && v1[0] < v2[0])
                return "No";
            if (a[0] == 'B' && v1[0] > v2[0])
                return "No";
            for (let i = 1; i < n; i++) {
                if (a[i] == 'A') {
                    if (v1[i] < v2[i])
                        return "No";
                }
                else {
                    if (v1[i] > v2[i])
                        return "No";
                }
            }
            return "Yes";
        }
        return "No";
    }
 
    // Driver code
 
    let s1 = "#B#A#";
    let s2 = "##BA#";
    document.write(moveRobots(s1, s2));
 
// This code is contributed by rakeshsahni
 

输出

Yes

时间复杂度： O(N)
辅助空间： O(N)