检查是否可以在给定约束下将一个字符串转换为另一个字符串

给定的两个字符串包含三个字符，即“ A”，“ B”和“＃”。通过首先对字符串执行以下操作，检查是否可以将第一个字符串转换为另一个字符串。
1-‘A’只能向左移动
2-‘B’只能向右移动
3-‘A’和’B’都不交叉
如果可能，则打印“是”，否则打印“否”。
例子：

Input : str1=” #A#B#B# “, str2=” A###B#B ” 
Output :Yes 
Explanation : 
‘A’ in str1 is right to the ‘A’ in str2 so ‘A’ of str1 can move easily towards the left because there is no ‘B’ on its left positions and for first ‘B’ in str1 is left to the ‘B’ in str2 so ‘B’ of str2 can move easily towards the right because there is no ‘A’ on its right positions and it is same for next ‘B’ so str1 can be easily converted into str2.
Input :str1=” #A#B# “, str2=” #B#A# ” 
Output :No 
Explanation : 
Here first ‘A’ in str1 is left to the ‘A’ in str2 and according to the condition ‘A’ can’tmove towards right. so str1 can’t be converted into str2. 

方法 ：
两个字符串的1长度必须相同
2号两个字符串中A和B的个数必须相等
两个字符串中A和B的3阶应该相同(例如：如果第二个字符串“ A”在“ B”之前，那么必须首先对字符串进行相同的排序)

C++

// C++ Program for above implementation
#include 
using namespace std;
 
// Function to check is it possible to convert
// first string into another string or not.
bool isItPossible(string str1, string str2, int m, int n)
{
 
    // To Check Length of Both String is Equal or Not
    if (m != n)
        return false;
 
    // To Check  Frequency of A's and  B's are
    // equal in both strings or not.
    if (count(str1.begin(), str1.end(), 'A') !=
           count(str2.begin(), str2.end(), 'A') ||
        count(str1.begin(), str1.end(), 'B') !=
            count(str2.begin(), str2.end(), 'B'))
        return false;
 
    // Start traversing
    for (int i = 0; i < m; i++) {
        if (str1[i] != '#') {
            for (int j = 0; j < n; j++) {
 
                // To Check no two elements cross each other.
                if ((str2[j] != str1[i]) && str2[j] != '#')
                    return false;
 
                if (str2[j] == str1[i]) {
                    str2[j] = '#';
 
                    // To Check Is it Possible to Move
                    // towards Left or not.
                    if (str1[i] == 'A' && i < j)
                        return false;
 
                    // To Check Is it Possible to Move
                    // towards Right or not.
                    if (str1[i] == 'B' && i > j)
                        return false;
 
                    break;
                }
            }
        }
    }
 
    return true;
}
 
// Drivers code
int main()
{
    string str1 = "A#B#";
    string str2 = "A##B";
 
    int m = str1.length();
    int n = str2.length();
 
    isItPossible(str1, str2, m, n) ? cout << "Yes\n"
                                   : cout << "No\n";
 
    return 0;
}

Java

// Java Program for above implementation
class GFG
{
 
// Function to check is it possible to convert
// first String into another String or not.
static boolean isItPossible(char[] str1, char[] str2,
                            int m, int n)
{
 
    // To Check Length of Both String is Equal or Not
    if (m != n)
        return false;
 
    // To Check Frequency of A's and B's are
    // equal in both Strings or not.
    if (count(str1, 'A') !=
        count(str2, 'A') ||
        count(str1, 'B') !=
            count(str2, 'B'))
        return false;
 
    // Start traversing
    for (int i = 0; i < m; i++) {
        if (str1[i] != '#') {
            for (int j = 0; j < n; j++) {
 
                // To Check no two elements cross each other.
                if ((str2[j] != str1[i]) && str2[j] != '#')
                    return false;
 
                if (str2[j] == str1[i]) {
                    str2[j] = '#';
 
                    // To Check Is it Possible to Move
                    // towards Left or not.
                    if (str1[i] == 'A' && i < j)
                        return false;
 
                    // To Check Is it Possible to Move
                    // towards Right or not.
                    if (str1[i] == 'B' && i > j)
                        return false;
 
                    break;
                }
            }
        }
    }
 
    return true;
}
 
private static int count(char[] str1, char c) {
    int count = 0;
    for(char temp : str1) {
        if(c == temp)
            count++;
    }
    return count;
}
 
// Drivers code
public static void main(String[] args)
{
    String str1 = "A#B#";
    String str2 = "A##B";
 
    int m = str1.length();
    int n = str2.length();
 
    System.out.print(isItPossible(str1.toCharArray(), str2.toCharArray(), m, n) ?
            "Yes\n":"No\n");
 
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python Program for above implementation
 
# Function to check is it possible to convert
# first string into another string or not.
def isItPossible(str1, str2, m, n):
 
    # To Check Length of Both String is Equal or Not
    if (m != n):
        return False
 
    # To Check Frequency of A's and B's are
    # equal in both strings or not.
    if str1.count('A') != str2.count('A') \
    or str1.count('B') != str2.count('B'):
        return False
 
    # Start traversing
    for i in range(m):
        if (str1[i] != '#'):
            for j in range(n):
                # To Check no two elements cross each other.
                if ((str2[j] != str1[i]) and str2[j] != '#'):
                    return False
 
                if (str2[j] == str1[i]):
                    str2[j] = '#'
 
                    # To Check Is it Possible to Move
                    # towards Left or not.
                    if (str1[i] == 'A' and i < j):
                        return False
 
                    # To Check Is it Possible to Move
                    # towards Right or not.
                    if (str1[i] == 'B' and i > j):
                        return False
 
                    break
                 
    return True
 
# Drivers code
 
str1 = "A#B#"
str2 = "A##B"
 
m = len(str1)
n = len(str2)
 
str1 = list(str1)
str2 = list(str2)
 
if(isItPossible(str1, str2, m, n)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by ankush_953

C#

// C# Program for above implementation
using System;
 
class GFG
{
  
// Function to check is it possible to convert
// first String into another String or not.
static bool isItPossible(char[] str1, char[] str2,
                            int m, int n)
{
  
    // To Check Length of Both String is Equal or Not
    if (m != n)
        return false;
  
    // To Check Frequency of A's and B's are
    // equal in both Strings or not.
    if (count(str1, 'A') !=
        count(str2, 'A') ||
        count(str1, 'B') !=
            count(str2, 'B'))
        return false;
  
    // Start traversing
    for (int i = 0; i < m; i++) {
        if (str1[i] != '#') {
            for (int j = 0; j < n; j++) {
  
                // To Check no two elements cross each other.
                if ((str2[j] != str1[i]) && str2[j] != '#')
                    return false;
  
                if (str2[j] == str1[i]) {
                    str2[j] = '#';
  
                    // To Check Is it Possible to Move
                    // towards Left or not.
                    if (str1[i] == 'A' && i < j)
                        return false;
  
                    // To Check Is it Possible to Move
                    // towards Right or not.
                    if (str1[i] == 'B' && i > j)
                        return false;
  
                    break;
                }
            }
        }
    }
  
    return true;
}
  
private static int count(char[] str1, char c) {
    int count = 0;
    foreach(char temp in str1) {
        if(c == temp)
            count++;
    }
    return count;
}
  
// Drivers code
public static void Main(String[] args)
{
    String str1 = "A#B#";
    String str2 = "A##B";
  
    int m = str1.Length;
    int n = str2.Length;
  
    Console.Write(isItPossible(str1.ToCharArray(), str2.ToCharArray(), m, n) ?
            "Yes\n":"No\n");
  
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// js Program for above implementation
function getFreq(string,chr) {
  let ans = 0;
    for (var i=0; i<string.length;i++) {
        if( chr == string.charAt(i))
          ans++; 
    }
    return ans;
};
// Function to check is it possible to convert
// first string into another string or not.
function isItPossible(str1, str2, m, n){
    // To Check Length of Both String is Equal or Not
    if (m != n)
        return false;
    // To Check  Frequency of A's and  B's are
    // equal in both strings or not.
    if (getFreq(str1, 'A') !=
           getFreq(str2, 'A') ||
        getFreq(str1, 'B') !=
            getFreq(str2, 'B'))
        return false;
 
    // Start traversing
    for (let i = 0; i < m; i++) {
        if (str1[i] != '#') {
            for (let j = 0; j < n; j++) {
 
                // To Check no two elements cross each other.
                if ((str2[j] != str1[i]) && str2[j] != '#')
                    return false;
 
                if (str2[j] == str1[i]) {
                    str2 = str2.substr(0,j)+'#'+str2.substr(j+1);
 
                    // To Check Is it Possible to Move
                    // towards Left or not.
                    if (str1[i] == 'A' && i < j)
                        return false;
 
                    // To Check Is it Possible to Move
                    // towards Right or not.
                    if (str1[i] == 'B' && i > j)
                        return false;
 
                    break;
                }
            }
        }
    }
 
    return true;
}
 
// Drivers code
let str1 = "A#B#";
let str2 = "A##B";
 
let m = str1.length;
let n = str2.length;
isItPossible(str1, str2, m, n) ?document.write( "Yes<br>")
                                   : document.write( "No<br>");
 
 

输出：

Yes

时间复杂度： O(n ^ 2)

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