通过替换将数字字符串修改为平衡括号

给定一个仅由字符‘1’、’2′和‘3’组成的数字字符串S ，任务是用开括号（ ‘(‘ ) 或闭括号 ( ‘)’ ）替换字符，使得新形成的字符串成为平衡括号序列。

注意：所有出现的字符必须替换为相同的括号。

例子：

Input: S = “1123”
Output: Yes, (())
Explanation: Replacing occurrences of character ‘1’ with ‘(‘, ‘2’ with ‘)’ and ‘3’ with ‘)’. Therefore, the obtained bracket sequence is “(())”, which is balanced.

Input: S = “1121”
Output: No

方法：根据以下观察可以解决给定的问题：

• 对于平衡支架序列，有必要为第一和最后一个字符是打开和关闭分别括号。因此，第一个和最后一个字符应该不同。
• 如果第一和字符串的最后一个字符是相同的，那么就不可能获得平衡的支架序列。
• 如果字符串的第一个和最后一个字符是不同的，那么它们被打开和关闭括号分别替换。第三个字符是通过打开或关闭托架任一替换。
• 检查剩余的第三个字符的两种替换方式。
• 如果剩下的第三个字符的两个替换都不能构成平衡括号序列，那么就不可能构成平衡括号序列。

请按照以下步骤解决给定的问题：

• 检查字符串S的第一个和最后一个字符是否相等。如果发现为真，则打印“否”并返回。
• 初始化两个变量，比如cntforOpen和cntforClose ，以存储开括号和闭括号的计数。
• 遍历字符串中的字符，然后执行以下操作：
  • 如果当前字符与字符串的第一个字符相同，则增加cntforOpen。
  • 如果当前字符与字符串的最后一个字符相同，则递减cntforOpen。
  • 对于剩余的第三个字符，增加cntforOpen ，即用‘(‘替换该字符。
  • 如果在任何时刻，发现cntforOpen为负，则无法获得平衡的括号序列。
• 同样，使用cntforClose变量进行检查，即将第三个字符替换为‘)’ 。
• 如果以上两种方法都没有生成平衡括号序列，则打印“No” 。否则，打印“是”。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
void balBracketSequence(string str)
{
    int n = str.size();
 
    // Check if the first and
    // last characters are equal
    if (str[0]
        == str[n - 1])
 
    {
        cout << "No" << endl;
    }
    else {
 
        // Initialize two variables to store
        // the count of open and closed brackets
        int cntForOpen = 0, cntForClose = 0;
        int check = 1;
        for (int i = 0; i < n; i++) {
 
            // If the current character is
            // same as the first character
            if (str[i] == str[0])
                cntForOpen++;
 
            // If the current character is
            // same as the last character
            else if (str[i] == str[n - 1])
                cntForOpen--;
            else
                cntForOpen++;
 
            // If count of open brackets
            // becomes less than 0
            if (cntForOpen < 0) {
                check = 0;
                break;
            }
        }
        if (check && cntForOpen == 0) {
            cout << "Yes, ";
 
            // Print the new string
            for (int i = 0; i < n; i++) {
                if (str[i] == str[n - 1])
                    cout << ')';
                else
                    cout << '(';
            }
            return;
        }
        else {
            for (int i = 0; i < n; i++) {
 
                // If the current character is
                // same as the first character
                if (str[i] == str[0])
                    cntForClose++;
                else
                    cntForClose--;
 
                // If bracket sequence
                // is not balanced
                if (cntForClose
                    < 0) {
                    check = 0;
                    break;
                }
            }
 
            // Check for unbalanced
            // bracket sequence
            if (check
                && cntForClose
                       == 0) {
                cout << "Yes, ";
 
                // Print the sequence
                for (int i = 0; i < n;
                     i++) {
                    if (str[i] == str[0])
                        cout << '(';
                    else
                        cout << ')';
                }
                return;
            }
        }
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    // Given Input
    string str = "123122";
 
    // Function Call
    balBracketSequence(str);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
static void balBracketSequence(String str)
{
    int n = str.length();
 
    // Check if the first and
    // last characters are equal
    if (str.charAt(0)
        == str.charAt(n - 1))
 
    {
        System.out.println("No");
    }
    else {
 
        // Initialize two variables to store
        // the count of open and closed brackets
        int cntForOpen = 0, cntForClose = 0;
        int check = 1;
        for (int i = 0; i < n; i++) {
 
            // If the current character is
            // same as the first character
            if (str.charAt(i) == str.charAt(0))
                cntForOpen++;
 
            // If the current character is
            // same as the last character
            else if (str.charAt(i) == str.charAt(n - 1))
                cntForOpen -= 1;
            else
                cntForOpen += 1;
 
            // If count of open brackets
            // becomes less than 0
            if (cntForOpen < 0) {
                check = 0;
                break;
            }
        }
        if (check != 0 && cntForOpen == 0) {
            System.out.print("Yes, ");
 
            // Print the new string
            for (int i = 0; i < n; i++) {
                if (str.charAt(i) == str.charAt(n - 1))
                    System.out.print(')');
                else
                   System.out.print('(');
            }
            return;
        }
        else {
            for (int i = 0; i < n; i++) {
 
                // If the current character is
                // same as the first character
                if (str.charAt(i) == str.charAt(0))
                    cntForClose++;
                else
                    cntForClose--;
 
                // If bracket sequence
                // is not balanced
                if (cntForClose
                    < 0) {
                    check = 0;
                    break;
                }
            }
 
            // Check for unbalanced
            // bracket sequence
            if (check != 0
                && cntForClose
                       == 0) {
                System.out.print("Yes, ");
 
                // Print the sequence
                for (int i = 0; i < n;
                     i++) {
                    if (str.charAt(i) == str.charAt(0))
                       System.out.print('(');
                    else
                        System.out.print(')');
                }
                return;
            }
        }
        System.out.print("No");
    }
}
 
// Driver Code
public static void main(String args[])
{
    // Given Input
    String str = "123122";
 
    // Function Call
    balBracketSequence(str);
}
}
 
// This code is contributed by ipg2016107.

Python3

# Python program for the above approach;
# Function to check if the given
# string can be converted to a
# balanced bracket sequence or not
def balBracketSequence(str):
    n = len(str)
 
    # Check if the first and
    # last characters are equal
    if (str[0] == str[n - 1]):
        print("No", end="")
    else:
 
        # Initialize two variables to store
        # the count of open and closed brackets
        cntForOpen = 0
        cntForClose = 0
        check = 1
 
        for i in range(n):
 
            # If the current character is
            # same as the first character
            if (str[i] == str[0]):
                cntForOpen += 1
 
            # If the current character is
            # same as the last character
            elif str[i] == str[n - 1] :
                cntForOpen -= 1
            else:
                cntForOpen += 1
 
            # If count of open brackets
            # becomes less than 0
            if (cntForOpen < 0):
                check = 0
                break
             
         
        if (check and cntForOpen == 0):
            print("Yes, ", end="")
 
            # Prlet the new string
            for i in range(n):
                if (str[i] == str[n - 1]):
                    print(')', end="")
                else:
                    print('(', end="")
             
            return
         
        else:
            for i in range(n):
 
                # If the current character is
                # same as the first character
                if (str[i] == str[0]):
                    cntForClose += 1
                else:
                    cntForClose -= 1
 
                # If bracket sequence
                # is not balanced
                if (cntForClose < 0):
                    check = 0
                    break
                 
             
 
            # Check for unbalanced
            # bracket sequence
            if (check and cntForClose == 0):
                print("Yes, ", end="")
 
                # Prlet the sequence
                for i in range(n):
                    if (str[i] == str[0]):
                        print('(', end="") 
                    else:
                        print(')', end="")
                 
                return
             
         
        print("NO", end="")
     
 
 
# Driver Code
 
# Given Input
str = "123122"
 
# Function Call
balBracketSequence(str)
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
static void balBracketSequence(string str)
{
    int n = str.Length;
     
    // Check if the first and
    // last characters are equal
    if (str[0] == str[n - 1])
    {
        Console.Write("No");
    }
    else
    {
         
        // Initialize two variables to store
        // the count of open and closed brackets
        int cntForOpen = 0, cntForClose = 0;
        int check = 1;
         
        for(int i = 0; i < n; i++)
        {
             
            // If the current character is
            // same as the first character
            if (str[i] == str[0])
                cntForOpen++;
 
            // If the current character is
            // same as the last character
            else if (str[i] == str[n - 1])
                cntForOpen--;
            else
                cntForOpen++;
 
            // If count of open brackets
            // becomes less than 0
            if (cntForOpen < 0)
            {
                check = 0;
                break;
            }
        }
        if (check != 0 && cntForOpen == 0)
        {
            Console.Write("Yes, ");
 
            // Print the new string
            for(int i = 0; i < n; i++)
            {
                if (str[i] == str[n - 1])
                    Console.Write(')');
                else
                    Console.Write('(');
            }
            return;
        }
        else
        {
            for(int i = 0; i < n; i++)
            {
                 
                // If the current character is
                // same as the first character
                if (str[i] == str[0])
                    cntForClose++;
                else
                    cntForClose--;
 
                // If bracket sequence
                // is not balanced
                if (cntForClose < 0)
                {
                    check = 0;
                    break;
                }
            }
 
            // Check for unbalanced
            // bracket sequence
            if (check != 0 && cntForClose == 0)
            {
                Console.Write("Yes, ");
 
                // Print the sequence
                for(int i = 0; i < n; i++)
                {
                    if (str[i] == str[0])
                        Console.Write('(');
                    else
                        Console.Write(')');
                }
                return;
            }
        }
        Console.Write("No");
    }
}
 
// Driver Code
public static void Main()
{
     
    // Given Input
    string str = "123122";
 
    // Function Call
    balBracketSequence(str);
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
        // JavaScript program for the above approach;
// Function to check if the given
// string can be converted to a
// balanced bracket sequence or not
function balBracketSequence(str)
{
    let n = str.length;
 
    // Check if the first and
    // last characters are equal
    if (str[0]
        == str[n - 1])
 
    {
        document.write( "No");
    }
    else {
 
        // Initialize two variables to store
        // the count of open and closed brackets
        let cntForOpen = 0, cntForClose = 0;
        let check = 1;
        for (let i = 0; i < n; i++) {
 
            // If the current character is
            // same as the first character
            if (str[i] == str[0])
                cntForOpen++;
 
            // If the current character is
            // same as the last character
            else if (str[i] == str[n - 1])
                cntForOpen--;
            else
                cntForOpen++;
 
            // If count of open brackets
            // becomes less than 0
            if (cntForOpen < 0) {
                check = 0;
                break;
            }
        }
        if (check && cntForOpen == 0) {
            document.write("Yes, ");
 
            // Prlet the new string
            for (let i = 0; i < n; i++) {
                if (str[i] == str[n - 1])
                    document.write(')');
                else
                    document.write('(');
            }
            return;
        }
        else {
            for (let i = 0; i < n; i++) {
 
                // If the current character is
                // same as the first character
                if (str[i] == str[0])
                    cntForClose++;
                else
                    cntForClose--;
 
                // If bracket sequence
                // is not balanced
                if (cntForClose
                    < 0) {
                    check = 0;
                    break;
                }
            }
 
            // Check for unbalanced
            // bracket sequence
            if (check
                && cntForClose
                       == 0) {
                document.write("Yes, ");
 
                // Prlet the sequence
                for (let i = 0; i < n;
                     i++) {
                    if (str[i] == str[0])
                        document.write('(');
                    else
                        document.write(')');
                }
                return;
            }
        }
       document.write("NO") ;
    }
}
 
// Driver Code
 
    // Given Input
    let str = "123122";
 
    // Function Call
    balBracketSequence(str);
    
   // This code is contributed by Potta Lokesh
    

输出：

Yes, ()(())

时间复杂度： O(N)
辅助空间： O(1)

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