检查方括号的顺序是否最多可以改变一个方括号的位置。套装2

给定一个括号序列作为字符串str ，任务是查找是否可以通过将最多一个括号从序列中的原始位置移动到任何其他位置来平衡给定的字符串。

例子：

Input: str = “)(()”
Output: Yes
As by moving s[0] to the end will make it valid.
“(())”

Input: str = “()))(()”
Output: No

方法：可以使用本文中讨论的堆栈解决问题。在本文中，将讨论一种不占用额外空间的方法。
如果’(’的频率小于”’的频率。如果上述差异大于1，则序列无法平衡，否则，如果总差异为零，则可以平衡序列。

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
// the string can be balanced
bool canBeBalanced(string s, int n)
{
  
    // Count to check the difference between
    // the frequencies of '(' and ')' and
    // count_1 is to find the minimum value
    // of freq('(') - freq(')')
    int count = 0, count_1 = 0;
  
    // Traverse the given string
    for (int i = 0; i < n; i++) {
  
        // Increase the count
        if (s[i] == '(')
            count++;
  
        // Decrease the count
        else
            count--;
  
        // Find the minimum value
        // of freq('(') - freq(')')
        count_1 = min(count_1, count);
    }
  
    // If the minimum difference is greater
    // than or equal to -1 and the overall
    // difference is zero
    if (count_1 >= -1 && count == 0)
        return true;
  
    return false;
}
  
// Driver code
int main()
{
    string s = "())()(";
    int n = s.length();
  
    if (canBeBalanced(s, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

// Java program to toggle K-th bit of a number N
class GFG
{
  
// Function that returns true if
// the string can be balanced
static boolean canBeBalanced(String s, int n)
{
  
    // Count to check the difference between
    // the frequencies of '(' and ')' and
    // count_1 is to find the minimum value
    // of freq('(') - freq(')')
    int count = 0, count_1 = 0;
  
    // Traverse the given string
    for (int i = 0; i < n; i++) 
    {
  
        // Increase the count
        if (s.charAt(i) == '(')
            count++;
  
        // Decrease the count
        else
            count--;
  
        // Find the minimum value
        // of freq('(') - freq(')')
        count_1 = Math.min(count_1, count);
    }
  
    // If the minimum difference is greater
    // than or equal to -1 and the overall
    // difference is zero
    if (count_1 >= -1 && count == 0)
        return true;
  
    return false;
}
  
// Driver code
public static void main(String []args)
{
    String s = "())()(";
    int n = s.length();
  
    if (canBeBalanced(s, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
  
# Function that returns true if
# the can be balanced
def canBeBalanced(s, n):
  
    # Count to check the difference between
    # the frequencies of '(' and ')' and
    # count_1 is to find the minimum value
    # of freq('(') - freq(')')
    count = 0
    count_1 = 0
  
    # Traverse the given string
    for i in range(n):
  
        # Increase the count
        if (s[i] == '('):
            count += 1
  
        # Decrease the count
        else:
            count -= 1
  
        # Find the minimum value
        # of freq('(') - freq(')')
        count_1 = min(count_1, count)
  
    # If the minimum difference is greater
    # than or equal to -1 and the overall
    # difference is zero
    if (count_1 >= -1 and count == 0):
        return True
  
    return False
  
# Driver code
s = "())()("
n = len(s)
  
if (canBeBalanced(s, n)):
    print("Yes")
else:
    print("No")
  
# This code is contributed by Mohit Kumar

C#

// C# program to toggle K-th bit of a number N
using System;
  
class GFG
{
  
// Function that returns true if
// the string can be balanced
static Boolean canBeBalanced(String s, int n)
{
  
    // Count to check the difference between
    // the frequencies of '(' and ')' and
    // count_1 is to find the minimum value
    // of freq('(') - freq(')')
    int count = 0, count_1 = 0;
  
    // Traverse the given string
    for (int i = 0; i < n; i++) 
    {
  
        // Increase the count
        if (s[i] == '(')
            count++;
  
        // Decrease the count
        else
            count--;
  
        // Find the minimum value
        // of freq('(') - freq(')')
        count_1 = Math.Min(count_1, count);
    }
  
    // If the minimum difference is greater
    // than or equal to -1 and the overall
    // difference is zero
    if (count_1 >= -1 && count == 0)
        return true;
  
    return false;
}
  
// Driver code
public static void Main(String []args)
{
    String s = "())()(";
    int n = s.Length;
  
    if (canBeBalanced(s, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
  
// This code is contributed by Rajput-Ji

输出：

Yes