Check if it is possible to obtain a Balanced Parenthesis by shifting brackets to either end at most K times (original) (raw)

Last Updated : 23 Jul, 2025

Given a string S of size N consisting of only '(' and ')' only and a positive integer K, the task is to check if the given string can be made a valid parenthesis sequence by moving any characters of the string S to either end of the string at most K number of times.

Examples:

Input: S = ")(", K = 1
Output: Yes
Explanation: Move S[0] to the end of the string.
Now, the modified string S is "()" which is balanced. Therefore, the number of moves required is 1( = K).

Input: S = "()()", K = 0
Output: Yes

Approach: The given problem can be solved based on the following observations:

• If N is odd or the count of opening and closing brackets are not equal, then it is not possible to make a valid parenthesis sequence.
• The idea is to traverse the given sequence and keep track of the difference of count of opening and closing brackets, and if the difference becomes negative at any index, then move some opening bracket after the current index and move it to the beginning.

Follow the steps below to solve the problem:

• If N is odd or the count of opening and closing brackets are not equal, then it is not possible to make a valid parenthesis sequence. Hence, print "No". Otherwise, perform the following steps:
• Initialize two variables, say count and ans as 0 that keeps track of the difference of opening and closing brackets and the required number of moves respectively.
• Traverse the given string S and perform the following steps:
  • If the current character S[i] is '(', then increment the value of count by 1.
  • Otherwise, decrement the value of count by 1.
  • If the count is less than 0, then update the count to 0, and increment the value of ans by 1.
• After completing the above steps, if the value of ans is at most K, then print "Yes". Otherwise, print "No".

Below is the implementation of the above approach:

C++ `

// C++ program for the above approach

#include <bits/stdc++.h> using namespace std;

// Function to check if a valid parenthesis // can be obtained by moving characters // to either end at most K number of times void minimumMoves(string s, int n, int k) { // Base Case 1 if (n & 1) { cout << "No"; return; }

// Count of '(' and ')'
int countOpen = count(s.begin(),
                      s.end(), '(');
int countClose = count(s.begin(),
                       s.end(), ')');

// Base Case 2
if (countOpen != countClose) {
    cout << "No";
    return;
}

// Store the count of moves required
// to make a valid parenthesis
int ans = 0;
int cnt = 0;

// Traverse the string
for (int i = 0; i < n; ++i) {

    // Increment cnt if opening
    // bracket has occurred
    if (s[i] == '(')
        ++cnt;

    // Otherwise, decrement cnt by 1
    else {

        // Decrement cnt by 1
        --cnt;

        // If cnt is negative
        if (cnt < 0) {

            // Update the cnt
            cnt = 0;

            // Increment the ans
            ++ans;
        }
    }
}

// If ans is at most K, then
// print Yes. Otherwise print No
if (ans <= k)
    cout << "Yes";
else
    cout << "No";

}

// Driver Code int main() { string S = ")("; int K = 1; minimumMoves(S, S.length(), K);

return 0;

}

Java

// Java program for the above approach import java.io.; import java.lang.; import java.util.*;

class GFG{

// Function to check if a valid parenthesis // can be obtained by moving characters // to either end at most K number of times static void minimumMoves(String s, int n, int k) {

// Base Case 1
if (n % 2 == 1) 
{
    System.out.println("No");
    return;
}

// Count of '(' and ')'
int countOpen = 0, countClose = 0;
for(char ch : s.toCharArray())
    if (ch == '(')
        countOpen++;
    else if (ch == ')')
        countClose++;

// Base Case 2
if (countOpen != countClose)
{
    System.out.println("No");
    return;
}

// Store the count of moves required
// to make a valid parenthesis
int ans = 0;
int cnt = 0;

// Traverse the string
for(int i = 0; i < n; ++i)
{
    
    // Increment cnt if opening
    // bracket has occurred
    if (s.charAt(i) == '(')
        ++cnt;

    // Otherwise, decrement cnt by 1
    else 
    {
        
        // Decrement cnt by 1
        --cnt;

        // If cnt is negative
        if (cnt < 0)
        {
            
            // Update the cnt
            cnt = 0;

            // Increment the ans
            ++ans;
        }
    }
}

// If ans is at most K, then
// print Yes. Otherwise print No
if (ans <= k)
    System.out.println("Yes");
else
    System.out.println("No");

}

// Driver Code public static void main(String[] args) { String S = ")("; int K = 1;

minimumMoves(S, S.length(), K);

} }

// This code is contributed by Kingash

Python3

python 3 program for the above approach

Function to check if a valid parenthesis

can be obtained by moving characters

to either end at most K number of times

def minimumMoves(s, n, k):

# Base Case 1
if (n & 1):
    print("No")
    return

# Count of '(' and ')'
countOpen = s.count('(')
countClose = s.count(')')

# Base Case 2
if (countOpen != countClose):
    print("No")
    return

# Store the count of moves required
# to make a valid parenthesis
ans = 0
cnt = 0

# Traverse the string
for i in range(n):

    # Increment cnt if opening
    # bracket has occurred
    if (s[i] == '('):
        cnt += 1

    # Otherwise, decrement cnt by 1
    else:

        # Decrement cnt by 1
        cnt -= 1

        # If cnt is negative
        if (cnt < 0):

            # Update the cnt
            cnt = 0

            # Increment the ans
            ans += 1

# If ans is at most K, then
# print Yes. Otherwise print No
if (ans <= k):
    print("Yes")
else:
    print("No")

Driver Code

if name == "main":

S = ")("
K = 1
minimumMoves(S, len(S), K)

# This code is contributed by ukasp.

C#

// C# program for the above approach using System;

class GFG{

// Function to check if a valid parenthesis // can be obtained by moving characters // to either end at most K number of times static void minimumMoves(string s, int n, int k) {

// Base Case 1
if (n % 2 == 1) 
{
     Console.WriteLine("No");
    return;
}

// Count of '(' and ')'
int countOpen = 0, countClose = 0;
foreach(char ch in s.ToCharArray())
    if (ch == '(')
        countOpen++;
    else if (ch == ')')
        countClose++;

// Base Case 2
if (countOpen != countClose)
{
    Console.WriteLine("No");
    return;
}

// Store the count of moves required
// to make a valid parenthesis
int ans = 0;
int cnt = 0;

// Traverse the string
for(int i = 0; i < n; ++i)
{
    
    // Increment cnt if opening
    // bracket has occurred
    if (s[i] == '(')
        ++cnt;

    // Otherwise, decrement cnt by 1
    else 
    {
        
        // Decrement cnt by 1
        --cnt;

        // If cnt is negative
        if (cnt < 0)
        {
            
            // Update the cnt
            cnt = 0;

            // Increment the ans
            ++ans;
        }
    }
}

// If ans is at most K, then
// print Yes. Otherwise print No
if (ans <= k)
     Console.WriteLine("Yes");
else
     Console.WriteLine("No");

}

// Driver Code static void Main() { string S = ")("; int K = 1;

minimumMoves(S, S.Length, K);

} }

// This code is contributed by SoumikMondal

JavaScript

`

Time Complexity: O(N)
Auxiliary Space: O(1)