Given an unbalanced bracket sequence as a string str, the task is to find whether the given string can be balanced by moving at most one bracket from its original place in the sequence to any other position.
Examples:
Input: str = “)(()”
Output: Yes
As by moving s[0] to the end will make it valid.
“(())”
Input: str = “()))(()”
Output: No
Approach: Consider X as a valid bracket then definitely (X) is also valid. If X is not valid and can be balanced with just one change of position in some bracket then it must be of type X = “)(“ where ‘)’ has been placed before ‘(‘.
Now, X can be replaced with (X) as it will not affect the balanced nature of X. The new string becomes X = “()()” which is balanced.
Hence, if (X) is balanced then we can say that X can be balanced with at most one change in the position of some bracket.
Below is the implementation of the above approach:
C++
// CPP implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if the sequence// can be balanced by changing the// position of at most one bracketbool canBeBalanced(string s, int n){ // Odd length string can // never be balanced if (n % 2 == 1) return false; // Add '(' in the beginning and ')' // in the end of the string string k = "("; k += s + ")"; vector<string> d; int cnt = 0; for (int i = 0; i < k.length(); i++) { // If its an opening bracket then // append it to the temp string if (k[i] == '(') d.push_back("("); // If its a closing bracket else { // There was an opening bracket // to match it with if (d.size() != 0) d.pop_back(); // No opening bracket to // match it with else return false; } } // Sequence is balanced if (d.empty()) return true; return false;}// Driver Codeint main(int argc, char const *argv[]){ string s = ")(()"; int n = s.length(); (canBeBalanced(s, n)) ? cout << "Yes" << endl : cout << "No" << endl; return 0;}// This code is contributed by// sanjeev2552 |
Java
// Java implementation of the approachimport java.util.Vector;class GFG { // Function that returns true if the sequence // can be balanced by changing the // position of at most one bracket static boolean canBeBalanced(String s, int n) { // Odd length string can // never be balanced if (n % 2 == 1) return false; // Add '(' in the beginning and ')' // in the end of the string String k = "("; k += s + ")"; Vector<String> d = new Vector<>(); for (int i = 0; i < k.length(); i++) { // If its an opening bracket then // append it to the temp string if (k.charAt(i) == '(') d.add("("); // If its a closing bracket else { // There was an opening bracket // to match it with if (d.size() != 0) d.remove(d.size() - 1); // No opening bracket to // match it with else return false; } } // Sequence is balanced if (d.isEmpty()) 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// sanjeev2552 |
Python3
# Python3 implementation of the approach# Function that returns true if the sequence # can be balanced by changing the # position of at most one bracketdef canBeBalanced(s, n): # Odd length string can # never be balanced if n % 2 == 1: return False # Add '(' in the beginning and ')' # in the end of the string k = "(" k = k + s+")" d = [] count = 0 for i in range(len(k)): # If its an opening bracket then # append it to the temp string if k[i] == "(": d.append("(") # If its a closing bracket else: # There was an opening bracket # to match it with if len(d)!= 0: d.pop() # No opening bracket to # match it with else: return False # Sequence is balanced if len(d) == 0: return True return False# Driver codeS = ")(()"n = len(S)if(canBeBalanced(S, n)): print("Yes")else: print("No") |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG { // Function that returns true if the sequence // can be balanced by changing the // position of at most one bracket static bool canBeBalanced(string s, int n) { // Odd length string can // never be balanced if (n % 2 == 1) return false; // Add '(' in the beginning and ')' // in the end of the string string k = "("; k += s + ")"; List<string> d = new List<string>(); for (int i = 0; i < k.Length; i++) { // If its an opening bracket then // append it to the temp string if (k[i] == '(') d.Add("("); // If its a closing bracket else { // There was an opening bracket // to match it with if (d.Count != 0) d.RemoveAt(d.Count - 1); // No opening bracket to // match it with else return false; } } // Sequence is balanced if (d.Count == 0) return true; return false; } // Driver Code public static void Main() { string s = ")(()"; int n = s.Length; if (canBeBalanced(s, n)) Console.Write("Yes"); else Console.Write("No"); }}// This code is contributed by// mohit kumar 29 |
Javascript
<script>// JavaScript implementation of the approach// Function that returns true if the sequence // can be balanced by changing the // position of at most one bracketfunction canBeBalanced(s,n){ // Odd length string can // never be balanced if (n % 2 == 1) return false; // Add '(' in the beginning and ')' // in the end of the string let k = "("; k += s + ")"; let d = []; for (let i = 0; i < k.length; i++) { // If its an opening bracket then // append it to the temp string if (k[i] == '(') d.push("("); // If its a closing bracket else { // There was an opening bracket // to match it with if (d.length != 0) d.pop(); // No opening bracket to // match it with else return false; } } // Sequence is balanced if (d.length==0) return true; return false;}// Driver Codelet s = ")(()"; let n = s.length; if (canBeBalanced(s, n)) document.write("Yes"); else document.write("No");// This code is contributed by unknown2108</script> |
Output:
Yes
Time Complexity : O(n) ,where n is size of given string
Space Complexity : O(n)
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