Maximum length of balanced string after swapping and removal of characters
Given a string str consisting of characters ‘(‘, ‘)’, ‘[‘, ‘]’, ‘{‘ and ‘}’ only. The task is to find the maximum length of the balanced string after removing any character and swapping any two adjacent characters.
Examples:
Input: str = “))[]]((”
Output: 6
The string can be converted to ()[]()Input: str = “{{{{{{{}”
Output: 2
Approach: The idea is to remove extra unmatched parentheses from string because we cannot generate a balanced pair for it and swap the remaining characters to balance the string. Therefore the answer is equal summation of pairs of all balanced parenthesis. Note that we can move a character to any other place by adjacent swappings.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the length of // the longest balanced sub-string int maxBalancedStr(string s) { // To store the count of parentheses int open1 = 0, close1 = 0; int open2 = 0, close2 = 0; int open3 = 0, close3 = 0; // Traversing the string for (int i = 0; i < s.length(); i++) { // Check type of parentheses and // incrementing count for it switch (s[i]) { case '(': open1++; break; case ')': close1++; break; case '{': open2++; break; case '}': close2++; break; case '[': open3++; break; case ']': close3++; break; } } // Sum all pair of balanced parentheses int maxLen = 2 * min(open1, close1) + 2 * min(open2, close2) + 2 * min(open3, close3); return maxLen; } // Driven code int main() { string s = "))[]](("; cout << maxBalancedStr(s); return 0; } |
Java
// Java implementation of the approach class GFG { // Function to return the length of // the longest balanced sub-string static int maxBalancedStr(String s) { // To store the count of parentheses int open1 = 0, close1 = 0; int open2 = 0, close2 = 0; int open3 = 0, close3 = 0; // Traversing the string for (int i = 0; i < s.length(); i++) { // Check type of parentheses and // incrementing count for it switch (s.charAt(i)) { case '(': open1++; break; case ')': close1++; break; case '{': open2++; break; case '}': close2++; break; case '[': open3++; break; case ']': close3++; break; } } // Sum all pair of balanced parentheses int maxLen = 2 * Math.min(open1, close1) + 2 * Math.min(open2, close2) + 2 * Math.min(open3, close3); return maxLen; } // Driven code public static void main(String[] args) { String s = "))[]](("; System.out.println(maxBalancedStr(s)); } } // This code is contributed by Code_Mech. |
Python3
# Python 3 implementation of the approach # Function to return the length of # the longest balanced sub-string def maxBalancedStr(s): # To store the count of parentheses open1 = 0 close1 = 0 open2 = 0 close2 = 0 open3 = 0 close3 = 0 # Traversing the string for i in range(len(s)): # Check type of parentheses and # incrementing count for it if(s[i] == '('): open1 += 1 continue if s[i] == ')': close1 += 1 continue if s[i] == '{': open2 += 1 continue if s[i] == '}': close2 += 1 continue if s[i] == '[': open3 += 1 continue if s[i] == ']': close3 += 1 continue # Sum all pair of balanced parentheses maxLen = (2 * min(open1, close1) + 2 * min(open2, close2) + 2 * min(open3, close3)) return maxLen # Driven code if __name__ == '__main__': s = "))[]]((" print(maxBalancedStr(s)) # This code is contributed by # Surendra_Gangwar |
C#
// C# implementation of the approach using System; class GFG { // Function to return the length of // the longest balanced sub-string static int maxBalancedStr(string s) { // To store the count of parentheses int open1 = 0, close1 = 0; int open2 = 0, close2 = 0; int open3 = 0, close3 = 0; // Traversing the string for (int i = 0; i < s.Length; i++) { // Check type of parentheses and // incrementing count for it switch (s[i]) { case '(': open1++; break; case ')': close1++; break; case '{': open2++; break; case '}': close2++; break; case '[': open3++; break; case ']': close3++; break; } } // Sum all pair of balanced parentheses int maxLen = 2 * Math.Min(open1, close1) + 2 * Math.Min(open2, close2) + 2 * Math.Min(open3, close3); return maxLen; } // Driver code public static void Main() { string s = "))[]](("; Console.WriteLine(maxBalancedStr(s)); } } // This code is contributed by Code_Mech. |
PHP
<?php // PHP implementation of the approach // Function to return the length of // the longest balanced sub-string function maxBalancedStr($s) { // To store the count of parentheses $open1 = 0; $close1 = 0; $open2 = 0; $close2 = 0; $open3 = 0; $close3 = 0; // Traversing the string for ($i = 0; $i < strlen($s); $i++) { // Check type of parentheses and // incrementing count for it switch ($s[$i]) { case '(': $open1++; break; case ')': $close1++; break; case '{': $open2++; break; case '}': $close2++; break; case '[': $open3++; break; case ']': $close3++; break; } } // Sum all pair of balanced parentheses $maxLen = 2 * min($open1, $close1) + 2 * min($open2, $close2) + 2 * min($open3, $close3); return $maxLen; } // Driven code { $s = "))[]](("; echo(maxBalancedStr($s)); } // This code is contributed by Code_Mech. |
Output:
6
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