Given string str of length N, consisting of pairs of balanced parentheses, the task is to calculate the score of the given string based on the given rules:
- “()” has a score of 1.
- “a b” has a score of a + b, where a and b are individual pairs of balanced parentheses.
- “(a)” has a score twice of a i.e., the score is 2 * score of a.
Examples:
Input: str = “()()”
Output: 2
Explanation: The string str is of the form “ab”, that makes the total score = (score of a) + (score of b) = 1 + 1 = 2.Input: str = “(()(()))”
Output: 6
Explanation: The string str is of the form “(a(b))” which makes the total score = 2 * ((score of a) + 2*(score of b)) = 2*(1 + 2*(1)) = 6.
Tree-based Approach: Refer to the previous post of this article for the tree-based approach.
Time Complexity: O(N)
Auxiliary Space: O(N)
Stack-based Approach: The idea is to traverse the string and while traversing the string str, if the parenthesis ‘)’ is encountered, then calculate the score of this pair of parentheses. Follow the steps below to solve the problem:
- Initialize a stack, say S, to keep track of the score and initially push 0 into the stack.
- Traverse the string str using the variable i and perform the following steps:
- If the value of str[i] is equal to ‘(‘, push 0 to the stack S.
- Otherwise, perform the following steps:
- Store the top of the stack S in a variable, say temp, and pop the element from the top of the stack.
- If the value of temp is non-zero, then inner parentheses exist. Add 2 * temp to the top of the stack. Otherwise, add 1 to the top of the stack.
- After completing the above steps, print the value of the top of the stack as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the score// of the parentheses using stackvoid scoreOfParentheses(string s){ // To keep track of the score stack<int> stack; // Initially, push 0 to stack stack.push(0); // Traverse the string s for (char c : s) { // If '(' is encountered, // then push 0 to stack if (c == '(') stack.push(0); // Otherwise else { // Balance the last '(', and store // the score of inner parentheses int tmp = stack.top(); stack.pop(); int val = 0; // If tmp is not zero, it means // inner parentheses exists if (tmp > 0) val = tmp * 2; // Otherwise, it means no // inner parentheses exists else val = 1; // Pass the score of this level // to parent parentheses stack.top() += val; } } // Print the score cout << stack.top();}// Driver Codeint main(){ string S = "(()(()))"; scoreOfParentheses(S); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Function to calculate the score // of the parentheses using stack static void scoreOfParentheses(String s) { // To keep track of the score Stack<Integer> stack = new Stack<>(); // Initially, push 0 to stack stack.push(0); // Traverse the string s for (char c : s.toCharArray()) { // If '(' is encountered, // then push 0 to stack if (c == '(') stack.push(0); // Otherwise else { // Balance the last '(', and store // the score of inner parentheses int tmp = stack.pop(); int val = 0; // If tmp is not zero, it means // inner parentheses exists if (tmp > 0) val = tmp * 2; // Otherwise, it means no // inner parentheses exists else val = 1; // Pass the score of this level // to parent parentheses stack.push(stack.pop() + val); } } // Print the score System.out.println(stack.peek()); } // Driver code public static void main(String[] args) { String S = "(()(()))"; // Function call scoreOfParentheses(S); }}// This code is contributed by Kingash. |
Python3
# Python 3 program for the above approach# Function to calculate the score# of the parentheses using stackdef scoreOfParentheses(s): # To keep track of the score stack = [] # Initially, push 0 to stack stack.append(0) # Traverse the string s for c in s: # If '(' is encountered, # then push 0 to stack if (c == '('): stack.append(0) # Otherwise else: # Balance the last '(', and store # the score of inner parentheses tmp = stack[len(stack) - 1] stack = stack[:-1] val = 0 # If tmp is not zero, it means # inner parentheses exists if (tmp > 0): val = tmp * 2 # Otherwise, it means no # inner parentheses exists else: val = 1 # Pass the score of this level # to parent parentheses stack[len(stack) - 1] += val # Print the score print(stack[len(stack) - 1])# Driver Codeif __name__ == '__main__': S = "(()(()))" scoreOfParentheses(S) # This code is contributed by bgangwar59. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG { // Function to calculate the score // of the parentheses using stack static void scoreOfParentheses(String s) { // To keep track of the score Stack<int> stack = new Stack<int>(); // Initially, push 0 to stack stack.Push(0); // Traverse the string s foreach (char c in s.ToCharArray()) { // If '(' is encountered, // then push 0 to stack if (c == '(') stack.Push(0); // Otherwise else { // Balance the last '(', and store // the score of inner parentheses int tmp = stack.Pop(); int val = 0; // If tmp is not zero, it means // inner parentheses exists if (tmp > 0) val = tmp * 2; // Otherwise, it means no // inner parentheses exists else val = 1; // Pass the score of this level // to parent parentheses stack.Push(stack.Pop() + val); } } // Print the score Console.WriteLine(stack.Peek()); } // Driver code public static void Main(String[] args) { String S = "(()(()))"; // Function call scoreOfParentheses(S); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program for the above approach// Function to calculate the score// of the parentheses using stackfunction scoreOfParentheses(s){ // To keep track of the score var stack = []; // Initially, push 0 to stack stack.push(0); // Traverse the string s s.split('').forEach(c => { // If '(' is encountered, // then push 0 to stack if (c == '(') stack.push(0); // Otherwise else { // Balance the last '(', and store // the score of inner parentheses var tmp = stack[stack.length-1]; stack.pop(); var val = 0; // If tmp is not zero, it means // inner parentheses exists if (tmp > 0) val = tmp * 2; // Otherwise, it means no // inner parentheses exists else val = 1; // Pass the score of this level // to parent parentheses stack[stack.length-1] += val; } }); // Print the score document.write( stack[stack.length-1]);}// Driver Codevar S = "(()(()))";scoreOfParentheses(S);</script> |
Output:
6
Time Complexity: O(N)
Auxiliary Space: O(N)
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