Given an expression string exp, write a program to examine whether the pairs and the orders of “{“, “}”, “(“, “)”, “[“, “]” are correct in exp.
Example:
Input: exp = “[()]{}{[()()]()}”
Output: BalancedInput: exp = “[(])”
Output: Not Balanced
Algorithm:
- Declare a character stack S.
- Now traverse the expression string exp.
- If the current character is a starting bracket (‘(‘ or ‘{‘ or ‘[‘) then push it to stack.
- If the current character is a closing bracket (‘)’ or ‘}’ or ‘]’) then pop from stack and if the popped character is the matching starting bracket then fine else brackets are not balanced.
- After complete traversal, if there is some starting bracket left in stack then “not balanced”
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
C#
// C# program for checking // balanced Brackets using System; using System.Collections.Generic; public class BalancedBrackets { public class stack { public int top = -1; public char[] items = new char[100]; public void push(char x) { if (top == 99) { Console.WriteLine("Stack full"); } else { items[++top] = x; } } char pop() { if (top == -1) { Console.WriteLine("Underflow error"); return '�'; } else { char element = items[top]; top--; return element; } } Boolean isEmpty() { return (top == -1) ? true : false; } } // Returns true if character1 and character2 // are matching left and right brackets */ static Boolean isMatchingPair(char character1, char character2) { if (character1 == '(' && character2 == ')') return true; else if (character1 == '{' && character2 == '}') return true; else if (character1 == '[' && character2 == ']') return true; else return false; } // Return true if expression has balanced // Brackets static Boolean areBracketsBalanced(char[] exp) { // Declare an empty character stack */ Stack<char> st = new Stack<char>(); // Traverse the given expression to // check matching brackets for (int i = 0; i < exp.Length; i++) { // If the exp[i] is a starting // bracket then push it if (exp[i] == '{' || exp[i] == '(' || exp[i] == '[') st.Push(exp[i]); // If exp[i] is an ending bracket // then pop from stack and check if the // popped bracket is a matching pair if (exp[i] == '}' || exp[i] == ')' || exp[i] == ']') { // If we see an ending bracket without // a pair then return false if (st.Count == 0) { return false; } // Pop the top element from stack, if // it is not a pair brackets of // character then there is a mismatch. This // happens for expressions like {(}) else if (!isMatchingPair(st.Pop(), exp[i])) { return false; } } } // If there is something left in expression // then there is a starting bracket without // a closing bracket if (st.Count == 0) return true; // balanced else { // not balanced return false; } } // Driver code public static void Main(String[] args) { char[] exp = { '{', '(', ')', '}', '[', ']' }; // Function call if (areBracketsBalanced(exp)) Console.WriteLine("Balanced "); else Console.WriteLine("Not Balanced "); } } // This code is contributed by 29AjayKumar |
Output
Balanced
Time Complexity: O(n)
Auxiliary Space: O(n) for stack.
Please refer complete article on Check for Balanced Brackets in an expression (well-formedness) using Stack for more details!
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] There you can find 84623 more Information on that Topic: geeksforgeeks.org/c-program-to-check-for-balanced-brackets-in-an-expression-well-formedness-using-stack-2/ […]