Last Updated on October 10, 2022 by Gokul Kannan

Given an expression, find and mark matched and unmatched parenthesis in it. We need to replace all balanced opening parenthesis with 0, balanced closing parenthesis with 1, and all unbalanced with -1. This is one of the most important questions to learn while preparing for interviews.

### Examples:

**Input** : ((x)

**Output** : -10×1

**Input** : (x))

**Output** : 0x1-1

**Input** : (((abc))((z)))))

**Output**: 000abc1100z111-1-1

### Algorithm:-

We will use a Stack data structure to solve this question.

- Start traversing through the given string and every time when ‘(‘ is encountered then push the current index in the stack.
- If ‘)‘ is encountered then replace all opening brackets with 0 and closing brackets with 1 while doing this pop the index of the opening brackets from the stack because they refer to open ‘(‘ brackets.
- If the stack is empty, and we encounter a closing bracket ‘)’ we replace -1 at that index of the string because this is contributing the unmatched brackets.

### Dry Run

### Code Implementation

#includeusing namespace std; void identifyMarkParenthesis(string a){ // stack to store index stack index; // traverse through the whole string for (int i = 0; i < a.length(); i++) { // if a[i] is opening bracket '(' // then just push into stack if (a[i] == '('){ index.push(i); } // if a[i] is closing bracket ')' else if (a[i] == ')') { if (index.empty()){ // If this closing bracket is // unmatched then mark -1 a.replace(i, 1, "-1"); }else{ // replace all opening // brackets with 0 // and closing brackets with 1 a.replace(i, 1, "1"); a.replace(index.top(), 1, "0"); index.pop(); } } } // if stack is not empty then pop //out all elements from it and replace-1 // at that index of the string because // they are unmatched while (!index.empty()) { a.replace(index.top(), 1, "-1"); index.pop(); } // print final string cout << a << endl; } int main(){ string str = "(x))"; identifyMarkParenthesis(str); return 0; }

**Output**:

0x1-1

**Time Complexity**: O(n)

**Space Complexity**: O(n)

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