added problem 1106 - Parsing a Boolean String

Author: Raj Mhetar

Diff for: .DS_Store

@@ -0,0 +1,11 @@
+A boolean expression is an expression that evaluates to either true or false, and it can have the following forms:
+
+1. `'t'`: evaluates to true.
+2. `'f'`: evaluates to false.
+3. `'!(subExpr)'`: evaluates to the logical NOT of the inner expression `subExpr`.
+4. `'&(subExpr1, subExpr2, ..., subExprn)'`: evaluates to the logical AND of the inner expressions `subExpr1, subExpr2, ..., subExprn`, where `n >= 1`.
+5. `'|(subExpr1, subExpr2, ..., subExprn)'`: evaluates to the logical OR of the inner expressions `subExpr1, subExpr2, ..., subExprn`, where `n >= 1`.
+
+Given a string `expression` that represents a boolean expression, return its evaluation. It is guaranteed that the expression is valid and follows the rules described above.
+
+---

Diff for: Cpp/.DS_Store

@@ -0,0 +1,73 @@
+class Solution
+{
+public:
+    char solve(vector<char> &temp, char op)
+    {
+        if (op == '!')
+        {
+            return temp[0] == 'f' ? 't' : 'f'; // Negate the first element
+        }
+        else if (op == '&')
+        {
+            for (char &ch : temp)
+            {
+                if (ch == 'f')
+                {
+                    return 'f'; // If any 'f', the result is 'f'
+                    break;
+                }
+            }
+            return 't'; // Otherwise, the result is 't'
+        }
+        else if (op == '|')
+        {
+            for (char &ch : temp)
+            {
+                if (ch == 't')
+                {
+                    return 't'; // If any 't', the result is 't'
+                    break;
+                }
+            }
+            return 'f'; // Otherwise, the result is 'f'
+        }
+        return 'f'; // Default case (shouldn't reach here)
+    }
+
+    bool parseBoolExpr(string exp)
+    {
+        stack<char> st;
+
+        for (char &ch : exp)
+        {
+            if (ch == ',')
+            {
+                continue; // Ignore commas
+            }
+            else if (ch == ')')
+            {
+                vector<char> temp;
+
+                // Collect all characters until '('
+                while (st.top() != '(')
+                {
+                    temp.push_back(st.top());
+                    st.pop();
+                }
+                st.pop(); // Remove the '('
+
+                char op = st.top(); // Get the operator ('!', '&', or '|')
+                st.pop();           // Remove the operator
+
+                // Solve the current expression and push the result
+                st.push(solve(temp, op));
+            }
+            else
+            {
+                st.push(ch); // Push characters ('t', 'f', operators, '(') onto
+                             // the stack
+            }
+        }
+        return st.top() == 't'; // The final result is at the top of the stack
+    }
+};