chore(CPlusPlus): add infix to prefix (#858)

algorithms/CPlusPlus/Stacks/infix-to-prefix.cpp

@@ -0,0 +1,148 @@
+/*
+    Program to convert a infix expression to prefix expression a.k.a Polish Notation.
+    Infix  expression : An expression in which the operator appears between the two operands.
+    Prefix expression : An expression in which the operator appears before the operands. 
+*/
+
+/*
+   LOGIC
+   1. Initialize the Stack.
+   2. Scan the operator from left to right in the infix expression.
+   3. Reverse the infix expression i.e A+B*C will become C*B+A. Note while reversing each ‘(‘ will become ‘)’ and each ‘)’ becomes ‘(‘.
+   4. Obtain the “nearly” postfix expression of the modified expression i.e CB*A+.
+   5. Reverse the postfix expression. Hence in our example prefix is +A*BC.
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+bool isOperator(char c)
+{
+  return (!isalpha(c) && !isdigit(c));
+}
+
+int getPriority(char C)
+{
+  if (C == '-' || C == '+')
+    return 1;
+  else if (C == '*' || C == '/')
+    return 2;
+  else if (C == '^')
+    return 3;
+  return 0;
+}
+
+string infixToPostfix(string infix)
+{
+  infix = '(' + infix + ')';
+  int l = infix.size();
+  stack<char> char_stack;
+  string output;
+
+  for (int i = 0; i < l; i++) {
+
+    // If the scanned character is an operand, add it to output.
+    if (isalpha(infix[i]) || isdigit(infix[i]))
+      output += infix[i];
+
+    // If the scanned character is an ‘(‘, push it to the stack.
+    else if (infix[i] == '(')
+      char_stack.push('(');
+
+    // If the scanned character is an‘)’, pop and output from the stack until an ‘(‘ is encountered.
+    else if (infix[i] == ')') {
+      while (char_stack.top() != '(') {
+        output += char_stack.top();
+        char_stack.pop();
+      }
+      // Remove '(' from the stack
+      char_stack.pop();
+    }
+
+    // Operator found
+    else
+    {
+      if (isOperator(char_stack.top()))
+      {
+        if(infix[i] == '^')
+        {
+          while (getPriority(infix[i]) <= getPriority(char_stack.top()))
+          {
+            output += char_stack.top();
+            char_stack.pop();
+          }
+          
+        }
+        else
+        {
+          while (getPriority(infix[i]) < getPriority(char_stack.top()))
+          {
+            output += char_stack.top();
+            char_stack.pop();
+          }
+          
+        }
+        // Push current Operator on stack
+        char_stack.push(infix[i]);
+      }
+    }
+  }
+  while(!char_stack.empty()){
+    output += char_stack.top();
+    char_stack.pop();
+  }
+  return output;
+}
+
+string infixToPrefix(string infix)
+{
+  /* Reverse String
+  * Replace ( with ) and vice versa
+  * Get Postfix
+  * Reverse Postfix * */
+  
+  int l = infix.size();
+
+  // Reverse infix
+  reverse(infix.begin(), infix.end());
+
+  // Replace ( with ) and vice versa
+  for (int i = 0; i < l; i++) {
+
+    if (infix[i] == '(') {
+      infix[i] = ')';
+    }
+    else if (infix[i] == ')') {
+      infix[i] = '(';
+    }
+  }
+
+  string prefix = infixToPostfix(infix);
+
+  // Reverse postfix
+  reverse(prefix.begin(), prefix.end());
+
+  return prefix;
+}
+
+int main()
+{
+    int n;
+    string infix;
+    cout << "Enter the number of test cases: ";
+    cin >> n; 
+    while(n--)
+    {
+        cout << "Enter the Infix expression: ";
+        cin >> infix;
+        cout << "Prefix Expression of " << infix << " is: " << infixToPrefix(infix)  << endl;
+    }
+
+}
+/*
+infixToPrefix("A*B+C/D") // Should return "+*AB/CD "
+infixToPrefix("(A-B/C)*(A/K-L)"") // Should return "*-A/BC-/AKL"
+
+time complexity : T(n)
+space complexity : O(n)
+*/