Calculate sum of the array generated by given operations

Given an array arr[] consisting of N strings, the task is to find the total sum of the array brr[] (initially empty) constructed by performing following operations while traversing the given array arr[]:

• If the array arr[] contains an integer, then insert that integer into the array brr[].
• If the array arr[] has the string "+", then insert the sum of the last two elements in the array brr[] into the array brr[].
• If the array arr[] has the string "D", then insert the value which is twice the last element of the array brr[] to the array brr[].
• If the array arr[] has the string "C", then remove the last element of the array brr[] to the array brr[].

Examples:

Input: arr[] = {"5", "2", "C", "D", "+"}
Output: 30
Explanation:
While traversing the array arr[], the array brr[] is modified as:

• "5" - Add 5 to the array brr[]. Now, the array brr[] modifies to {5}.
• "2" - Add 2 to the array brr[]. Now, the array brr[] modifies to {5, 2}.
• "C" - Remove the last element from the array brr[]. Now, the array brr[] modifies to {5}.
• "D" - Add twice the last element of the array brr[] to the array brr[]. Now, the array brr[] modifies to {5, 10}.
• "+" - Add the sum of the last two elements of the array brr[] to the array brr[]. Now the array brr[] modifies to {5, 10, 15}.

After performing the above operations, the total sum of the array brr[] is 5 + 10 + 15 = 30.

Input: arr[] = {"5", "-2", "4", "C", "D", "9", "+", "+"}
Output: 27

Approach: The idea to solve the given problem is to use a Stack. Follow the steps below to solve the problem:

• Initialize a stack of integers, say S, and initialize a variable, say ans as 0, to store the resultant sum of the array formed.
• Traverse the given array arr[] and perform the following steps:
  • If the value of arr[i] is "C", then subtract the top element of the stack from the ans and pop it from S.
  • If the value of arr[i] is "D", then push twice the top element of the stack S in the stack S and then add its value to ans.
  • If the value of arr[i] is "+", then push the value of the sum of the top two elements of the stack S and add their sum to ans.
  • Otherwise, push arr[i] to the stack S, and add its value to ans. 
• After the loop, print the value of ans as the result.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the sum of the array
// formed by performing given set of
// operations while traversing the array ops[]
void findTotalSum(vector<string>& ops)
{
    // If the size of array is 0
    if (ops.empty()) {
        cout << 0;
        return;
    }

    stack<int> pts;

    // Stores the required sum
    int ans = 0;

    // Traverse the array ops[]
    for (int i = 0; i < ops.size(); i++) {

        // If the character is C, remove
        // the top element from the stack
        if (ops[i] == "C") {

            ans -= pts.top();
            pts.pop();
        }

        // If the character is D, then push
        // 2 * top element into stack
        else if (ops[i] == "D") {

            pts.push(pts.top() * 2);
            ans += pts.top();
        }

        // If the character is +, add sum
        // of top two elements from the stack
        else if (ops[i] == "+") {

            int a = pts.top();
            pts.pop();
            int b = pts.top();
            pts.push(a);
            ans += (a + b);
            pts.push(a + b);
        }

        // Otherwise, push x
        // and add it to ans
        else {
            int n = stoi(ops[i]);
            ans += n;
            pts.push(n);
        }
    }

    // Print the resultant sum
    cout << ans;
}

// Driver Code
int main()
{
    vector<string> arr = { "5", "-2", "C", "D", "+" };
    findTotalSum(arr);

    return 0;
}

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG 
{

  // Function to find the sum of the array
  // formed by performing given set of
  // operations while traversing the array ops[]
  static void findTotalSum(String ops[])
  {

    // If the size of array is 0
    if (ops.length == 0)
    {
      System.out.println(0);
      return;
    }

    Stack<Integer> pts = new Stack<>();

    // Stores the required sum
    int ans = 0;

    // Traverse the array ops[]
    for (int i = 0; i < ops.length; i++) {

      // If the character is C, remove
      // the top element from the stack
      if (ops[i] == "C") {

        ans -= pts.pop();
      }

      // If the character is D, then push
      // 2 * top element into stack
      else if (ops[i] == "D") {

        pts.push(pts.peek() * 2);
        ans += pts.peek();
      }

      // If the character is +, add sum
      // of top two elements from the stack
      else if (ops[i] == "+") {

        int a = pts.pop();
        int b = pts.peek();
        pts.push(a);
        ans += (a + b);
        pts.push(a + b);
      }

      // Otherwise, push x
      // and add it to ans
      else {
        int n = Integer.parseInt(ops[i]);
        ans += n;
        pts.push(n);
      }
    }

    // Print the resultant sum
    System.out.println(ans);
  }

  // Driver Code
  public static void main(String[] args)
  {

    String arr[] = { "5", "-2", "C", "D", "+" };
    findTotalSum(arr);
  }
}

// This code is contributed by Kingash.

# Python3 program for the above approach

# Function to find the sum of the array
# formed by performing given set of
# operations while traversing the array ops[]
def findTotalSum(ops):

    # If the size of array is 0
    if (len(ops) == 0):
        print(0)
        return

    pts = []

    # Stores the required sum
    ans = 0

    # Traverse the array ops[]
    for i in range(len(ops)):

        # If the character is C, remove
        # the top element from the stack
        if (ops[i] == "C"):

            ans -= pts[-1]
            pts.pop()

        # If the character is D, then push
        # 2 * top element into stack
        elif (ops[i] == "D"):

            pts.append(pts[-1] * 2)
            ans += pts[-1]

        # If the character is +, add sum
        # of top two elements from the stack
        elif (ops[i] == "+"):

            a = pts[-1]
            pts.pop()
            b = pts[-1]
            pts.append(a)
            ans += (a + b)
            pts.append(a + b)

        # Otherwise, push x
        # and add it to ans
        else:
            n = int(ops[i])
            ans += n
            pts.append(n)

    # Print the resultant sum
    print(ans)

# Driver Code
if __name__ == "__main__":

    arr = ["5", "-2", "C", "D", "+"]
    findTotalSum(arr)

    # This code is contributed by ukasp.

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find the sum of the array
// formed by performing given set of
// operations while traversing the array ops[]
static void findTotalSum(string []ops)
{

    // If the size of array is 0
    if (ops.Length == 0)
    {
        Console.WriteLine(0);
        return;
    }
    
    Stack<int> pts = new Stack<int>();
    
    // Stores the required sum
    int ans = 0;
    
    // Traverse the array ops[]
    for(int i = 0; i < ops.Length; i++) 
    {
        
        // If the character is C, remove
        // the top element from the stack
        if (ops[i] == "C") 
        {
            ans -= pts.Pop();
        }
        
        // If the character is D, then push
        // 2 * top element into stack
        else if (ops[i] == "D")
        {
            pts.Push(pts.Peek() * 2);
            ans += pts.Peek();
        }
        
        // If the character is +, add sum
        // of top two elements from the stack
        else if (ops[i] == "+") 
        {
            int a = pts.Pop();
            int b = pts.Peek();
            pts.Push(a);
            ans += (a + b);
            pts.Push(a + b);
        }
        
        // Otherwise, push x
        // and add it to ans
        else 
        {
            int n = Int32.Parse(ops[i]);
            ans += n;
            pts.Push(n);
        }
    }
    
    // Print the resultant sum
    Console.WriteLine(ans);
}

// Driver Code
public static void Main()
{
    string []arr = { "5", "-2", "C", "D", "+" };
    
    findTotalSum(arr);
}
}

// This code is contributed by ipg2016107

<script>

// JavaScript program for the above approach

// Function to find the sum of the array
// formed by performing given set of
// operations while traversing the array ops[]
function findTotalSum(ops)
{
    // If the size of array is 0
    if (ops.length==0) {
        document.write( 0);
        return;
    }

    var pts = [];

    // Stores the required sum
    var ans = 0;

    // Traverse the array ops[]
    for (var i = 0; i < ops.length; i++) {

        // If the character is C, remove
        // the top element from the stack
        if (ops[i] == "C") {

            ans -= pts[pts.length-1];
            pts.pop();
        }

        // If the character is D, then push
        // 2 * top element into stack
        else if (ops[i] == "D") {

            pts.push(pts[pts.length-1] * 2);
            ans += pts[pts.length-1];
        }

        // If the character is +, add sum
        // of top two elements from the stack
        else if (ops[i] == "+") {

            var a = pts[pts.length-1];
            pts.pop();
            var b = pts[pts.length-1];
            pts.push(a);
            ans += (a + b);
            pts.push(a + b);
        }

        // Otherwise, push x
        // and add it to ans
        else {
            var n = parseInt(ops[i]);
            ans += n;
            pts.push(n);
        }
    }

    // Print the resultant sum
    document.write( ans);
}

// Driver Code

var arr = ["5", "-2", "C", "D", "+" ];
findTotalSum(arr);


</script> 

30

Time Complexity: O(N)
Auxiliary Space: O(N)