# Calculate sum of the array generated by given operations

• Difficulty Level : Medium
• Last Updated : 10 Jun, 2021

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[]:

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++

 `// C++ program for the above approach``#include ``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& 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 arr = { ``"5"``, ``"-2"``, ``"C"``, ``"D"``, ``"+"` `};``    ``findTotalSum(arr);` `    ``return` `0;``}`

## Java

 `// 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 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

 `# 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#

 `// 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`

## Javascript

 ``

Output:

`30`

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

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