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Alkesh Ghorpade
Alkesh Ghorpade

Posted on • Originally published at alkeshghorpade.me

LeetCode - Evaluate Reverse Polish Notation

Problem statement

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, , and */. Each operand may be an integer or another expression.

Note that division between two integers should truncate toward zero.

It is guaranteed that the given RPN expression is always valid. That means the expression would always evaluate to a result, and there will not be any division by zero operation.

Problem statement taken from: https://leetcode.com/problems/evaluate-reverse-polish-notation

Example 1:

Input: tokens = ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9
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Example 2:

Input: tokens = ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6
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Example 3:

Input: tokens = ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22
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Constraints:

- 1 <= tokens.length <= 10^4
- tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200]
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Explanation

Reverse Polish Notation is a mathematical notation in which operators follow their operands.

The basic approach is using the stack. Let's check the algorithm directly:

- create stack st
  initialize int op1, op2

- loop for i = 0; i < tokens.size; i++
  - if tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/"
    - push to stack st.push(tokens[i])
  - else
    - set op2 = st.pop()
      set op1 = st.pop()

    - if tokens[i] == "+"
      - st.push(op1 + op2)
    - else if tokens[i] == "-"
      - st.push(op1 - op2)
    - else if tokens[i] == "*"
      - st.push(op1 * op2)
    - else
      - st.push(op1 / op2)
- for end

- return st.top()
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This function's time complexity is O(N), and the space complexity is O(N).

Let's check our solutions in C++, Golang, and Javascript.

C++ solution

typedef long long ll;

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<ll> st;
        ll op1, op2;

        for(int i = 0; i < tokens.size(); i++) {
            if(tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/") {
                st.push(stol(tokens[i]));
            } else {
                op2 = st.top();
                st.pop();
                op1 = st.top();
                st.pop();

                if(tokens[i] == "+")
                    st.push(op1 + op2);
                else if(tokens[i] == "-")
                    st.push(op1 - op2);
                else if(tokens[i] == "*")
                    st.push(op1 * op2);
                else
                    st.push(op1 / op2);
            }
        }

        return (int) st.top();
    }
};
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Golang solution

func evalRPN(tokens []string) int {
    st := []int{}

    for _, token := range tokens {
        if token == "+" {
            st[len(st) - 2] += st[len(st) - 1]
            st = st[:len(st) - 1]
        } else if token == "-" {
            st[len(st) - 2] -= st[len(st) - 1]
            st = st[:len(st) - 1]
        } else if token == "*" {
            st[len(st) - 2] *= st[len(st) - 1]
            st = st[:len(st) - 1]
        } else if token == "/" {
            st[len(st) - 2] /= st[len(st) - 1]
            st = st[:len(st) - 1]
        } else {
            num, _ := strconv.Atoi(token)
            st = append(st, num)
        }
    }

    return st[0]
}
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Javascript solution

var evalRPN = function(tokens) {
    let st = [];
    let op1, op2;

    for(let i = 0; i < tokens.length; i++) {
        if(tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/") {
                st.push(parseInt(tokens[i]));
        } else {
            op2 = st.pop();
            op1 = st.pop();

            if(tokens[i] === "+")
                st.push(op1 + op2);
            else if(tokens[i] === "-")
                st.push(op1 - op2);
            else if(tokens[i] === "*")
                st.push(op1 * op2);
            else
                st.push(op1 / op2 | 0);
        }
    }

    return st[0];
};
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Let's dry-run our algorithm to see how the solution works.

Input: tokens = ["2", "1", "+", "3", "*"]

Step 1: stack<ll> st
        int op1, op2

Step 2: loop for int i = 0; i < tokens.size()
          i < tokens.size()
          0 < 5
          true

          if tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/"
             tokens[0] is "2"
             true

             st.push(stoi(tokens[i]))
             st.push(stoi(tokens[0]))
             st.push(2)

             st = [2]

          i++
          i = 1

Step 3: loop for i < tokens.size()
          1 < 5
          true

          if tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/"
             tokens[1] is "1"
             true

             st.push(stoi(tokens[i]))
             st.push(stoi(tokens[1]))
             st.push(1)

             st = [2, 1]

          i++
          i = 2

Step 4: loop for i < tokens.size()
          2 < 5
          true

          if tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/"
             tokens[1] is "+"
             false

          else
            op2 = st.top()
                = 1

            st.pop()
            st = [2]

            op1 = st.top()
                = 2

            st.pop()
            st = []

            if tokens[i] == "+"
               true

               st.push(op1 + op2)
               st.push(2 + 1)
               st.push(3)

               st = [3]

          i++
          i = 3

Step 5: loop for i < tokens.size()
          3 < 5
          true

          if tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/"
             tokens[3] is "3"
             true

             st.push(stoi(tokens[i]))
             st.push(stoi(tokens[3]))
             st.push(3)

             st = [3, 3]

          i++
          i = 4

Step 6: loop for i < tokens.size()
          4 < 5
          true

          if tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/"
             tokens[4] is "*"
             false

          else
            op2 = st.top()
                = 3

            st.pop()
            st = [3]

            op1 = st.top()
                = 3

            st.pop()
            st = []

            if tokens[i] == "+"
              false
            else if tokens[i] == "-"
              false
            else if tokens[i] == "*"
               st.push(op1 * op2)
               st.push(3 * 3)
               st.push(9)

               st = [9]

          i++
          i = 5

Step 7: loop for i < tokens.size()
          5 < 5
          false

Step 8: return st[0]

We return the answer as 9.
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