Merge pull request #1535 from katarianikita2003/main

Create-0150-Evaluate-Reverse-Polish_Notation

Author: ajay-dhangar

dsa-solutions/lc-solutions/0100-0199/0150-Evaluate-Reverse-Polish-Notation.md

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+---
+id: Evaluate-Reverse-Polish_Notation
+title: Evaluate-Reverse-Polish_Notation
+sidebar_label: Evaluate-Reverse-Polish_Notation
+tags: 
+    - Reverse Polish Notation
+    - Stack
+    - Evaluation
+---
+
+## Problem Description
+
+| Problem Statement                                       | Solution Link                                                              | LeetCode Profile                                        |
+| :------------------------------------------------------ | :------------------------------------------------------------------------- | :------------------------------------------------------ |
+| [Evaluate-Reverse-Polish_Notation](https://leetcode.com/problems/Evaluate-Reverse-Polish_Notation/description/) | [Evaluate-Reverse-Polish_Notation Solution on LeetCode](https://leetcode.com/problems/Evaluate-Reverse-Polish_Notation/solutions/) | [Nikita Saini](https://leetcode.com/u/Saini_Nikita/) |
+
+## Problem Description
+
+You are given an array of strings `tokens` that represents an arithmetic expression in Reverse Polish Notation.
+
+Evaluate the expression and return an integer that represents the value of the expression.
+
+Note that:
+- The valid operators are '+', '-', '*', and '/'.
+- Each operand may be an integer or another expression.
+- The division between two integers always truncates toward zero.
+- There will not be any division by zero.
+- The input represents a valid arithmetic expression in reverse polish notation.
+- The answer and all the intermediate calculations can be represented in a 32-bit integer.
+
+### Examples
+
+#### Example 1
+**Input**: `tokens = ["2", "1", "+", "3", "*"]`  
+**Output**: `9`  
+**Explanation**: ((2 + 1) * 3) = 9
+
+#### Example 2
+**Input**: `tokens = ["4", "13", "5", "/", "+"]`  
+**Output**: `6`  
+**Explanation**: (4 + (13 / 5)) = 6
+
+#### 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
+
+## Constraints
+
+- `1 <= tokens.length <= 10^4`
+- `tokens[i]` is either an operator: `"+"`, `"-"`, `"*"` or `"/"`, or an integer in the range `[-200, 200]`.
+
+## Approach
+
+1. Initialize an empty stack.
+2. Iterate over each token in the `tokens` array:
+   - If the token is an operand (integer), push it onto the stack.
+   - If the token is an operator, pop the necessary number of operands from the stack, apply the operator, and push the result back onto the stack.
+3. The final result will be the only element left in the stack.
+
+## Solution in different languages:
+
+### Solution in Python
+```python
+def evalRPN(tokens):
+    stack = []
+    for token in tokens:
+        if token in "+-*/":
+            b = stack.pop()
+            a = stack.pop()
+            if token == '+':
+                stack.append(a + b)
+            elif token == '-':
+                stack.append(a - b)
+            elif token == '*':
+                stack.append(a * b)
+            elif token == '/':
+                stack.append(int(a / b))  # use int() for truncating towards zero
+        else:
+            stack.append(int(token))
+    return stack[0]
+```
+
+### Solution in Java
+```java
+import java.util.Stack;
+
+class Solution {
+    public int evalRPN(String[] tokens) {
+        Stack<Integer> stack = new Stack<>();
+        for (String token : tokens) {
+            if ("+-*/".contains(token)) {
+                int b = stack.pop();
+                int a = stack.pop();
+                switch (token) {
+                    case "+":
+                        stack.push(a + b);
+                        break;
+                    case "-":
+                        stack.push(a - b);
+                        break;
+                    case "*":
+                        stack.push(a * b);
+                        break;
+                    case "/":
+                        stack.push(a / b);
+                        break;
+                }
+            } else {
+                stack.push(Integer.parseInt(token));
+            }
+        }
+        return stack.pop();
+    }
+}
+```
+
+### Solution in C++
+```cpp
+#include <vector>
+#include <string>
+#include <stack>
+
+class Solution {
+public:
+    int evalRPN(std::vector<std::string>& tokens) {
+        std::stack<int> stack;
+        for (const std::string& token : tokens) {
+            if (token == "+" || token == "-" || token == "*" || token == "/") {
+                int b = stack.top(); stack.pop();
+                int a = stack.top(); stack.pop();
+                if (token == "+") stack.push(a + b);
+                else if (token == "-") stack.push(a - b);
+                else if (token == "*") stack.push(a * b);
+                else if (token == "/") stack.push(a / b);
+            } else {
+                stack.push(std::stoi(token));
+            }
+        }
+        return stack.top();
+    }
+};
+```
+
+### Solution in C
+```c
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+#define STACK_SIZE 10000
+
+typedef struct {
+    int data[STACK_SIZE];
+    int top;
+} Stack;
+
+void init(Stack* s) {
+    s->top = -1;
+}
+
+void push(Stack* s, int value) {
+    s->data[++s->top] = value;
+}
+
+int pop(Stack* s) {
+    return s->data[s->top--];
+}
+
+int evalRPN(char ** tokens, int tokensSize){
+    Stack stack;
+    init(&stack);
+    for (int i = 0; i < tokensSize; i++) {
+        char *token = tokens[i];
+        if (strcmp(token, "+") == 0 || strcmp(token, "-") == 0 || strcmp(token, "*") == 0 || strcmp(token, "/") == 0) {
+            int b = pop(&stack);
+            int a = pop(&stack);
+            if (strcmp(token, "+") == 0) push(&stack, a + b);
+            else if (strcmp(token, "-") == 0) push(&stack, a - b);
+            else if (strcmp(token, "*") == 0) push(&stack, a * b);
+            else if (strcmp(token, "/") == 0) push(&stack, a / b);
+        } else {
+            push(&stack, atoi(token));
+        }
+    }
+    return pop(&stack);
+}
+```
+
+### Solution in JavaScript
+```js
+var evalRPN = function(tokens) {
+    const stack = [];
+    for (const token of tokens) {
+        if ("+-*/".includes(token)) {
+            const b = stack.pop();
+            const a = stack.pop();
+            switch (token) {
+                case '+':
+                    stack.push(a + b);
+                    break;
+                case '-':
+                    stack.push(a - b);
+                    break;
+                case '*':
+                    stack.push(a * b);
+                    break;
+                case '/':
+                    stack.push(Math.trunc(a / b));
+                    break;
+            }
+        } else {
+            stack.push(Number(token));
+        }
+    }
+    return stack[0];
+};
+```
+
+### Step-by-step Algorithm
+1. Initialize an empty stack.
+2. Iterate through each token in the `tokens` list.
+3. If the token is an operand (number), push it onto the stack.
+4. If the token is an operator, pop two operands from the stack.
+    - Apply the operator on the two operands.
+    - Push the result back onto the stack.
+5. Continue this process until all tokens are processed.
+6. The final result will be the last item remaining in the stack.
+
+### Conclusion
+The Reverse Polish Notation evaluation is efficiently handled using a stack. This approach ensures that we can process the expression in a single pass through the tokens, leading to a time complexity of O(n) and a space complexity of O(n) where n is the number of tokens.