Pemdas Calculator With Solution

var input1 = document.getElementById(‘input1’).value;\nvar input2 = document.getElementById(‘input2’).value;\nvar input3 = document.getElementById(‘input3’).value;\nvar result = … // calculation based on pemdas\n\n# pemdas calculator with solution\n\n## Problem\n\nYou are given an arithmetic expression containing integers, the four basic arithmetic operations (+, -, *, /), and parentheses. Your task is to evaluate this expression correctly and return the result.\n\n## Example 1\n\n**Input:** \”3 + 2 * 2\”\n\n**Solution:**\n1. According to the order of operations (PEMDAS/BODMAS), multiplication takes precedence over addition.\n2. First, calculate the multiplication: $2 \\\\times 2 = 4$.\n3. Then, perform the addition: $3 + 4 = 7$.\n\n**Result:** 7\n\n## Example 2\n\n**Input:** \”(3 + 5) * 2\”\n\n**Solution:**\n1. According to PEMDAS, parentheses must be evaluated first.\n2. Calculate the expression inside the parentheses: $3 + 5 = 8$.\n3. Then, perform the multiplication: $8 \\\\times 2 = 16$.\n\n**Result:** 16\n\n## Example 3\n\n**Input:** \”10 – 4 / 2\”\n\n**Solution:**\n1. Evaluate division before subtraction: $4 / 2 = 2$.\n2. Perform the subtraction: $10 – 2 = 8$.\n\n**Result:** 8\n\n## Algorithm Description\n\nTo solve this problem efficiently, we can use a stack-based approach that handles the order of operations correctly.\n\n### Using Two Stacks\n\nWe can use two stacks: one for numbers (operands) and one for operators. We iterate through the expression character by character:\n\n1. **Numbers**: If we encounter a digit, we parse the complete number (it might be multi-digit) and push it onto the `values` stack.\n\n2. **Opening Parenthesis**: If we see ‘(‘, we push it onto the `ops` stack.\n\n3. **Closing Parenthesis**: If we see ‘)’, we evaluate the expression inside the parentheses. We pop operators from `ops` and the corresponding operands from `values`, apply the operation, and push the result back onto `values` until we encounter the matching ‘(‘.\n\n4. **Operators**: If we encounter an operator (+, -, *, /), we check the top of the `ops` stack. While the operator at the top has higher or equal precedence than the current operator, we pop it, apply it to the top two values from `values`, and push the result back.\n\nAfter parsing the entire expression, we evaluate any remaining operators in the `ops` stack.\n\n### Single Stack Approach (Simplified)\n\nA slightly cleaner approach is to use a single stack and handle the logic smartly:\n\n1. Initialize an empty stack.\n2. Iterate through the expression. Maintain a `currentNumber` and the `operation` (defaulting to ‘+’).\n3. If we hit an operator or the end of the string:\n – Apply the previous `operation` to `currentNumber` and the stack.\n – If `operation` was ‘+’: push `currentNumber`.\n – If `operation` was ‘-‘: push `-currentNumber`.\n – If `operation` was ‘*’: push `currentNumber * stack.top()`.\n – If `operation` was ‘/’: push `stack.top() / currentNumber` (integer division).\n – Reset `currentNumber` to 0.\n – Update `operation` to the current character.\n4. Finally, sum all the values in the stack to get the result.\n\n### Handling Division by Zero\n\nIn any division operation, we must check if the divisor (the second operand) is zero. If it is, the operation is undefined. Depending on the requirements:\n- We can return an error message.\n- Or, if specified, skip the operation or treat the result as infinity/NaN.\n\n### Time and Space Complexity\n\n**Time Complexity**: $O(n)$, where $n$ is the length of the expression. We iterate through the expression once.\n\n**Space Complexity**: $O(n)$ in the worst case for the stack (e.g., an expression like (((((1))))).)\n\n”
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Evaluate arithmetic expressions with parentheses using the order of operations (PEMDAS/BODMAS).

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\n \n \n Enter a mathematical expression with integers, +, -, *, /, and parentheses.\n

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