0173MediumMathStringStack

Stockroom Quantity Formula

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Stockroom staff type quantity formulas into a handheld terminal instead of doing the arithmetic themselves. A formula is a string formula holding whole counts joined by the four symbols +, -, * and /, with any number of spaces sprinkled between the pieces.

Evaluate the formula the way the terminal does:

* and / bind tighter than + and -.
Symbols of equal binding are applied from left to right.
/ keeps only the whole part of the quotient, discarding the fraction, so the result is truncated toward zero.
Spaces carry no meaning.

Return the resulting count. You may not hand the string to a language-provided expression evaluator.

Examples

Example 1

Input: formula = "14+6*3-8/3"
Output: 30

The tighter symbols run first: `6*3` is 18 and `8/3` truncates to 2, leaving 14 + 18 - 2.

Example 2

Input: formula = " 42 - 7*4 "
Output: 14

Spaces are dropped and `7*4` is folded before the subtraction, so the terminal computes 42 - 28.

Example 3

Input: formula = "100/7/2"
Output: 7

Equal binding runs left to right: 100/7 truncates to 14, and that 14 is then halved to 7.

Constraints

1 <= formula.length <= 3 * 10^5
formula holds whole counts and the symbols '+', '-', '*', '/', separated by any number of spaces
formula is always a well-formed expression
Every count in formula lies in [0, 2^31 - 1]
No divisor is zero
The result fits in a signed 32-bit integer

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def evaluate_formula(formula: str) -> int:

Java

public int evaluateFormula(String formula)

The intended approach

Time O(n)Space O(n)

Keep a list of finished additive terms. Scan the characters once, building each count digit by digit, and when the next symbol arrives decide what to do with the count just finished: `+` and `-` push it as a new term (negated for `-`), while `*` and `/` fold it into the term already on top, because that term is still the pending product. A sentinel character after the last digit flushes the final count, and the answer is the sum of the terms.

Pythonentry point: evaluate_formula

def evaluate_formula(formula):
    # One left-to-right pass over n characters, so O(n) time. The term list can
    # hold up to one entry per count, which is O(n) space in the worst case.
    terms = []
    count = 0
    symbol = '+'
    size = len(formula)
    index = 0
    while index <= size:
        char = formula[index] if index < size else '#'
        if char.isdigit():
            count = count * 10 + int(char)
        elif char != ' ':
            if symbol == '+':
                terms.append(count)
            elif symbol == '-':
                terms.append(-count)
            elif symbol == '*':
                terms.append(terms.pop() * count)
            else:
                top = terms.pop()
                whole = abs(top) // abs(count)
                terms.append(whole if top >= 0 else -whole)
            count = 0
            symbol = char
        index += 1
    return sum(terms)

The obvious version

Tokenise into a flat list, then make two sweeps over that list: the first replaces every '*' and '/' with its value, the second adds and subtracts what is left. No stack.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_evaluate_formula(formula):
    tokens = []
    count = None
    for char in formula:
        if char.isdigit():
            count = (count if count is not None else 0) * 10 + int(char)
        elif char != ' ':
            tokens.append(count)
            tokens.append(char)
            count = None
    tokens.append(count)
    folded = [tokens[0]]
    step = 1
    while step < len(tokens):
        symbol = tokens[step]
        value = tokens[step + 1]
        if symbol == '*':
            folded.append(folded.pop() * value)
        elif symbol == '/':
            left = folded.pop()
            whole = abs(left) // abs(value)
            folded.append(whole if left >= 0 else -whole)
        else:
            folded.append(symbol)
            folded.append(value)
        step += 2
    total = folded[0]
    step = 1
    while step < len(folded):
        if folded[step] == '+':
            total += folded[step + 1]
        else:
            total -= folded[step + 1]
        step += 2
    return total

How this problem was checked

Reference solution executed against 18 test cases — 18 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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