0981MediumMathStringDynamic ProgrammingRecursionMemoizationBracket Sequences

Every Value a Sum Can Be Grouped Into

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Trains the technique from

LeetCode 241Different Ways to Add Parentheses

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.

An arithmetic line is given as expression, made of whole numbers and the signs '+', '-' and '*', with no spaces and no signs marking a number as negative.

Brackets may be put anywhere, so long as the result is a valid arithmetic line, which fixes the order the operations happen in. Different bracketings may give different values.

Return the values from every possible bracketing, in increasing order. A value appears as many times as there are bracketings producing it.

Examples

Example 1

Input: expression = "2-3-4"
Output: [-5, 3]

Bracketing the first subtraction gives minus five, and bracketing the second gives three.

Example 2

Input: expression = "99"
Output: [99]

There is no sign to bracket, so the line takes its own value.

Example 3

Input: expression = "1+1"
Output: [2]

One sign means one bracketing.

Constraints

1 <= expression.length <= 20
expression consists of digits and the signs '+', '-' and '*' only
Every whole number in expression is between 0 and 99
No number in expression is marked with a leading sign

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 diff_ways_to_compute(expression: str) -> list[int]:

Java

public List<Integer> diffWaysToCompute(String expression)

The intended approach

Time O(C * n) for the C bracketings of a line of length n, the count of bracketings being the usual bracket-sequence count.Space O(C) for the values, plus the remembered answers per part.

A bracketing is settled by which operation happens last, because that operation splits the line into a left part and a right part, each bracketed on its own. So for every sign in the line, work out all the values the part before it can take and all the values the part after it can take, then combine every pairing with that sign. A line with no sign at all is a single number and takes one value, which ends the recursion. Because the same part gets asked about repeatedly, remembering the answers per part keeps the work down. Sorting the collected values at the end puts them in the order asked for, and repeats are kept because different bracketings genuinely count separately.

Pythonentry point: diff_ways_to_compute

def diff_ways_to_compute(expression):
    memo = {}

    def values(text):
        if text in memo:
            return memo[text]
        out = []
        for at in range(len(text)):
            ch = text[at]
            if ch not in "+-*":
                continue
            for a in values(text[:at]):
                for b in values(text[at + 1:]):
                    if ch == "+":
                        out.append(a + b)
                    elif ch == "-":
                        out.append(a - b)
                    else:
                        out.append(a * b)
        if not out:
            out = [int(text)]
        memo[text] = out
        return out

    answer = list(values(expression))
    answer.sort()
    return answer

The obvious version

Write out every fully bracketed version of the line as its own string, then work each one out with a small reader that handles the brackets directly. Builds the strings and parses them rather than combining values, and keeps nothing between them.

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_diff_ways_to_compute(expression):
    def spellings(text):
        out = []
        for at in range(len(text)):
            ch = text[at]
            if ch not in "+-*":
                continue
            for a in spellings(text[:at]):
                for b in spellings(text[at + 1:]):
                    out.append("(" + a + ch + b + ")")
        if not out:
            out = [text]
        return out

    def read(text):
        # a fully bracketed line, worked out with two stacks
        numbers = []
        signs = []
        at = 0
        while at < len(text):
            ch = text[at]
            if ch == "(":
                at += 1
            elif ch in "+-*":
                signs.append(ch)
                at += 1
            elif ch == ")":
                b = numbers.pop()
                a = numbers.pop()
                sign = signs.pop()
                if sign == "+":
                    numbers.append(a + b)
                elif sign == "-":
                    numbers.append(a - b)
                else:
                    numbers.append(a * b)
                at += 1
            else:
                start = at
                while at < len(text) and text[at].isdigit():
                    at += 1
                numbers.append(int(text[start:at]))
        return numbers[0]

    out = []
    for spelling in spellings(expression):
        out.append(read(spelling))
    out.sort()
    return out

How this problem was checked

Reference solution executed against 11 test cases — 11 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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