0586HardMathStringBacktracking

Insert Signs to Hit a Reading

Tracked in this browser only

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.

A paper tape carries the digit run num. A technician writes an arithmetic line by walking the tape from left to right and, in each gap between two neighbouring digits, either writing one of the signs +, -, * or leaving the gap blank. The digits are never reordered and none is dropped.

Digits separated by blank gaps run together into a single number, so the blanks decide how the tape is cut into numbers. A number of two or more digits may not start with 0, though a number that is the single digit 0 is fine.

The line is worth what ordinary arithmetic says it is worth: every * is applied before any + or -, and otherwise the line reads left to right. There are no brackets and no sign in front of the first number.

Return every line worth exactly target, as strings, in any order. Work the arithmetic out yourself rather than handing the text to a built-in expression evaluator.

Examples

Example 1

Input: num = "105", target = 5
Output: ["1*0+5", "10-5"]

Both lines are worth 5: in "1*0+5" the product 1*0 is worked out first and 0+5 follows, and "10-5" runs the first two digits together. The cut "1+05" is not allowed because 05 starts with a zero.

Example 2

Input: num = "234", target = 14
Output: ["2+3*4"]

"2+3*4" is worth 14 because 3*4 is worked out before the addition. No other way of filling the two gaps is worth 14.

Example 3

Input: num = "00", target = 0
Output: ["0*0", "0+0", "0-0"]

The single gap can hold any of the three signs and each line is worth 0. Leaving the gap blank would make the number 00, which is not allowed.

Constraints

1 <= num.length <= 10
num holds decimal digits only and may start with 0.
-2^31 <= target <= 2^31 - 1
Every number and every partial value stays within 10^10 in absolute size.

The values you return may be in any order.

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 add_operators(num: str, target: int) -> list[str]:

Java

public List<String> addOperators(String num, int target)

The intended approach

Time O(4^n * n) — three signs or a blank in each of the n - 1 gaps, and building each line's text costs O(n).Space O(n) for the recursion and the text under construction, plus the returned list.

Build the line left to right by backtracking. At each position, try every length of the next number, stopping the moment a longer number would start with `0`, and try each of the three signs before it. Two values are carried along the search: the value of the line so far, and the signed value of the number most recently attached. When the next sign is `*`, the last number has to bind to the new one instead of standing alone, so subtract it from the running value and add back its product with the new number, and remember that product as the new last number. The first number is placed on its own with no sign. Every line reaching the end of the tape with the running value equal to `target` is recorded.

Pythonentry point: add_operators

def add_operators(num, target):
    # O(4^n * n) time, O(n) space beyond the returned list: extend the tape one
    # operand at a time. Carrying the running value together with the signed
    # value of the operand that was last attached is what lets multiplication be
    # applied late: undo that last operand from the running value, multiply it by
    # the new operand, and put the product back.
    n = len(num)
    found = []

    def walk(pos, text, running, last):
        if pos == n:
            if running == target:
                found.append(text)
            return
        for end in range(pos + 1, n + 1):
            piece = num[pos:end]
            if len(piece) > 1 and piece[0] == "0":
                break
            value = int(piece)
            if pos == 0:
                walk(end, piece, value, value)
            else:
                walk(end, text + "+" + piece, running + value, value)
                walk(end, text + "-" + piece, running - value, -value)
                walk(end, text + "*" + piece, running - last + last * value,
                     last * value)

    walk(0, "", 0, 0)
    return found

The obvious version

Enumerate the gap decisions directly by counting in base four, one digit per gap. Each count gives a cut of the tape into numbers and a list of signs. Throw the cut away if any number of two or more digits starts with `0`. Otherwise value the line in two layers: multiply neighbouring numbers joined by `*` into running products, then add or subtract those products in order. Keep the cuts whose value matches.

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_add_operators(num, target):
    n = len(num)
    gaps = n - 1
    out = []
    for mask in range(4 ** gaps):
        code = mask
        pieces = [num[0]]
        signs = []
        for g in range(gaps):
            choice = code % 4
            code //= 4
            if choice == 0:
                pieces[-1] += num[g + 1]
            else:
                signs.append("+-*"[choice - 1])
                pieces.append(num[g + 1])
        bad = False
        for piece in pieces:
            if len(piece) > 1 and piece[0] == "0":
                bad = True
                break
        if bad:
            continue
        total = 0
        sign = 1
        product = int(pieces[0])
        for idx in range(len(signs)):
            value = int(pieces[idx + 1])
            if signs[idx] == "*":
                product *= value
            else:
                total += sign * product
                sign = 1 if signs[idx] == "+" else -1
                product = value
        total += sign * product
        if total == target:
            text = pieces[0]
            for idx in range(len(signs)):
                text += signs[idx] + pieces[idx + 1]
            out.append(text)
    return out

How this problem was checked

Reference solution executed against 14 test cases — 14 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in