1012MediumArrayGreedy

Filling the Blanks So Two Tallies Match

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Trains the technique from

LeetCode 2918Minimum Equal Sum of Two Arrays After Replacing Zeros

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.

Two sheets of figures read left and right. A 0 on either sheet is a blank that has to be filled in with a whole number of 1 or more; every other figure stays as it is.

Fill in all the blanks so that the two sheets come to the same total. Return the smallest total both can be brought to, or -1 when no filling makes them meet.

Examples

Example 1

Input: left = [3, 2, 0, 1, 0], right = [6, 5, 0]
Output: 12

The left sheet has six written and two blanks, so its floor is eight. The right sheet has eleven written and one blank, so its floor is twelve. Twelve is the larger, and the left sheet reaches it by writing 1 in one blank and 5 in the other.

Example 2

Input: left = [2, 0, 2, 0], right = [1, 4]
Output: -1

The left sheet has four written and two blanks, so it can never total less than six. The right sheet has no blanks at all and is stuck at five, so the two can never meet.

Example 3

Input: left = [6], right = [6]
Output: 6

Neither sheet has a blank and both already total six, so six is the only total either can show.

Constraints

1 <= left.length <= 10^5
1 <= right.length <= 10^5
0 <= left[i] <= 10^6
0 <= right[i] <= 10^6

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 min_sum(left: list[int], right: list[int]) -> int:

Java

public long minSum(int[] left, int[] right)

The intended approach

Time O(a + b) over the two sheets' lengths.Space O(1).

For each sheet, add up the figures already written and count the blanks. Writing a `1` in every blank gives that sheet's floor, the smallest total it can reach. Past the floor a sheet can reach any larger total at all, by piling the extra into a single blank — but only if it has a blank. A sheet with no blanks is pinned to its written total and cannot move. The two sheets can only meet at a total that is at least both floors, so the smallest candidate is the larger of the two floors. When the floors are equal, both sheets sit there already and that is the answer, whether or not either has a blank. When the floors differ, the sheet with the smaller floor has to stretch up to the larger one, which it can do exactly when it has a blank to pile the difference into. If it has none, no total works at all and the answer is `-1`.

Pythonentry point: min_sum

def min_sum(left, right):
    sum_left = 0
    gaps_left = 0
    for value in left:
        if value == 0:
            gaps_left += 1
        else:
            sum_left += value
    sum_right = 0
    gaps_right = 0
    for value in right:
        if value == 0:
            gaps_right += 1
        else:
            sum_right += value
    floor_left = sum_left + gaps_left
    floor_right = sum_right + gaps_right
    if floor_left == floor_right:
        return floor_left
    if floor_left > floor_right:
        return floor_left if gaps_right else -1
    return floor_right if gaps_left else -1

The obvious version

Work out every total each sheet can actually be filled to, by walking the blanks and trying each whole number of `1` or more in turn, stopping once the running total passes a cap that is comfortably above both floors. Then take the smallest total the two collections share. Costs an enumeration of fillings rather than a pair of floors.

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_min_sum(left, right):
    def totals(sheet, cap):
        base = 0
        gaps = 0
        for value in sheet:
            if value == 0:
                gaps += 1
            else:
                base += value
        out = set()
        def place(spot, running):
            if running > cap:
                return
            if spot == gaps:
                out.add(running)
                return
            for value in range(1, cap - running + 1):
                place(spot + 1, running + value)
        place(0, base)
        return out
    cap = 1
    for sheet in (left, right):
        for value in sheet:
            cap += value if value else 1
    shared = totals(left, cap) & totals(right, cap)
    if not shared:
        return -1
    best = None
    for value in shared:
        if best is None or value < best:
            best = value
    return best

How this problem was checked

Reference solution executed against 16 test cases — 16 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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