1142MediumStringDynamic ProgrammingStack

Tidying a Line of Two Crate Kinds

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Trains the technique from

LeetCode 1653Minimum Deletions to Make String Balanced

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 line line holds only the characters a and b. The line is tidy when no b stands anywhere before an a, so every a comes before every b.

Return the fewest characters that have to be removed to leave the line tidy.

Examples

Example 1

Input: line = "bab"
Output: 1

Removing the single a leaves two b letters, which is tidy. Removing both b letters instead would cost two.

Example 2

Input: line = "bbbaaa"
Output: 3

Three b letters stand before three a letters, so either all the a letters or all the b letters have to go, and each costs three.

Example 3

Input: line = "ab"
Output: 0

The line is already tidy, so nothing is removed.

Constraints

1 <= line.length <= 10^5
every character of line is a or b

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 minimum_deletions(line: str) -> int:

Java

public int minimumDeletions(String line)

The intended approach

Time O(n) over the line.Space O(1).

A tidy line is settled by one thing: where the `a` letters stop and the `b` letters start. So walk the line keeping the fewest removals needed to leave everything read so far tidy, together with how many `b` letters have gone past. A `b` never spoils anything read so far, since it can always sit at the tidy end, so it only bumps the `b` count. An `a` is where the choice lies. Either it goes, costing one more removal on top of the best so far, or it stays, which means every `b` seen before it has to go instead. Taking the cheaper of those two settles the line up to that point. That is one pass with two running numbers and no need to try the split points one at a time.

Pythonentry point: minimum_deletions

def minimum_deletions(line):
    removed = 0
    bees = 0
    for ch in line:
        if ch == "b":
            bees += 1
        else:
            step = removed + 1
            if bees < step:
                step = bees
            removed = step
    return removed

The obvious version

Try every place the `a` letters could stop. For each split, count the `b` letters before it and the `a` letters after it, since exactly those have to go, and keep the smallest count. Costs the line length squared.

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_minimum_deletions(line):
    n = len(line)
    best = None
    for cut in range(n + 1):
        removed = 0
        for i in range(cut):
            if line[i] == "b":
                removed += 1
        for i in range(cut, n):
            if line[i] == "a":
                removed += 1
        if best is None or removed < best:
            best = removed
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 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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