0825EasyHash TableStringCounting

Pull One Stem to Even the Bunch

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Trains the technique from

LeetCode 2423Remove Letter To Equalize Frequency

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 florist writes down a bunch of flowers as the string bunch, one lowercase letter per stem, where the letter is the kind of flower.

Before the bunch is wrapped, exactly one stem must be pulled out. Pulling a stem is not optional and no more than one may go. The bunch counts as even when every kind of flower still in it appears the same number of times. A kind whose last stem was the one pulled is no longer in the bunch at all, so it is not compared against anything.

Return true if some single stem can be pulled to leave an even bunch, and false otherwise.

Examples

Example 1

Input: bunch = "ffgh"
Output: true

Pulling one of the two `f` stems leaves `fgh`, in which each of the three kinds still present appears once.

Example 2

Input: bunch = "ttyy"
Output: false

Whichever stem is pulled, one kind is left with two stems and the other with one, so the two kinds present do not appear the same number of times.

Example 3

Input: bunch = "wxxx"
Output: true

Pulling the `w` stem leaves `xxx`. The `w` kind is gone from the bunch, so only `x` is compared, and it appears three times.

Constraints

2 <= bunch.length <= 100
bunch consists of lowercase English letters only.

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 bunch_evens_out(bunch: str) -> bool:

Java

public boolean bunchEvensOut(String bunch)

The intended approach

Time O(n + k^2) for a bunch of n stems over k = 26 kinds of flower.Space O(k) for the tallies.

Two stems of the same kind are interchangeable, so the bunch reduces to a tally per kind and there are at most twenty-six different pulls to consider. Take each kind that is present, lower its tally by one, and check that the kinds still on a tally above zero all carry the same one, putting the stem back before moving on.

Pythonentry point: bunch_evens_out

def bunch_evens_out(bunch):
    # Which stem of a given kind is pulled makes no difference, so only the tally of
    # each kind matters: try lowering each kind that is present by one and see
    # whether the kinds still present all share one tally. A kind that drops to zero
    # is simply gone and is not compared.
    # Time: O(n + k^2) for a bunch of n stems over k = 26 kinds.
    # Space: O(k) for the tallies.
    tally = [0] * 26
    for ch in bunch:
        tally[ord(ch) - 97] += 1
    for pulled in range(26):
        if tally[pulled] == 0:
            continue
        tally[pulled] -= 1
        target = 0
        even = True
        for kind in range(26):
            if tally[kind] == 0:
                continue
            if target == 0:
                target = tally[kind]
            elif tally[kind] != target:
                even = False
                break
        tally[pulled] += 1
        if even:
            return True
    return False

The obvious version

Pull the stem at each position of the string in turn, rebuild the counts of what is left from scratch, and check whether those counts are all one number.

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_bunch_evens_out(bunch):
    """Pull out the stem at each position in turn, count the kinds left from scratch,
    and check whether the counts that survive are all the same number."""
    n = len(bunch)
    for i in range(n):
        rest = bunch[:i] + bunch[i + 1:]
        counts = {}
        for ch in rest:
            counts[ch] = counts.get(ch, 0) + 1
        sizes = list(counts.values())
        same = True
        for s in sizes:
            if s != sizes[0]:
                same = False
                break
        if same:
            return True
    return False

How this problem was checked

Reference solution executed against 24 test cases — 24 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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