This article breaks down the Huawei OD coding problem “Count the Geese.” The core task is to identify complete
quacksequences from a mixed string and compute the minimum number of geese required to produce overlapping calls. The main challenge lies in interleaved concurrent sounds, strict order validation, and reliable detection of invalid input. Keywords: finite-state machine, greedy reuse, string simulation.
Technical Specification Snapshot
| Parameter | Details |
|---|---|
| Problem Type | String Simulation / State Counting |
| Implementation Language | Python |
| Input Constraints | 1 <= string length <= 1000 |
| Valid Character Set | q, u, a, c, k |
| Core Idea | Advance states through the quack stages and count concurrency |
| Time Complexity | O(n) |
| Space Complexity | O(1) |
| Source Context | Huawei OD / Reconstructed from a real coding test problem |
| Star Rating | Not provided in the original data |
| Core Dependencies | Python standard library only, no third-party dependencies |
This problem is fundamentally about finite-state progression
The input is a string composed of q, u, a, c, and k. You need to determine whether it can be decomposed into one or more complete quack calls produced by geese.
The difficulty is not matching a single word. The real challenge is that multiple quack sequences can be interleaved. In other words, several geese may be calling at the same time, and the character stream represents the merged result of concurrent sounds.
A typical example makes the concurrency pattern clear
quqauackckThis string is not two consecutive quackquack substrings, but it can still be formed by two geese calling in an interleaved way.
The key takeaway is that you cannot solve this with simple substring matching. You must track which stage of the call each goose is currently in.
A correct solution should track how many geese are waiting at each stage
You can break a goose call into five stages:
q: starts a callu: advances fromqa: advances fromuc: advances fromak: completes a full call fromc
As you scan the string, every incoming character must correspond to at least one goose that is ready to advance from the previous stage.
The transition rules can be compressed into a counting array
def count_geese(sound: str) -> int:
stages = [0, 0, 0, 0] # Counts of geese currently waiting for u/a/c/k
active = 0 # Number of geese currently calling
answer = 0 # Historical maximum concurrency, i.e. the minimum required geese
for ch in sound:
if ch == 'q':
# A new quack starts, so increment the active goose count
stages[0] += 1
active += 1
answer = max(answer, active)
elif ch == 'u':
if stages[0] == 0: # No goose is waiting for u, so the sequence is invalid
return -1
stages[0] -= 1
stages[1] += 1
elif ch == 'a':
if stages[1] == 0: # No goose is waiting for a, so the sequence is invalid
return -1
stages[1] -= 1
stages[2] += 1
elif ch == 'c':
if stages[2] == 0: # No goose is waiting for c, so the sequence is invalid
return -1
stages[2] -= 1
stages[3] += 1
elif ch == 'k':
if stages[3] == 0: # No goose is waiting for k, so the sequence is invalid
return -1
stages[3] -= 1
active -= 1 # One goose finishes a complete call
else:
return -1 # Invalid character encountered
# If any quack is still incomplete after the scan, the input is invalid
if active != 0 or any(stages):
return -1
return answer if answer > 0 else -1This code uses constant space to simulate multiple geese calling concurrently and returns the minimum number of geese required.
The key to this problem is not counting calls but reusing geese
Once a goose finishes one quack, it can theoretically produce another quack later. So the problem does not ask for the total number of quack occurrences. It asks for the peak number of geese calling at the same time.
That is why the algorithm increments active when it reads q and decrements active when it reads k. The maximum value reached by active during the scan is exactly the number of simultaneously calling geese.
Invalid input mainly falls into three categories
- The string contains characters outside
quack. - A character arrives, but no goose exists in the required previous stage.
- The scan ends while one or more calls are still incomplete.
samples = [
"quack", # 1 goose
"quqauackck", # 2 geese calling concurrently
"quaack", # Invalid: the order does not hold
"quac", # Invalid: the call does not complete
]
for s in samples:
print(s, "->", count_geese(s))This test snippet verifies the expected outputs for complete, interleaved, and invalid inputs.
The complexity is stable and ideal for live coding interviews
The entire algorithm only needs one linear scan. Each character is processed exactly once, with no backtracking and no need to explicitly maintain object-level state for each goose.
Therefore, the time complexity is O(n) and the space complexity is O(1). Under the maximum length constraint of 1000, this is more than efficient enough.
The original page illustration does not provide useful sample-level detail
AI Visual Insight: In the original page, this image appears to be more of a site illustration or promotional graphic than a technical diagram. It does not show input-output mappings, character state transitions, or concurrency timing, so it adds no direct value to the algorithmic reasoning and can be ignored during review.
In an interview, it is best to present the decision framework directly
First, explain that quack is a strictly ordered five-stage state chain. Then explain that interleaved calls require you to count how many calls are currently in progress. Finally, conclude that the maximum number of active calls is the answer.
This presentation path is ideal for helping an interviewer quickly verify that you truly understand the problem, rather than just memorizing a template.
A concise version you can repeat directly
# Core conclusions:
# 1. q means one new active goose starts calling
# 2. u/a/c/k must always advance from the previous stage
# 3. k means one active goose finishes a full call
# 4. The maximum active value is the minimum number of geese
# 5. If any stage cannot advance, or the final state is not cleared, return -1This summary works well as a pre-coding explanation and can significantly improve the clarity of your problem-solving communication.
FAQ
1. Why is the answer the maximum concurrency instead of the number of quack occurrences?
Because the same goose can call multiple times. The problem asks for the minimum number of geese that could have produced the sound stream, so you must count the peak number of geese simultaneously in the calling process.
2. Why can’t you simply split the string into groups of five characters?
Because multiple quack sequences can be interleaved. The global character order may not form contiguous blocks. Fixed-size slicing will incorrectly handle concurrent calling scenarios.
3. If the string contains only q/u/a/c/k, is it always valid?
No. A valid character set does not guarantee a valid order. For example, in qqauck, the second q may start a new sequence, but the later progression still needs stage-by-stage validation. If any sequence remains incomplete at the end, the function must return -1.
Core Summary: This article reconstructs the Huawei OD coding problem “Count the Geese” and focuses on how to determine the minimum number of geese from a mixed quack character stream. It provides a state-counting method, invalid-sequence validation logic, complexity analysis, and a Python reference implementation, making it useful for both interview prep and algorithm review.