Hey there! Let's figure out why your PermCheck solution is hitting a snag on double-element test cases—even though your local tests seem fine, Codility's edge cases are probably catching something you missed.

First, let's recap the core requirements of the PermCheck task: we need to verify if the input array is a permutation of integers from 1 to N, where N is the length of the array. That means three non-negotiable conditions must all be true:

1. Every element is between 1 and N (inclusive).
2. No element is repeated.
3. All integers from 1 to N are present exactly once.

Common Double-Element Test Cases That Trip Up Solutions #

Codility likely runs these edge cases that you might not have tested (or might have overlooked):

• [1,1]: Should return 0 (duplicate element, missing 2)
• [2,3]: Should return 0 (elements exceed N=2)
• [0,1]: Should return 0 (element 0 is out of the 1~2 range)
• [1,3]: Should return 0 (missing 2, element 3 exceeds N=2)
• [2,1]: Should return 1 (valid permutation)

Why Your Code Might Be Failing #

Here are the most common mistakes that cause correctness issues on these cases:

1. Only checking the sum of elements: You might have used the formula N*(N+1)/2 to verify the total sum, but this doesn't account for duplicates or out-of-range elements. For example, [3,0] sums to 3 (which equals 2*3/2) but is not a valid permutation.
2. Forgetting to validate element ranges: If your code only checks for presence of 1~N but doesn't flag elements outside this range, cases like [2,3] will slip through as valid (which they're not).
3. Incomplete set checks: If you're using a set to track elements, you might have made a mistake in verifying coverage. For example, using range(1, N) instead of range(1, N+1) would miss checking for the number N itself—so [1,3] with N=2 would incorrectly pass.
4. Ignoring duplicates: If your code doesn't check for repeated values, cases like [1,1] would be incorrectly marked as valid.

Fixes to Implement #

Let's outline a bulletproof approach that covers all edge cases:

• Track seen elements: Use a boolean array (or a set) to mark which numbers from 1~N have been encountered.
• Validate element ranges on the fly: For every element, if it's less than 1 or greater than N, immediately return 0.
• Check for duplicates: If an element is already marked as seen, return 0.
• Ensure full coverage: After processing all elements, confirm every number from 1 to N was seen (though the above checks might make this redundant if implemented correctly).

Here's an example Python solution that hits 100% correctness and performance:

def solution(A):
    n = len(A)
    seen = [False] * (n + 1)  # Indexes 0 to n; we'll use 1 to n
    
    for num in A:
        # Check if number is out of valid range
        if num < 1 or num > n:
            return 0
        # Check for duplicate
        if seen[num]:
            return 0
        seen[num] = True
    
    # If we made it here, all conditions are satisfied
    return 1

Next Steps for You #

1. Run the double-element test cases listed above through your current code.
2. Identify which case returns an incorrect result—this will point directly to the missing condition in your code.
3. Add the necessary check (range validation, duplicate detection, or full coverage verification) to fix the gap.

内容的提问来源于stack exchange，提问作者Muhammad Rahil Rafiq