Leetcode 217: Contains Duplicate - Last Part Answer Not Working

The standard solution for Leetcode 217 uses a HashSet to track seen numbers, but this fails if you incorrectly return true inside the loop without checking all elements or mishandle the logic flow. The most common bug occurs when developers confuse the condition for returning false (no duplicates found) with early returns that skip remaining checks. This article provides the correct Java implementation using a Set to ensure every element is validated against previously seen values, guaranteeing accurate results for all input arrays.

Understanding the Problem Constraints

Leetcode 217 requires determining if any value appears at least twice in an integer array. If every element is distinct, the function must return false. The challenge lies in efficiently detecting duplicates without resorting to nested loops, which would result in O(n^2) time complexity. The optimal approach leverages hash-based data structures to achieve O(n) time complexity by storing elements as they are encountered. A frequent mistake among beginners is assuming that sorting the array is always the best first step. While sorting allows adjacent comparison, it modifies the original order and takes O(n log n) time. Using a HashSet is generally preferred because it handles the duplicate check in linear time relative to the number of elements, provided the hash collisions are managed well by the underlying implementation.

Common Implementation Errors

Many developers write code that seems logical but fails specific test cases due to flawed control flow. A typical erroneous pattern involves iterating through the array and adding each element to a set, but then checking for duplicates in a separate pass or using incorrect boolean flags. Another common error is returning true immediately upon finding a match, which is actually correct, but failing to handle the case where no matches are found until the loop completes. Consider the scenario where the array has only one element. The loop might not execute the duplicate check correctly if the initialization is wrong. Furthermore, some users attempt to use primitive arrays with binary search, forgetting that the array must be sorted first. If the array is not sorted, binary search will yield incorrect results. This leads to the "last part" failure mentioned in the title, where edge cases like empty arrays or single-element arrays cause runtime errors or incorrect outputs.

The Correct HashSet Approach

The robust solution involves creating a HashSet<Integer> to store unique elements. As you iterate through the input array, you check if the current number already exists in the set. If it does, you have found a duplicate and can immediately return true. If the loop finishes without finding any duplicates, you return false. This logic is straightforward but requires precise syntax in Java. Here is the correct implementation:

public class Solution {
    public boolean containsDuplicate(int[] nums) {
        if (nums == null || nums.length < 2) {
            return false;
        }
        
        java.util.Set seen = new java.util.HashSet<>();
        
        for (int num : nums) {
            if (seen.contains(num)) {
                return true;
            }
            seen.add(num);
        }
        
        return false;
    }
}

This code explicitly handles null or small arrays at the start, preventing unnecessary processing. The core loop uses enhanced for-each syntax for clarity. The critical line is if (seen.contains(num)), which checks for existence before adding. If you add first and then check, you might miss the immediate detection or complicate the logic. This approach ensures that we catch the first occurrence of a duplicate instantly.

Why Sorting Can Be Risky

While sorting is a valid alternative, it introduces overhead. In Java, Arrays.sort(nums) uses Dual-Pivot Quicksort for primitives, which is fast but still O(n log n). After sorting, you only need to check if nums[i] == nums[i+1]. However, this method mutates the input array, which might be undesirable if the caller needs the original order preserved. Additionally, for very large datasets, the constant factors associated with sorting can make it slower than hashing in practice, despite the better theoretical time complexity class. If you choose the sorting route, ensure you do not create a copy unless necessary. Modifying the input directly saves memory but changes state outside the function scope. For interview settings, clarifying whether mutation is allowed is crucial. If not, you must clone the array first, adding O(n) space complexity. The HashSet approach naturally handles this by using extra space for the set, keeping the trade-off predictable.

Performance Considerations

Time complexity for the HashSet solution is O(n) on average. Each insertion and lookup operation is O(1) amortized. However, worst-case scenarios exist where all elements hash to the same bucket, degrading performance to O(n^2). In competitive programming contexts, this is rare with integer keys in Java's HashSet, but it is worth noting. Space complexity is O(n) because, in the worst case (all unique elements), the set stores every item from the input array. For extremely large inputs where memory is constrained, the sorting method might be preferable due to lower space overhead (O(1) auxiliary if mutation is allowed). However, for most Leetcode constraints, the HashSet solution is the expected answer due to its simplicity and speed. Always profile your solution if the constraints allow for O(n^2) solutions to pass, but aim for the optimal approach.

Debugging the "Last Part" Failure

The "last part" failure often stems from missing the final return false statement. Developers sometimes forget that if the loop completes, no duplicates were found, so the function must explicitly return false. Without this, the compiler may throw an error, or the logic may fall through incorrectly depending on the language. Another subtle bug is initializing the set inside the loop or resetting it incorrectly. Ensure the set persists across iterations. Also, verify that you are handling the input array correctly. If the array is empty, the loop should not run, and the function should return false. The initial check if (nums.length < 2) covers this efficiently. Skipping this check might lead to unnecessary iteration over an empty set, though it usually doesn't break functionality, it wastes cycles. Removing print statement affects functionality... corruption at play? discusses similar issues where side effects or missing returns cause unexpected behavior in other contexts.

Frequently Asked Questions

Why does my code return true for arrays with all unique elements?

This usually happens if you mistakenly check the size of the set after adding all elements instead of during iteration. You must check for existence before adding. If you add first, the set size increases, but you lose the ability to detect the exact moment a duplicate was introduced.

Can I use a boolean array instead of a HashSet?

Only if the range of integers is known and small. Since Leetcode 217 allows any integer, a boolean array would require massive memory allocation for negative numbers or large values. HashSet handles sparse and wide ranges efficiently.

Is the HashSet solution faster than sorting for large inputs?

Yes, typically. O(n) average time is faster than O(n log n) for sorting. The overhead of hashing is minimal compared to the comparisons required for sorting, making it the preferred choice for performance-critical applications.

What happens if the input array is null?

You must handle null explicitly. Accessing nums.length on a null reference throws a NullPointerException. Always check for null at the beginning of your function to ensure safe execution.

Does the order of elements matter in the HashSet solution?

No, the HashSet does not maintain insertion order. It only cares about uniqueness. Therefore, the position of duplicates in the original array does not affect the correctness of the result, only their presence.