Problem Description

You are given an integer array nums with specific properties:

• The array has a length of 2 * n (an even number of elements)
• The array contains exactly n + 1 unique elements
• Exactly one element appears n times, while all other elements appear exactly once

Your task is to find and return the element that appears n times.

For example, if n = 2, the array would have 4 elements total, with 3 unique values. One value would appear 2 times, and the other two values would each appear once. You need to identify which value appears 2 times.

The key insight is that since there are 2n total elements and n + 1 unique elements, with one element appearing n times, the remaining n elements must be the other n unique values appearing once each. This means any element that appears more than once must be the element that appears n times.

The solution uses a set to track seen elements. As we iterate through the array, if we encounter an element that's already in the set, it must be the repeated element (since all other elements appear only once), so we can immediately return it.

Intuition

Let's think about the mathematical constraints of this problem. We have 2n total elements and n + 1 unique elements. If one element appears n times, that accounts for n positions in the array. The remaining n positions must be filled by the other n unique elements.

Since we have exactly n remaining positions and n remaining unique elements, each of these other elements can only appear exactly once. This is the critical observation: every element appears either exactly once or exactly n times.

This means the moment we see any element for the second time while scanning through the array, we can be certain it's the element that appears n times. Why? Because if an element appears more than once, it cannot be one of the elements that appears exactly once, so it must be the element that appears n times.

This leads us to a simple approach: keep track of elements we've seen before using a set. As we iterate through the array:

• If we encounter an element we haven't seen before, add it to our set
• If we encounter an element that's already in our set, this must be our answer - return it immediately

We don't need to count occurrences or track frequencies. The first duplicate we find is guaranteed to be the element that appears n times, making this a very efficient O(n) time solution with potentially early termination.

Solution Approach

The implementation uses a hash set to efficiently track elements we've encountered while iterating through the array.

Here's the step-by-step implementation:

1. Initialize a set: Create an empty set s to store elements we've seen.

2. Iterate through the array: For each element x in nums:

   • Check if element exists: First check if x is already in the set s
   • If found: If x is in the set, this means we've seen it before. Since any repeated element must be the one that appears n times (as all others appear exactly once), we immediately return x
   • If not found: Add x to the set and continue to the next element

class Solution:
    def repeatedNTimes(self, nums: List[int]) -> int:
        s = set()
        for x in nums:
            if x in s:
                return x
            s.add(x)

Time Complexity: O(n) in the worst case, where we might need to check up to n + 1 elements before finding the duplicate. However, the algorithm often terminates early since the repeated element appears n times in 2n positions.

Space Complexity: O(n) in the worst case for storing up to n unique elements in the set before finding the duplicate.

The beauty of this solution is its simplicity - we don't need to count frequencies or use complex data structures. The mathematical constraints of the problem guarantee that the first duplicate we encounter is our answer.

Example Walkthrough

Let's walk through a concrete example to see how the solution works.

Example Input: nums = [5, 1, 5, 2, 5, 3, 5, 4]

In this case:

• Array length = 8, so 2n = 8, meaning n = 4
• We have 5 unique elements: {1, 2, 3, 4, 5}
• Element 5 appears 4 times (which is n times)
• Elements 1, 2, 3, 4 each appear once

Step-by-step execution:

1. Initialize: s = {} (empty set)

2. Iteration 1: x = 5

   • Is 5 in set? No
   • Add 5 to set: s = {5}
   • Continue

3. Iteration 2: x = 1

   • Is 1 in set? No
   • Add 1 to set: s = {5, 1}
   • Continue

4. Iteration 3: x = 5

   • Is 5 in set? Yes!
   • We found our duplicate!
   • Return 5

The algorithm terminates early after checking only 3 elements out of 8. We didn't need to count how many times 5 appears - we just needed to detect that it appears more than once. Since the problem guarantees that only one element appears multiple times (exactly n times), the first duplicate we find must be our answer.

Why this works: The mathematical constraints ensure that:

• If an element appears twice, it can't be one of the elements that appears exactly once
• Therefore, it must be the element that appears n times
• No need to verify it appears exactly n times - the problem guarantees this

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the input array nums.

The algorithm iterates through the array once in the worst case. However, due to the problem's constraints (one element appears n/2 times in an array of length 2n), the algorithm will typically terminate much earlier. By the pigeonhole principle, we're guaranteed to find a duplicate within the first n/2 + 1 elements, making the expected time complexity O(1) but the worst-case remains O(n).

Space Complexity: O(n) in the worst case.

The algorithm uses a set s to store unique elements encountered during iteration. In the worst-case scenario, we might need to store up to n/2 elements before finding the repeated element (though practically this would be much less due to the problem constraints). Therefore, the space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting to Calculate N First

A common mistake is trying to determine the value of n first by calculating len(nums) // 2, then attempting to count frequencies of all elements to find which one appears n times. This approach is unnecessarily complex and inefficient.

Incorrect approach:

def repeatedNTimes(self, nums: List[int]) -> int:
    n = len(nums) // 2
    freq = {}
    for num in nums:
        freq[num] = freq.get(num, 0) + 1
        if freq[num] == n:
            return num

Why it's problematic: While this works, it's overly complicated and potentially slower since you might count all elements even after finding the answer.

2. Using List Instead of Set for Tracking

Some might use a list to track seen elements, which significantly degrades performance.

Inefficient approach:

def repeatedNTimes(self, nums: List[int]) -> int:
    seen = []
    for num in nums:
        if num in seen:  # O(n) operation for list lookup
            return num
        seen.append(num)

Why it's problematic: Checking membership in a list is O(n) for each lookup, making the overall time complexity O(n²) instead of O(n).

3. Edge Case: Assuming Element Distribution

Some solutions might assume the repeated element is evenly distributed or appears in specific positions, leading to solutions that work for most cases but fail on edge cases.

Risky optimization attempt:

def repeatedNTimes(self, nums: List[int]) -> int:
    # Assuming the repeated element appears within first few elements
    for i in range(4):
        for j in range(i):
            if nums[i] == nums[j]:
                return nums[i]

Why it's problematic: While the repeated element often appears early due to probability, it's not guaranteed. For example, in [1,2,3,4,5,5], all repeated elements could be clustered at the end.

4. Not Returning Immediately Upon Finding Duplicate

Continuing to process after finding the duplicate wastes computational resources.

Suboptimal approach:

def repeatedNTimes(self, nums: List[int]) -> int:
    seen = set()
    repeated = None
    for num in nums:
        if num in seen:
            repeated = num  # Stores but doesn't return immediately
        seen.add(num)
    return repeated

Solution: Always return immediately when you find the first duplicate, as shown in the optimal solution. The mathematical constraints guarantee this first duplicate is the answer.