One of the intriguing array problems on Leetcode is the “N-Repeated Element in Size 2N Array”. This article takes you through an in-depth understanding of the problem, its potential solutions, and a step-by-step Python implementation.

### Problem Statement

In a 2N-sized array, there are N+1 unique elements, and exactly one of these elements is repeated N times. Your task is to find and return the element that is repeated N times.

### Initial Thoughts

Given that there’s a single element that appears more than once, one might be tempted to utilize a brute-force approach: possibly using nested loops to compare elements and find the repeated one. However, there’s a more efficient way to approach this, which we’ll explore later.

### Key Insight

With a total of 2N elements and N+1 unique ones, the repeated element will show up at least twice. Moreover, because it repeats N times, we can be sure that if we consider any set of three adjacent elements in the array, at least one of these sets will contain two identical numbers, which will be the repeated element. This insight greatly reduces the number of required comparisons.

### Algorithm to Solve the Problem

- Traverse the array and consider each set of three adjacent elements.
- If any of the three elements are equal, return one of the equal elements as the answer.

### Python Code Solution

Let’s now translate our algorithm into a Python function:

```
def repeatedNTimes(A):
for i in range(len(A) - 2):
if A[i] == A[i + 1] or A[i] == A[i + 2]:
return A[i]
if A[i + 1] == A[i + 2]:
return A[i + 1]
```

### Complexity Analysis

**Time Complexity**: O(N) – We traverse the array only once, considering every set of three adjacent elements.**Space Complexity**: O(1) – Our solution doesn’t make use of any additional data structures that scale with input size. The space usage is constant.

### Testing the Solution

Testing ensures that the solution works as expected across varied test cases:

```
print(repeatedNTimes([1, 2, 3, 3])) # Expected output: 3
print(repeatedNTimes([2, 1, 2, 5, 3, 2])) # Expected output: 2
print(repeatedNTimes([5, 1, 5, 2, 5, 3])) # Expected output: 5
```

### Alternative Approach: Using a Set

Another efficient method to solve this problem is by utilizing a set data structure. By iterating through the array and adding elements to the set, the first number we encounter that’s already in the set is our answer.

```
def repeatedNTimes(A):
seen = set()
for num in A:
if num in seen:
return num
seen.add(num)
```

### Conclusion

The “N-Repeated Element in Size 2N Array” problem offers a profound lesson in algorithmic efficiency. It demonstrates that, sometimes, understanding the unique characteristics of the data can lead to a highly optimized solution. Instead of going for the more intuitive, brute-force solutions, it’s always beneficial to pause, reflect on the problem’s specifics, and seek a more efficient path.