How do you solve N-Repeated Element in Size 2N Array on LeetCode?

N-Repeated Element in Size 2N Array (LeetCode #961) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The problem guarantees that there is exactly one element that is repeated n times, which simplifies our search.

What is the time complexity of N-Repeated Element in Size 2N Array?

The optimal solution for N-Repeated Element in Size 2N Array runs in O(n) time and uses O(n) space. This complexity is due to the single pass through the array to count occurrences, and the space is used to store counts in the HashMap.

What is the brute force solution for N-Repeated Element in Size 2N Array?

The brute force approach for N-Repeated Element in Size 2N Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each element against every other element to find the one that appears more than once. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve N-Repeated Element in Size 2N Array?

N-Repeated Element in Size 2N Array uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for N-Repeated Element in Size 2N Array?

Always clarify the problem constraints and guarantees before diving into coding. Consider edge cases, like the smallest possible input, to ensure your solution handles all scenarios. Explain your thought process while coding; it shows your understanding and helps interviewers follow your logic.

What are common mistakes when solving N-Repeated Element in Size 2N Array?

Not recognizing that the problem guarantees one element is repeated n times, leading to unnecessary checks. Confusing the counts of unique elements versus the repeated element.

#961

N-Repeated Element in Size 2N Array

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a HashMap allows us to count occurrences of each element in a single pass through the array, making it much more efficient.

⚙️

Algorithm

3 steps

1Step 1: Initialize a HashMap to keep track of the counts of each element.
2Step 2: Loop through the array and populate the HashMap with counts.
3Step 3: Check the HashMap for the element that has a count of n and return it.

solution.py8 lines

1# Full working Python code
2
3def repeatedNTimes(nums):
4    counts = {}
5    for num in nums:
6        counts[num] = counts.get(num, 0) + 1
7        if counts[num] == len(nums) // 2:
8            return num

ℹ

Complexity note: This complexity is due to the single pass through the array to count occurrences, and the space is used to store counts in the HashMap.

1The problem guarantees that there is exactly one element that is repeated n times, which simplifies our search.
2Using a HashMap allows us to efficiently count occurrences without needing nested loops.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.