#961

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

N-Repeated Element in Size 2N Array

Find the element that is repeated exactly N times in a 2N-sized array using array scanning and hash lookups.

Problem Statement

You are given an integer array, nums, with a length of 2N. The array contains N unique elements, and one of the elements is repeated exactly N times. Your task is to find the repeated element in the array.

For example, in the array nums = [1, 2, 3, 3], the number 3 is repeated twice. Since the length of the array is 4, which is 2N, the output is 3. Your goal is to determine the repeated element efficiently.

Examples

Example 1

Input: nums = [1,2,3,3]

Output: 3

Example details omitted.

Example 2

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

Output: 2

Example details omitted.

Example 3

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

Output: 5

Example details omitted.

Constraints

2 <= n <= 5000
nums.length == 2 * n
0 <= nums[i] <= 104
nums contains n + 1 unique elements and one of them is repeated exactly n times.

Solution Approach

Array Scanning with Hash Lookup

Iterate through the array while maintaining a hash map to store the frequency of each element. Once an element's count reaches N, it is identified as the repeated element.

Optimized Lookup with Early Termination

While scanning the array, you can stop as soon as you find the element repeated N times. This approach optimizes both time and space complexity.

Space-Efficient Hash Table Approach

Instead of storing the entire array's frequency, track only the repeated element using a hash map. This avoids unnecessary space consumption while ensuring a time-efficient solution.

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(1)

The solution runs in O(N) time, where N is the number of unique elements in the array. The space complexity is O(1) as we only track the repeated element in the hash map, not the entire array.

What Interviewers Usually Probe

Can the candidate explain the space-time trade-off in hash-based solutions?
Does the candidate consider stopping the scan early upon identifying the repeated element?
Is the candidate aware of the space complexity and its optimization?

Common Pitfalls or Variants

Common pitfalls

Using unnecessary space by tracking all elements instead of just the repeated one.
Failing to stop the array scan early once the repeated element is found.
Misunderstanding the time complexity of hash lookups and scanning.

Follow-up variants

What if the array is sorted? How would the solution change?
Can we solve this problem without using a hash table? What other techniques can be applied?
What if there are multiple repeated elements with different frequencies?

FAQ

What is the time complexity of the N-Repeated Element problem?

The time complexity is O(N), where N is the number of unique elements in the array, due to the array scan and hash lookup.

Can I solve this problem without using a hash table?

Yes, it is possible to solve the problem with other techniques like sorting or using a set, but hash tables provide the most efficient solution.

How does array scanning help in this problem?

Array scanning allows us to efficiently check for repetitions while keeping track of the count using a hash map, enabling quick identification of the repeated element.

What are the common mistakes when solving the N-Repeated Element problem?

Common mistakes include not optimizing space by tracking only the repeated element and failing to terminate the scan early once the answer is found.

What makes the N-Repeated Element problem unique?

This problem requires understanding both array scanning and hash table lookups, making it a great example of optimizing both time and space in solving problems efficiently.

terminal

Solution

Solution 1: Hash Table

Since the array $\textit{nums}$ has a total of $2n$ elements, with $n + 1$ distinct elements, and one element repeated $n$ times, this means the remaining $n$ elements in the array are all distinct.

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

Solution 2: Mathematics

According to the problem description, half of the elements in the array $\textit{nums}$ are the same. If we view the array as a circular arrangement, then there is at most $1$ other element between two identical elements.

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

Continue Practicing

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