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

The optimal solution has a time complexity of O(n).

What data structures are used to solve N-Repeated Element in Size 2N Array?

This problem uses Array, Hash Table. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving N-Repeated Element in Size 2N Array?

Insight 1: Since only one element is repeated N times, and there are n+1 unique elements in the array, the repeated element is guaranteed to be present. Insight 2: The repeated element must occur at least twice in the array since n >= 2.

Which companies ask LeetCode 961. N-Repeated Element in Size 2N Array in interviews?

This problem is frequently asked at Akamai.

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N-Repeated Element in Size 2N Array - LeetCode 961 Solution

Q: How to solve N-Repeated Element in Size 2N Array?

The Python3 solution uses a hash table (dictionary) to store the counts of each number encountered in the input array `nums`. It iterates through the array, and for each number, it checks if the number is already present in the hash table. If it is, it means we have found the element that is repeated N times, and we return it. Otherwise, we add the number to the hash table with a count of 1. The Java/C/C++ solutions try to detect repeated elements by comparing adjacent elements and elements two indices apart. Since the repeated number appears n times and the array has 2*n size, the repeated number will likely appear adjacent or with one number in between. The algorithm returns the first number found this way.

N-Repeated Element in Size 2N Array - Complete Solution Guide

N-Repeated Element in Size 2N Array is LeetCode problem 961, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer array nums with the following properties: nums.length == 2 * n . nums contains n + 1 unique elements. Exactly one element of nums is repeated n times. Return the element that is repeated n times . Example 1: Input: nums = [1,2,3,3] Output: 3 Example 2: Input: nums = [2,1,2,5,3,2] Output: 2 Example 3: Input: nums = [5,1,5,2,5,3,5,4] Output: 5 Constraints: 2 <= n <= 5000 nums.length == 2 * n 0 <= nums[i] <= 10 4 nums contains n + 1 unique elements and one of them is repeat

Detailed Explanation

The problem asks us to find the element that appears N times in an array 'nums' of size 2N. The array contains n+1 unique elements, meaning one element is repeated exactly n times. The input is an integer array 'nums', and the output is the integer that is repeated N times.

Solution Approach

The Python3 solution uses a hash table (dictionary) to store the counts of each number encountered in the input array `nums`. It iterates through the array, and for each number, it checks if the number is already present in the hash table. If it is, it means we have found the element that is repeated N times, and we return it. Otherwise, we add the number to the hash table with a count of 1. The Java/C/C++ solutions try to detect repeated elements by comparing adjacent elements and elements two indices apart. Since the repeated number appears n times and the array has 2*n size, the repeated number will likely appear adjacent or with one number in between. The algorithm returns the first number found this way.

Step-by-Step Algorithm

Step 1: Initialize an empty hash table (dictionary) to store the counts of each element.
Step 2: Iterate through the input array `nums`.
Step 3: For each element `num` in `nums`:
Step 4: Check if `num` is present as a key in the hash table.
Step 5: If `num` is present, return `num` as it is the element repeated N times.
Step 6: If `num` is not present, add `num` as a key to the hash table with a value of 1 (representing its count).
Step 7: The algorithm will return the result within the loop. If it reaches the end of the array without finding a repeated number using Python, that means there is an issue with the given array (but that would contradict with the problem statement). The C/Java/C++ solutions will return the first element if no adjacent elements are found.

Key Insights

Insight 1: Since only one element is repeated N times, and there are n+1 unique elements in the array, the repeated element is guaranteed to be present.
Insight 2: The repeated element must occur at least twice in the array since n >= 2.
Insight 3: A simple hash table (dictionary) can be used to efficiently count the occurrences of each element.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: Akamai.

N-Repeated Element in Size 2N Array | LeetCode 961

Easy

Array Hash Table

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int repeatedNTimes(int[] nums) {
        for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i] == nums[i + 1]) {
                return nums[i];
            }
            if (i < nums.length - 2 && nums[i] == nums[i + 2]) {
                return nums[i];
            }
        }
        return nums[0];
    }
}

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Problem Info

Description

Problem 961: N-Repeated Element in Size 2N Array - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

Akamai

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:961

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize an empty hash table (dictionary) to store the counts of each element.
2
Step 2: Iterate through the input array `nums`.
3
Step 3: For each element `num` in `nums`:
4
Step 4: Check if `num` is present as a key in the hash table.
5
Step 5: If `num` is present, return `num` as it is the element repeated N times.
6
Step 6: If `num` is not present, add `num` as a key to the hash table with a value of 1 (representing its count).
7
Step 7: The algorithm will return the result within the loop. If it reaches the end of the array without finding a repeated number using Python, that means there is an issue with the given array (but that would contradict with the problem statement). The C/Java/C++ solutions will return the first element if no adjacent elements are found.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the single loop iterating through the array. The space complexity depends on the size of the hash table, which, in the worst case, can store up to n unique elements.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting

Sort the array and then iterate through it. Since n instances of a number exist, after sorting all those n instances are next to each other. We can look at neighbors and find the repeated one.

Time: O(n log n) due to sorting.Space: O(1) or O(n) depending on the sorting algorithm used (in-place vs. out-of-place sorting).

Trade-offs: Sorting is generally slower than the hash table approach due to the O(n log n) time complexity. However, it may use less space if an in-place sorting algorithm is used.

Checking every 3 elements

Based on the pigeonhole principle and the constraints of the problem, it is guaranteed that at least two out of the first three elements must be the same. We can return the duplicate in the first three elements of the array.

Time: O(1)Space: O(1)

Trade-offs: Extremely fast, but relies on specific problem constraints. Not generalizable to other array repetition problems.

✨

Key Insights

1Insight 1: Since only one element is repeated N times, and there are n+1 unique elements in the array, the repeated element is guaranteed to be present.
2Insight 2: The repeated element must occur at least twice in the array since n >= 2.
3Insight 3: A simple hash table (dictionary) can be used to efficiently count the occurrences of each element.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Empty array. The problem statement indicates n >= 2, so the array will never be empty. Therefore this case does not need handling.
Edge case 2: Array with only two elements. The algorithm will correctly identify the repeated element if the two elements are same.
Edge case 3: All elements are the same. The algorithm will immediately find the repeated number.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Assuming that the array will always have adjacent repeated elements. The repeated elements might be separated by other unique elements.

✗Common mistake 2: Not considering the case where the repeated element appears at the beginning or end of the array.

✗Common mistake 3: Using less efficient data structures, like searching the entire array for each element which will result in O(n^2) time complexity

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Identifying frequently accessed data in a cache. We can use a similar counting approach to track which data items are accessed most often and prioritize them for caching.

🏦Real-world scenario 2: Finding popular products in an e-commerce store. We can track the number of times each product is viewed or added to a cart, and identify the most popular products based on their counts.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "N-Repeated Element in Size 2N Array"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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