What is the time complexity of Reduce Array Size to The Half?

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

What data structures are used to solve Reduce Array Size to The Half?

This problem uses Array, Hash Table, Greedy, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Reduce Array Size to The Half?

Insight 1: Counting the frequency of each number is crucial because we want to remove the most frequent numbers first to minimize the set size. Insight 2: Sorting the frequencies in descending order allows us to greedily select the most frequent numbers for removal.

Which companies ask LeetCode 1338. Reduce Array Size to The Half in interviews?

This problem is frequently asked at Akuna Capital.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Reduce Array Size to The Half - LeetCode 1338 Solution

Reduce Array Size to The Half - Complete Solution Guide

Reduce Array Size to The Half is LeetCode problem 1338, a Medium 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 arr . You can choose a set of integers and remove all the occurrences of these integers in the array. Return the minimum size of the set so that at least half of the integers of the array are removed . Example 1: Input: arr = [3,3,3,3,5,5,5,2,2,7] Output: 2 Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array). Possible sets of size 2 are {3,5},{3,2},{5,2}. Choosing set {2,7} is not possibl

Detailed Explanation

The problem asks us to find the minimum number of distinct integers we need to remove from an array `arr` such that at least half of the elements in the original array are removed. The array `arr` always has an even number of elements. We need to find the smallest set of integers that, when removed from `arr`, leave us with an array of size `arr.length / 2` or less.

Solution Approach

The solution involves counting the occurrences of each number in the array using a hash map (or similar data structure), sorting the frequencies of these numbers in descending order, and then greedily removing numbers from the array based on their frequency. We iterate through the sorted frequencies, keeping track of the total number of elements removed and the size of the set of distinct numbers removed. As soon as the total number of removed elements reaches or exceeds half the size of the original array, we return the size of the set.

Step-by-Step Algorithm

Step 1: Count the occurrences of each number in the input array `arr`. Store these counts in a hash map (e.g., `counts` in Python/Java/C++ or using an array representation in C).
Step 2: Extract the frequencies (counts) from the hash map into a separate list or array.
Step 3: Sort the frequencies in descending order.
Step 4: Initialize `removed_count` to 0 and `set_size` to 0. `removed_count` keeps track of the total number of elements removed, and `set_size` represents the number of distinct elements removed.
Step 5: Iterate through the sorted frequencies. In each iteration, add the current frequency to `removed_count` and increment `set_size`.
Step 6: Check if `removed_count` is greater than or equal to `arr.length / 2`. If it is, return `set_size`.
Step 7: If the loop completes without `removed_count` reaching `arr.length / 2`, return 0 (This should not occur given the problem constraints, but it's good to have for completeness).

Key Insights

Insight 1: Counting the frequency of each number is crucial because we want to remove the most frequent numbers first to minimize the set size.
Insight 2: Sorting the frequencies in descending order allows us to greedily select the most frequent numbers for removal.
Insight 3: Once the total number of removed elements is greater than or equal to half the original array size, we can stop removing elements, as we've met the problem's requirement.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Greedy, Sorting, Heap (Priority Queue).

Companies

Asked at: Akuna Capital.

Reduce Array Size to The Half | LeetCode 1338

Medium

Array Hash Table Greedy Sorting Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public int minSetSize(int[] arr) {
        int n = arr.length;
        int target = n / 2;

        Map<Integer, Integer> counts = new HashMap<>();
        for (int num : arr) {
            counts.put(num, counts.getOrDefault(num, 0) + 1);
        }

        List<Integer> frequencies = new ArrayList<>(counts.values());
        Collections.sort(frequencies, Collections.reverseOrder());

        int removedCount = 0;
        int setSize = 0;

        for (int freq : frequencies) {
            removedCount += freq;
            setSize += 1;
            if (removedCount >= target) {
                return setSize;
            }
        }

        return 0;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1338: Reduce Array Size to The Half - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy Sorting Heap (Priority Queue)

Companies

Akuna Capital

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1338

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: Count the occurrences of each number in the input array `arr`. Store these counts in a hash map (e.g., `counts` in Python/Java/C++ or using an array representation in C).
2
Step 2: Extract the frequencies (counts) from the hash map into a separate list or array.
3
Step 3: Sort the frequencies in descending order.
4
Step 4: Initialize `removed_count` to 0 and `set_size` to 0. `removed_count` keeps track of the total number of elements removed, and `set_size` represents the number of distinct elements removed.
5
Step 5: Iterate through the sorted frequencies. In each iteration, add the current frequency to `removed_count` and increment `set_size`.
6
Step 6: Check if `removed_count` is greater than or equal to `arr.length / 2`. If it is, return `set_size`.
7
Step 7: If the loop completes without `removed_count` reaching `arr.length / 2`, return 0 (This should not occur given the problem constraints, but it's good to have for completeness).

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the sorting step. The space complexity is determined by the hash map (or array) and the frequency list/array. The counting operation iterates through the array once which is O(n). The sorting operation is O(n log n). The last loop to find the set size iterates at most n times, so it is O(n). Therefore the overall time complexity is O(n log n). The space complexity comes from the hash table that stores counts for each number which is O(n) in the worst case, where all the numbers are distinct.

📖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

Heap (Priority Queue)

Instead of sorting the frequencies, we can use a max heap to keep track of the largest frequencies. We repeatedly extract the maximum frequency from the heap until the removed count reaches half the size of the array.

Time: O(n log k), where n is the size of the array and k is the number of distinct elements. Building the heap takes O(k) time. Extracting elements from the heap until the target is reached takes O(m log k) time, where 'm' is the number of elements removed. In the worst case where all elements are distinct, m would be approximately n/2.Space: O(k), where k is the number of distinct elements. The heap stores the frequencies of the distinct elements.

Trade-offs: The heap approach can be more efficient than sorting if the number of distinct elements is significantly smaller than the total number of elements in the array. Sorting is generally easier to implement.

Bucket Sort (Counting Sort)

If the range of frequencies is relatively small, bucket sort or counting sort could be used to sort frequencies in linear time. This would optimize the time complexity to O(n+k), but requires knowing the upper bound of frequencies in advance.

Time: O(n + max_frequency), where max_frequency is the largest count value. If max_frequency is O(n), this becomes O(n).Space: O(max_frequency)

Trade-offs: Bucket Sort is very efficient (O(n)) when the maximum frequency is small. When the maximum frequency is much larger than n, the time complexity will grow past O(n log n), therefore sorting might be more appropriate.

✨

Key Insights

1Insight 1: Counting the frequency of each number is crucial because we want to remove the most frequent numbers first to minimize the set size.
2Insight 2: Sorting the frequencies in descending order allows us to greedily select the most frequent numbers for removal.
3Insight 3: Once the total number of removed elements is greater than or equal to half the original array size, we can stop removing elements, as we've met the problem's requirement.

📚

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!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Array with all same elements (e.g., [7,7,7,7,7,7]). The algorithm correctly identifies that removing only one distinct element (7) is sufficient to reduce the array size to zero.
Edge case 2: Array with all distinct elements (e.g., [1,2,3,4,5,6]). The algorithm correctly handles this by removing the largest possible number of distinct elements (n/2) to reduce array size to half.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the frequency of each number and simply removing random numbers, which may not result in the minimum set size.

✗Common mistake 2: Forgetting to sort the frequencies in descending order, leading to an incorrect set size because less frequent elements might be removed before more frequent ones.

✗Common mistake 3: Prematurely stopping the set size calculation before reaching at least half the array size in removed elements.

✅

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: Data compression. Identifying and removing the most frequent characters or words can reduce the size of text or data files.

🏦Real-world scenario 2: Resource allocation. Prioritizing and allocating resources based on the frequency of requests can optimize system performance. For example, caching the most frequently accessed data in a content delivery network (CDN).

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 18, 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

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

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Reduce Array Size to The Half"?

For this medium-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.

🤖 Need help with this problem?