What is the time complexity of Least Number of Unique Integers after K Removals?

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

What data structures are used to solve Least Number of Unique Integers after K Removals?

This problem uses Array, Hash Table, Greedy, Sorting, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Least Number of Unique Integers after K Removals?

Insight 1: The key is to remove elements based on their frequency in the array. Removing elements with lower frequencies first is a greedy approach that minimizes the number of unique integers. Insight 2: Using a hash table (or dictionary) to count the occurrences of each number is efficient for determining the frequency of each element.

Which companies ask LeetCode 1481. Least Number of Unique Integers after K Removals in interviews?

This problem is frequently asked at Expedia, IBM, J.P. Morgan.

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Least Number of Unique Integers after K Removals - LeetCode 1481 Solution

Q: How to solve Least Number of Unique Integers after K Removals?

The solution uses a greedy approach. First, it counts the frequency of each element in the input array using a hash table. Then, it sorts these frequencies in ascending order. Finally, it iterates through the sorted frequencies, removing elements from the array by decrementing `k` by the frequency of each number as long as `k` is greater than or equal to the current frequency. If `k` becomes less than the current frequency, it stops removing elements. The remaining number of frequencies is the least number of unique integers left in the array.

Least Number of Unique Integers after K Removals - Complete Solution Guide

Least Number of Unique Integers after K Removals is LeetCode problem 1481, 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

Given an array of integers arr and an integer k . Find the least number of unique integers after removing exactly k elements . Example 1: Input: arr = [5,5,4], k = 1 Output: 1 Explanation : Remove the single 4, only 5 is left. Example 2: Input: arr = [4,3,1,1,3,3,2], k = 3 Output: 2 Explanation : Remove 4, 2 and either one of the two 1s or three 3s. 1 and 3 will be left. Constraints: 1 <= arr.length <= 10^5 1 <= arr[i] <= 10^9 0 <= k <= arr.length

Detailed Explanation

The problem asks us to find the least number of unique integers remaining in an array after removing exactly `k` elements. We are given an array `arr` of integers and an integer `k`. The goal is to minimize the number of distinct elements left in the array after deleting `k` elements in total. We must remove *exactly* `k` elements, no more and no less.

Solution Approach

The solution uses a greedy approach. First, it counts the frequency of each element in the input array using a hash table. Then, it sorts these frequencies in ascending order. Finally, it iterates through the sorted frequencies, removing elements from the array by decrementing `k` by the frequency of each number as long as `k` is greater than or equal to the current frequency. If `k` becomes less than the current frequency, it stops removing elements. The remaining number of frequencies is the least number of unique integers left in the array.

Step-by-Step Algorithm

Step 1: Count the frequency of each element in the input array `arr`. Use a hash table (dictionary) to store the element as a key and its frequency as the value.
Step 2: Extract the frequencies from the hash table and store them in a list (or array).
Step 3: Sort the list of frequencies in ascending order.
Step 4: Iterate through the sorted frequencies. For each frequency, check if `k` is greater than or equal to the frequency. If it is, subtract the frequency from `k` and decrement the number of unique integers (initialized to the size of the frequency list).
Step 5: If `k` becomes less than the current frequency, stop the iteration because we cannot completely remove this element and hence, it will remain unique. Return the remaining number of unique integers.

Key Insights

Insight 1: The key is to remove elements based on their frequency in the array. Removing elements with lower frequencies first is a greedy approach that minimizes the number of unique integers.
Insight 2: Using a hash table (or dictionary) to count the occurrences of each number is efficient for determining the frequency of each element.
Insight 3: Sorting the frequencies allows us to strategically remove elements from the array, prioritizing those that appear least frequently.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Greedy, Sorting, Counting.

Companies

Asked at: Expedia, IBM, J.P. Morgan.

Least Number of Unique Integers after K Removals | LeetCode 1481

Medium

Array Hash Table Greedy Sorting Counting

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Map;

class Solution {
    public int findLeastNumOfUniqueInts(int[] arr, int k) {
        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);
        
        int uniqueCount = frequencies.size();
        
        for (int freq : frequencies) {
            if (k >= freq) {
                k -= freq;
                uniqueCount--;
            } else {
                break;
            }
        }
        
        return uniqueCount;
    }
}

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Problem Info

Description

Problem 1481: Least Number of Unique Integers after K Removals - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy Sorting Counting

Companies

ExpediaIBMJ.P. Morgan

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1481

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

5 steps

1
Step 1: Count the frequency of each element in the input array `arr`. Use a hash table (dictionary) to store the element as a key and its frequency as the value.
2
Step 2: Extract the frequencies from the hash table and store them in a list (or array).
3
Step 3: Sort the list of frequencies in ascending order.
4
Step 4: Iterate through the sorted frequencies. For each frequency, check if `k` is greater than or equal to the frequency. If it is, subtract the frequency from `k` and decrement the number of unique integers (initialized to the size of the frequency list).
5
Step 5: If `k` becomes less than the current frequency, stop the iteration because we cannot completely remove this element and hence, it will remain unique. Return the remaining number of unique integers.

💡

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. Hash table operations (insertion, lookup) are O(1) on average, leading to O(n) overall for frequency counting. Space complexity stems from storing frequency counts and sorted frequencies, both potentially needing space proportional to the number of unique elements in the input array.

📖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

Using a Priority Queue (Heap)

Instead of sorting the frequencies, a min-heap (priority queue) can be used. Insert all frequencies into the heap. Then, repeatedly remove the smallest frequency from the heap and decrement `k` until `k` becomes 0 or the heap is empty. The number of elements remaining in the heap is the least number of unique integers.

Time: O(n log u), where n is the length of the array and u is the number of unique elements in the array. Building heap takes O(n), and popping takes log(u) time.Space: O(u), where u is the number of unique elements in the array, to store the heap.

Trade-offs: The heap-based approach can be slightly more efficient if the number of unique elements is significantly smaller than the total number of elements. However, the difference in performance is usually not significant.

✨

Key Insights

1Insight 1: The key is to remove elements based on their frequency in the array. Removing elements with lower frequencies first is a greedy approach that minimizes the number of unique integers.
2Insight 2: Using a hash table (or dictionary) to count the occurrences of each number is efficient for determining the frequency of each element.
3Insight 3: Sorting the frequencies allows us to strategically remove elements from the array, prioritizing those that appear least frequently.

📚

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: `k` is 0. In this case, no elements are removed, and the number of unique integers remains the same. The algorithm correctly handles this by not entering the removal loop.
Edge case 2: `k` is greater than or equal to the total number of elements in the array. In this case, all elements can be removed, and the number of unique integers becomes 0. The algorithm correctly handles this because the loop continues until either all frequencies are processed or `k` becomes less than a frequency.
Edge case 3: Empty input array: If the array is empty, the number of unique integers is zero. The algorithm handles this correctly, as it initializes the count to 0 and nothing gets counted.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the frequencies. Removing elements randomly or based on the element value instead of frequency will not guarantee the minimal number of unique integers.

✗Common mistake 2: Incorrectly calculating the remaining `k` value. Ensuring that `k` is updated correctly after each removal is crucial for the algorithm's correctness.

✗Common mistake 3: Forgetting to decrement the unique count after successfully removing a frequency. Failing to do so will result in an incorrect answer.

✅

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. Reducing the number of unique values in a dataset can improve compression ratios. The algorithm can identify and remove the least frequent values, resulting in a smaller dataset with minimal information loss.

🏦Real-world scenario 2: Resource allocation. Suppose you have a limited budget (`k`) to maintain different types of items (`arr`). You want to minimize the variety of items you keep. This algorithm helps determine the minimum number of item types you can maintain after spending your budget to get rid of the least frequent items first.

🎮Real-world scenario 3: Recommendation systems: Filtering out less popular items to focus on a smaller, more relevant set of recommendations. The 'k removals' represent a constraint on the number of items that can be excluded.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: July 12, 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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Least Number of Unique Integers after K Removals"?

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.

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