What is the time complexity of Minimum Array Length After Pair Removals?

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

What data structures are used to solve Minimum Array Length After Pair Removals?

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

What are the key insights for solving Minimum Array Length After Pair Removals?

Insight 1: Since the array is sorted, we should try to pair the smallest elements with larger elements to maximize removals. Insight 2: A greedy approach of pairing elements from the first half of the array with elements from the second half is optimal.

Which companies ask LeetCode 2856. Minimum Array Length After Pair Removals in interviews?

This problem is frequently asked at Snowflake.

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Minimum Array Length After Pair Removals - LeetCode 2856 Solution

Q: How to solve Minimum Array Length After Pair Removals?

The provided solution employs a greedy approach. It divides the array conceptually into two halves. It then iterates through the first half (starting from index 0) and the second half (starting from index `(n + 1) // 2`). For each element in the first half, it tries to find a corresponding element in the second half that is strictly greater. If such a pair is found, it increments the pair count and moves to the next elements in both halves. If no suitable element is found in the second half for the current element in the first half, it skips that element in the first half and moves to the next element in the second half. The final array length is the original length minus twice the number of pairs found.

Minimum Array Length After Pair Removals - Complete Solution Guide

Minimum Array Length After Pair Removals is LeetCode problem 2856, 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 integer array num sorted in non-decreasing order. You can perform the following operation any number of times: Choose two indices, i and j , where nums[i] < nums[j] . Then, remove the elements at indices i and j from nums . The remaining elements retain their original order, and the array is re-indexed. Return the minimum length of nums after applying the operation zero or more times. Example 1: Input: nums = [1,2,3,4] Output: 0 Explanation: Example 2: Input: nums = [1,1,2,2,3,3] Output

Detailed Explanation

The problem asks us to find the minimum possible length of an integer array `nums` after repeatedly removing pairs of elements. The crucial condition for removing a pair is that `nums[i] < nums[j]`, where `i` and `j` are the indices of the elements being removed. The array `nums` is sorted in non-decreasing order. The goal is to maximize the number of pairs removed.

Solution Approach

The provided solution employs a greedy approach. It divides the array conceptually into two halves. It then iterates through the first half (starting from index 0) and the second half (starting from index `(n + 1) // 2`). For each element in the first half, it tries to find a corresponding element in the second half that is strictly greater. If such a pair is found, it increments the pair count and moves to the next elements in both halves. If no suitable element is found in the second half for the current element in the first half, it skips that element in the first half and moves to the next element in the second half. The final array length is the original length minus twice the number of pairs found.

Step-by-Step Algorithm

Step 1: Initialize `n` to the length of the input array `nums`.
Step 2: Initialize `i` to 0, representing the index for the first half of the array.
Step 3: Initialize `j` to `(n + 1) // 2`, representing the starting index for the second half of the array.
Step 4: Initialize `pairs` to 0, representing the number of successful pairs removed.
Step 5: Iterate while `i < n // 2` and `j < n`.
Step 6: Inside the loop, check if `nums[i] < nums[j]`. If it is, increment `pairs`, `i`, and `j`.
Step 7: If `nums[i] >= nums[j]`, only increment `j` (attempt to find a larger value to pair with nums[i]).
Step 8: After the loop finishes, return `n - 2 * pairs`.

Key Insights

Insight 1: Since the array is sorted, we should try to pair the smallest elements with larger elements to maximize removals.
Insight 2: A greedy approach of pairing elements from the first half of the array with elements from the second half is optimal.
Insight 3: We can count the number of successful pairs, and the final length will be the initial length minus twice the number of pairs.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Two Pointers, Binary Search, Greedy, Counting.

Companies

Asked at: Snowflake.

Minimum Array Length After Pair Removals | LeetCode 2856

Medium

Array Hash Table Two Pointers Binary Search Greedy Counting

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

Solution Code

Available in:

import java.util.List;

class Solution {
    public int minLengthAfterRemovals(List<Integer> nums) {
        int n = nums.size();
        
        int i = 0;
        int j = (n + 1) / 2;
        
        int pairs = 0;
        
        while (i < n / 2 && j < n) {
            if (nums.get(i) < nums.get(j)) {
                pairs += 1;
                i += 1;
                j += 1;
            } else {
                j += 1;
            }
        }
        
        return n - 2 * pairs;
    }
}

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Problem Info

Description

Problem 2856: Minimum Array Length After Pair Removals - Detailed explanation and solution approaches.

Topics

Array Hash Table Two Pointers Binary Search Greedy Counting

Companies

Snowflake

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2856

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

8 steps

1
Step 1: Initialize `n` to the length of the input array `nums`.
2
Step 2: Initialize `i` to 0, representing the index for the first half of the array.
3
Step 3: Initialize `j` to `(n + 1) // 2`, representing the starting index for the second half of the array.
4
Step 4: Initialize `pairs` to 0, representing the number of successful pairs removed.
5
Step 5: Iterate while `i < n // 2` and `j < n`.
6
Step 6: Inside the loop, check if `nums[i] < nums[j]`. If it is, increment `pairs`, `i`, and `j`.
7
Step 7: If `nums[i] >= nums[j]`, only increment `j` (attempt to find a larger value to pair with nums[i]).
8
Step 8: After the loop finishes, return `n - 2 * pairs`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the while loop. In the worst case, the `j` pointer will traverse the entire second half of the array for each element in the first half. However, both `i` and `j` are incrementing which means the runtime is O(n). The space complexity is constant because only a fixed number of variables (`n`, `i`, `j`, `pairs`) are used irrespective of the input array size.

📖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 HashMap to count frequencies

Count the frequency of each number in the array. Then the problem becomes finding out how many of the most frequent element you have, and comparing that to the remaining elements.

Time: O(n)Space: O(n) in the worst case if all numbers are distinct. O(k) where k is the number of unique elements

Trade-offs: If the array is not sorted, this might be simpler to code. However, it requires more memory.

✨

Key Insights

1Insight 1: Since the array is sorted, we should try to pair the smallest elements with larger elements to maximize removals.
2Insight 2: A greedy approach of pairing elements from the first half of the array with elements from the second half is optimal.
3Insight 3: We can count the number of successful pairs, and the final length will be the initial length minus twice the number of pairs.

📚

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!

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `nums` has length 1. The result should be 1, since no pairs can be removed. The solution correctly handles this since `i < n // 2` evaluates to false immediately and returns n - 2*0 = n = 1.
Edge case 2: `nums` has length 2 and `nums[0] >= nums[1]`. The result should be 2, since no pairs can be removed. The solution correctly handles this since `nums[i] < nums[j]` is false in the first comparison and returns n - 2 * 0 = n = 2.
Edge case 3: `nums` has all identical values. The result should be n. The solution correctly handles this, because no pair will be found.
Edge case 4: `nums` has only two elements and `nums[0] < nums[1]`. Result should be 0. Correctly handled as `pairs` becomes 1, and returned result is n-2 or 2-2 = 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `n` is odd correctly when calculating the starting index `j` for the second half. Using integer division `(n+1)//2` ensures the correct starting point.

✗Common mistake 2: Assuming the two halves of the array must be of equal size. When `n` is odd, the second half is larger. The loop conditions need to handle this difference.

✗Common mistake 3: Not realizing the problem can be solved greedily. Trying to use dynamic programming or recursion would lead to a more complex and less efficient solution.

✅

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: Resource allocation: You have a sorted list of tasks with different priorities (represented by numbers). You want to pair lower-priority tasks with higher-priority tasks to optimize resource usage. The minimum remaining tasks after this allocation could be relevant.

🏦Real-world scenario 2: Data compression: You have a sorted list of data elements. You want to pair smaller elements with larger ones to reduce the overall size of the dataset based on some predefined rule.

💡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: November 4, 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 13

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

@InterviewProJul 6

❓

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 "Minimum Array Length After Pair 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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