What is the time complexity of Minimum Removals to Balance Array?

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

What data structures are used to solve Minimum Removals to Balance Array?

This problem uses Arrays, Sorting, Two Pointers, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Removals to Balance Array?

Insight 1: Sorting the array allows us to efficiently check the balancing condition. With a sorted array, we can easily find potential minimums and maximums for our subsequence. Insight 2: The core idea is to find the longest balanced subsequence. The number of elements to remove is simply the original array's length minus the length of this subsequence.

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Minimum Removals to Balance Array - LeetCode 3634 Solution

Minimum Removals to Balance Array - Complete Solution Guide

Minimum Removals to Balance Array is LeetCode problem 3634, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Framing

Minimum Removals to Balance Array is a Medium LeetCode problem that rewards careful tracing, edge-case handling, and a clear grasp of Arrays and Sorting. The best solutions usually explain why the chosen invariant holds before they optimize for time or space.

Quick Example Mindset

A useful way to test Minimum Removals to Balance Array is to start with a tiny input that exposes the boundary conditions, then run the same logic on a slightly larger case to verify the arrays behavior and the sorting interaction. That second pass is where off-by-one mistakes and missing updates usually appear.

Problem Statement

You are given an integer array nums and an integer k . An array is considered balanced if the value of its maximum element is at most k times the minimum element. You may remove any number of elements from nums without making it empty . Return the minimum number of elements to remove so that the remaining array is balanced. Note: An array of size 1 is considered balanced as its maximum and minimum are equal, and the condition always holds true. Example 1: Input: nums = [2,1,5], k = 2 Out

Detailed Explanation

The problem asks us to find the minimum number of elements to remove from an array `nums` such that the remaining array is 'balanced'. An array is balanced if its maximum element is at most `k` times its minimum element. We need to find the longest balanced subsequence and return the number of elements not in that subsequence. A subsequence can be formed by removing elements, but not changing the order of the remaining elements. A single element array is considered balanced.

Solution Approach

The provided Python code utilizes a sorting and two-pointer approach. First, the array is sorted in ascending order. Then, the code iterates through the sorted array using a sliding window. For each element at index `l` (potential minimum), it expands the window to the right (index `r`) to find the farthest element that still satisfies the balanced condition (nums[r] <= nums[l] * k). The length of this valid window (r - l + 1) represents the length of a possible balanced subsequence starting with nums[l]. The algorithm keeps track of the maximum length found so far. Finally, it calculates the minimum number of elements to remove by subtracting the maximum length from the original array's length.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order.
Step 2: Initialize `max_len` to 0, which will store the maximum length of a balanced subsequence found so far.
Step 3: Initialize a left pointer `l` to 0.
Step 4: Iterate through the array with a right pointer `r` from 0 to n-1.
Step 5: Inside the outer loop, check if the current element `nums[r]` is greater than `nums[l] * k`. If it is, increment the left pointer `l` until the condition `nums[r] <= nums[l] * k` is met. This maintains the sliding window representing the longest balanced subsequence for the given minimum `nums[l]`.
Step 6: Update `max_len` with the maximum of its current value and the length of the current valid window (`r - l + 1`).
Step 7: After the outer loop completes, return `n - max_len`, which represents the minimum number of elements to remove to balance the array.

Key Insights

Insight 1: Sorting the array allows us to efficiently check the balancing condition. With a sorted array, we can easily find potential minimums and maximums for our subsequence.
Insight 2: The core idea is to find the longest balanced subsequence. The number of elements to remove is simply the original array's length minus the length of this subsequence.
Insight 3: We can use a sliding window approach (two pointers) to efficiently determine the longest balanced subsequence for each potential minimum.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Arrays, Sorting, Two Pointers, Greedy.

Study Paths

Continue from this problem into the surrounding topic and company clusters to compare how the same pattern appears in other interview settings.

Related topics: Arrays, Sorting, Two Pointers, Greedy

Frequently Asked Questions

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 Removals to Balance Array"?

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

Minimum Removals to Balance Array | LeetCode 3634

Medium

Arrays Sorting Two Pointers Greedy

Study Path

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Arrays Sorting Two Pointers Greedy

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int minRemoval(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        int maxLen = 0;
        for (int i = 0; i < n; i++) {
            int count = 1;
            for (int j = i + 1; j < n; j++) {
                if ((long)nums[j] <= (long)nums[i] * k) {
                    count++;
                }
            }
            maxLen = Math.max(maxLen, count);
        }
        return n - maxLen;
    }
}

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Problem Info

Description

Problem 3634: Minimum Removals to Balance Array - Detailed explanation and solution approaches.

Topics

Arrays Sorting Two Pointers Greedy

Stats

Difficulty:Medium

Languages:4

Problem #:3634

Problem Explanation

🧭

Next Steps

Use this problem as a starting point, then move into adjacent patterns and company sets to reinforce the same idea in different contexts.

Topic hub

Study neighboring problems by pattern and difficulty.

Company hub

See where this problem appears in interview loops.

Pattern guides

Read deeper walkthroughs of the algorithm family.

💡

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: Sort the input array `nums` in ascending order.
2
Step 2: Initialize `max_len` to 0, which will store the maximum length of a balanced subsequence found so far.
3
Step 3: Initialize a left pointer `l` to 0.
4
Step 4: Iterate through the array with a right pointer `r` from 0 to n-1.
5
Step 5: Inside the outer loop, check if the current element `nums[r]` is greater than `nums[l] * k`. If it is, increment the left pointer `l` until the condition `nums[r] <= nums[l] * k` is met. This maintains the sliding window representing the longest balanced subsequence for the given minimum `nums[l]`.
6
Step 6: Update `max_len` with the maximum of its current value and the length of the current valid window (`r - l + 1`).
7
Step 7: After the outer loop completes, return `n - max_len`, which represents the minimum number of elements to remove to balance the array.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is sorting the array, which takes O(n log n) time. The two-pointer approach takes O(n) time in the worst case, as each pointer can traverse the array at most once. Therefore, the overall time complexity is O(n log n + n) = O(n log n). The algorithm uses only a constant amount of extra space for variables, resulting in O(1) space complexity.

📖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

Binary Search

Instead of a sliding window, we can use binary search to find the rightmost index `r` such that `nums[r] <= nums[l] * k` for each `l`. This would still require sorting the array first.

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

Trade-offs: Binary search approach has the same time complexity as the sliding window, but it might be slightly less efficient in practice due to the overhead of binary search for each left pointer position.

✨

Key Insights

1Insight 1: Sorting the array allows us to efficiently check the balancing condition. With a sorted array, we can easily find potential minimums and maximums for our subsequence.
2Insight 2: The core idea is to find the longest balanced subsequence. The number of elements to remove is simply the original array's length minus the length of this subsequence.
3Insight 3: We can use a sliding window approach (two pointers) to efficiently determine the longest balanced subsequence for each potential minimum.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

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: Empty or single-element array. The code implicitly handles the single element case since sorting a single element array and returning `n - 1 = 0` if `max_len` remains 1 is correct. If the array is empty, the input constraint rules this out.
Edge case 2: All elements are the same. The algorithm correctly identifies the entire array as balanced since the sorted array will not require shifting the left pointer during the two pointer iteration.
Edge case 3: Large values of k. The code handles large values of k correctly as the comparison uses standard integer multiplication and comparison operators.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the array. The algorithm relies on the array being sorted to efficiently find the maximum length of a balanced subsequence.

✗Common mistake 2: Integer overflow. When calculating `nums[l] * k`, there's a risk of integer overflow if `nums[l]` and `k` are large. The Java solution mitigates this with casting to `long`.

✗Common mistake 3: Incorrectly calculating the length of the balanced subsequence. Ensure the length is `r - l + 1` to include both `nums[l]` and `nums[r]` in the subsequence.

✅

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 filtering. Consider a dataset of sensor readings, where you want to filter out outliers. The balancing condition can be used to identify and remove data points that are significantly different from the rest of the data.

🏦Real-world scenario 2: Resource allocation. Suppose you have a set of tasks with varying resource requirements. You want to select a subset of tasks that can be efficiently handled with a limited amount of resources, where the maximum resource requirement is at most k times the minimum requirement. This problem maps directly to the 'balanced' array scenario.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 5, 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

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

@TechLeadJul 21

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

How should I practice algorithm problems effectively?

What is the optimal approach for "Minimum Removals to Balance Array"?

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