What is the time complexity of Sorting Three Groups?

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

What data structures are used to solve Sorting Three Groups?

This problem uses Array, Binary Search, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Sorting Three Groups?

Insight 1: The problem can be solved by finding the longest non-decreasing subsequence of the input array. The minimum number of removals is then the length of the input array minus the length of the longest non-decreasing subsequence. Insight 2: Dynamic programming can be used to efficiently calculate the length of the longest non-decreasing subsequence. We only need to keep track of the longest non-decreasing subsequence ending with 1, 2, or 3.

Which companies ask LeetCode 2826. Sorting Three Groups in interviews?

This problem is frequently asked at UiPath.

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Sorting Three Groups - LeetCode 2826 Solution

Sorting Three Groups - Complete Solution Guide

Sorting Three Groups is LeetCode problem 2826, 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 nums . Each element in nums is 1, 2 or 3. In each operation, you can remove an element from nums . Return the minimum number of operations to make nums non-decreasing . Example 1: Input: nums = [2,1,3,2,1] Output: 3 Explanation: One of the optimal solutions is to remove nums[0] , nums[2] and nums[3] . Example 2: Input: nums = [1,3,2,1,3,3] Output: 2 Explanation: One of the optimal solutions is to remove nums[1] and nums[2] . Example 3: Input: nums = [2,2,2,2,3,3] O

Detailed Explanation

The problem asks us to find the minimum number of operations (removals) needed to make a given array `nums` non-decreasing. The array `nums` contains only the numbers 1, 2, and 3. A non-decreasing array means that each element is greater than or equal to the element before it. The input is an integer array `nums`, and the output is the minimum number of removals needed to make the array non-decreasing.

Solution Approach

The solution uses dynamic programming to find the length of the longest non-decreasing subsequence of the input array. It maintains three variables: `dp1`, `dp2`, and `dp3`. `dp1` stores the length of the longest non-decreasing subsequence ending with 1, `dp2` stores the length of the longest non-decreasing subsequence ending with 2, and `dp3` stores the length of the longest non-decreasing subsequence ending with 3. For each element `x` in the input array, we update the values of `dp1`, `dp2`, and `dp3` based on the value of `x`. Finally, the maximum of `dp1`, `dp2`, and `dp3` gives the length of the longest non-decreasing subsequence, and the minimum number of removals is the length of the input array minus this value.

Step-by-Step Algorithm

Step 1: Initialize `dp1`, `dp2`, and `dp3` to 0. These variables will store the lengths of the longest non-decreasing subsequences ending with 1, 2, and 3, respectively.
Step 2: Iterate through the input array `nums`. For each element `x`:
Step 3: If `x` is 1, increment `dp1` by 1.
Step 4: If `x` is 2, update `dp2` to `max(dp1, dp2) + 1`. This is because if we append 2 to a subsequence, we can either append it to the longest subsequence ending with 1, or the longest subsequence ending with 2.
Step 5: If `x` is 3, update `dp3` to `max(dp1, dp2, dp3) + 1`. This is because if we append 3 to a subsequence, we can either append it to the longest subsequence ending with 1, 2 or 3.
Step 6: After iterating through the entire array, find the maximum of `dp1`, `dp2`, and `dp3`. This represents the length of the longest non-decreasing subsequence.
Step 7: Return the length of the input array minus the maximum of `dp1`, `dp2`, and `dp3`. This is the minimum number of removals needed to make the array non-decreasing.

Key Insights

Insight 1: The problem can be solved by finding the longest non-decreasing subsequence of the input array. The minimum number of removals is then the length of the input array minus the length of the longest non-decreasing subsequence.
Insight 2: Dynamic programming can be used to efficiently calculate the length of the longest non-decreasing subsequence. We only need to keep track of the longest non-decreasing subsequence ending with 1, 2, or 3.
Insight 3: Since the input array only contains 1, 2, and 3, we can use a constant amount of space to store the lengths of the longest non-decreasing subsequences.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Dynamic Programming.

Companies

Asked at: UiPath.

Sorting Three Groups | LeetCode 2826

Medium

Array Binary Search Dynamic Programming

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

Solution Code

Available in:

import java.util.List;

class Solution {
    public int minimumOperations(List<Integer> nums) {
        int n = nums.size();
        int dp1 = 0;
        int dp2 = 0;
        int dp3 = 0;

        for (int x : nums) {
            if (x == 1) {
                dp1 += 1;
            } else if (x == 2) {
                dp2 = Math.max(dp1, dp2) + 1;
            } else {
                dp3 = Math.max(Math.max(dp1, dp2), dp3) + 1;
            }
        }

        int max_kept = Math.max(Math.max(dp1, dp2), dp3);

        return n - max_kept;
    }
}

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Problem Info

Description

Problem 2826: Sorting Three Groups - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming

Companies

UiPath

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2826

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 `dp1`, `dp2`, and `dp3` to 0. These variables will store the lengths of the longest non-decreasing subsequences ending with 1, 2, and 3, respectively.
2
Step 2: Iterate through the input array `nums`. For each element `x`:
3
Step 3: If `x` is 1, increment `dp1` by 1.
4
Step 4: If `x` is 2, update `dp2` to `max(dp1, dp2) + 1`. This is because if we append 2 to a subsequence, we can either append it to the longest subsequence ending with 1, or the longest subsequence ending with 2.
5
Step 5: If `x` is 3, update `dp3` to `max(dp1, dp2, dp3) + 1`. This is because if we append 3 to a subsequence, we can either append it to the longest subsequence ending with 1, 2 or 3.
6
Step 6: After iterating through the entire array, find the maximum of `dp1`, `dp2`, and `dp3`. This represents the length of the longest non-decreasing subsequence.
7
Step 7: Return the length of the input array minus the maximum of `dp1`, `dp2`, and `dp3`. This is the minimum number of removals needed to make the array non-decreasing.

💡

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 algorithm iterates through the input array `nums` only once. The number of operations in the loop is constant per iteration. Therefore the time complexity is O(n). We are using only a few integer variables, so the space complexity is constant, or O(1).

📖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 Approach

Instead of the dp solution, one can solve using binary search on tails of increasing subsequence. This would have same O(N) time complexity given the numbers are only 1, 2, and 3.

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

Trade-offs: Potentially more complex to implement than the DP solution. The DP solution is straightforward and easy to understand for this specific problem due to the limited range of numbers.

✨

Key Insights

1Insight 1: The problem can be solved by finding the longest non-decreasing subsequence of the input array. The minimum number of removals is then the length of the input array minus the length of the longest non-decreasing subsequence.
2Insight 2: Dynamic programming can be used to efficiently calculate the length of the longest non-decreasing subsequence. We only need to keep track of the longest non-decreasing subsequence ending with 1, 2, or 3.
3Insight 3: Since the input array only contains 1, 2, and 3, we can use a constant amount of space to store the lengths of the longest non-decreasing subsequences.

📚

Key Terms Explained

For Students

Array

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

Binary Search

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array. The code handles this case correctly because the loop will not execute, and the function will return 0.
Edge case 2: Input array is already non-decreasing (e.g., [1, 2, 3]). The code will correctly calculate the length of the longest non-decreasing subsequence, which will be equal to the length of the input array, and the function will return 0.
Edge case 3: Input array contains only one element. The loop executes once, the relevant `dp` value is incremented, and the function returns 0, which is correct.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering all possible cases when updating `dp2` and `dp3`. For example, when `x` is 2, it's important to take the maximum of `dp1` and `dp2` before adding 1.

✗Common mistake 2: Incorrectly calculating the minimum number of removals. The minimum number of removals is the length of the input array minus the length of the longest non-decreasing subsequence, *not* the length of the longest non-decreasing subsequence itself.

✗Common mistake 3: Confusing longest non-decreasing subsequence with longest increasing subsequence. The problem requires a non-decreasing sequence, where elements can be equal.

✅

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: Optimizing resource allocation. Suppose you have a sequence of tasks with priorities 1, 2, and 3. You want to execute the tasks in a non-decreasing order of priority to minimize disruptions. This problem can help you determine the minimum number of tasks that need to be rescheduled.

🏦Real-world scenario 2: Data cleaning and preprocessing. In some datasets, you might have categorical features with ordinal values (e.g., low, medium, high, represented as 1, 2, 3). This problem can be used to identify and correct inconsistencies or errors in the ordering of these values.

💡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: October 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

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.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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 "Sorting Three Groups"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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