What is the time complexity of Make Array Empty?

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

What data structures are used to solve Make Array Empty?

This problem uses Array, Binary Search, Greedy, Binary Indexed Tree, Segment Tree, Sorting, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Make Array Empty?

Insight 1: The core of the problem lies in efficiently tracking the positions of the elements as they are being removed or shifted within the array. Insight 2: Using sorting to determine the order in which elements are removed is crucial. However, directly simulating array rotations is inefficient.

Which companies ask LeetCode 2659. Make Array Empty in interviews?

This problem is frequently asked at Zepto.

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Make Array Empty - LeetCode 2659 Solution

Make Array Empty - Complete Solution Guide

Make Array Empty is LeetCode problem 2659, a Hard 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 containing distinct numbers, and you can perform the following operations until the array is empty : If the first element has the smallest value, remove it Otherwise, put the first element at the end of the array. Return an integer denoting the number of operations it takes to make nums empty. Example 1: Input: nums = [3,4,-1] Output: 5 Operation Array 1 [4, -1, 3] 2 [-1, 3, 4] 3 [3, 4] 4 [4] 5 [] Example 2: Input: nums = [1,2,4,3] Output: 5 Operation Array 1

Detailed Explanation

The problem presents a scenario where you're given an array of distinct integers, and you need to determine the number of operations to empty the array following a specific rule. If the first element of the array is the smallest, you remove it. Otherwise, you move it to the end of the array. The objective is to count the total operations required to completely empty the array.

Solution Approach

The solution leverages a Binary Indexed Tree (BIT) to keep track of the elements that have been removed from the array. First, we sort the array to determine the order in which the elements will be removed. Then, for each element in the sorted array, we determine its original position in the input array. We then calculate the number of moves it takes to bring that element to the front. The BIT is used to track which elements have been removed from the array. When considering a new position, we can use the BIT to calculate the number of already removed elements that are before the new position, thus helping to calculate the number of moves required.

Step-by-Step Algorithm

Step 1: Create a map to store the original index (position) of each number in the input array 'nums'.
Step 2: Sort the input array 'nums' and store it as `sorted_nums` to determine the order of removal.
Step 3: Initialize a Binary Indexed Tree (BIT) of the same size as the input array to keep track of which elements have been removed.
Step 4: Iterate through the `sorted_nums`. For each number, retrieve its original position (curr_pos) from the position map.
Step 5: Determine the number of moves required to bring the element at `curr_pos` to the front. If `curr_pos > last_pos`, the target is to remove all elements between `last_pos` and `curr_pos`. Otherwise, all elements after last_pos and before curr_pos must be moved before the element at curr_pos can be moved.
Step 6: Use BIT queries to determine how many of the elements in those ranges were already removed, and calculate the number of moves accordingly.
Step 7: Add 1 to the total number of moves for removing the element at the front.
Step 8: Update the BIT to mark the current element's position as removed.
Step 9: Update `last_pos` with the current element's position `curr_pos` for the next iteration.
Step 10: Return the total number of operations.

Key Insights

Insight 1: The core of the problem lies in efficiently tracking the positions of the elements as they are being removed or shifted within the array.
Insight 2: Using sorting to determine the order in which elements are removed is crucial. However, directly simulating array rotations is inefficient.
Insight 3: A Binary Indexed Tree (BIT), also known as a Fenwick Tree, is an ideal data structure to track removed elements and efficiently query the number of elements removed before a given position.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Greedy, Binary Indexed Tree, Segment Tree, Sorting, Ordered Set.

Companies

Asked at: Zepto.

Make Array Empty | LeetCode 2659

Hard

Array Binary Search Greedy Binary Indexed Tree Segment Tree Sorting Ordered Set

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

class Solution {
    public long countOperationsToEmptyArray(int[] nums) {
        int n = nums.length;
        Map<Integer, Integer> pos = new HashMap<>();
        for (int i = 0; i < n; i++) {
            pos.put(nums[i], i);
        }

        int[] bit = new int[n + 1];

        // Function to update the Binary Indexed Tree (BIT)
        class BIT {
            public void update(int idx, int delta) {
                idx += 1;
                while (idx <= n) {
                    bit[idx] += delta;
                    idx += idx & -idx;
                }
            }

            // Function to query the sum of elements up to index idx
            public int query(int idx) {
                if (idx < 0) {
                    return 0;
                }
                idx += 1;
                int s = 0;
                while (idx > 0) {
                    s += bit[idx];
                    idx -= idx & -idx;
                }
                return s;
            }
        }
        BIT bitObj = new BIT();

        long total_ops = 0;
        int last_pos = -1;

        int[] sorted_nums = Arrays.copyOf(nums, nums.length);
        Arrays.sort(sorted_nums);

        for (int i = 0; i < n; i++) {
            int curr_pos = pos.get(sorted_nums[i]);

            if (curr_pos > last_pos) {
                int num_elements_between = curr_pos - last_pos - 1;
                int removed_between = bitObj.query(curr_pos - 1) - bitObj.query(last_pos);
                long moves = num_elements_between - removed_between;
                total_ops += moves + 1;
            } else {
                long rem_after = (n - 1 - last_pos) - (bitObj.query(n - 1) - bitObj.query(last_pos));
                long rem_before = curr_pos - bitObj.query(curr_pos - 1);
                long moves = rem_after + rem_before;
                total_ops += moves + 1;
            }

            bitObj.update(curr_pos, 1);
            last_pos = curr_pos;
        }

        return total_ops;
    }
}

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Problem Info

Description

Problem 2659: Make Array Empty - Detailed explanation and solution approaches.

Topics

Array Binary Search Greedy Binary Indexed Tree Segment Tree Sorting Ordered Set

Companies

Zepto

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2659

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

10 steps

1
Step 1: Create a map to store the original index (position) of each number in the input array 'nums'.
2
Step 2: Sort the input array 'nums' and store it as `sorted_nums` to determine the order of removal.
3
Step 3: Initialize a Binary Indexed Tree (BIT) of the same size as the input array to keep track of which elements have been removed.
4
Step 4: Iterate through the `sorted_nums`. For each number, retrieve its original position (curr_pos) from the position map.
5
Step 5: Determine the number of moves required to bring the element at `curr_pos` to the front. If `curr_pos > last_pos`, the target is to remove all elements between `last_pos` and `curr_pos`. Otherwise, all elements after last_pos and before curr_pos must be moved before the element at curr_pos can be moved.
6
Step 6: Use BIT queries to determine how many of the elements in those ranges were already removed, and calculate the number of moves accordingly.
7
Step 7: Add 1 to the total number of moves for removing the element at the front.
8
Step 8: Update the BIT to mark the current element's position as removed.
9
Step 9: Update `last_pos` with the current element's position `curr_pos` for the next iteration.
10
Step 10: Return the total number of operations.

💡

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?

Sorting the array takes O(n log n) time. The loop iterates through the sorted array, which is O(n). Inside the loop, BIT operations (update and query) take O(log n) time. Therefore, the overall time complexity is O(n log n). The space complexity is O(n) because we store the position map, the sorted array, and the BIT, each requiring O(n) space.

📖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

Segment Tree

A segment tree can also be used in place of a binary indexed tree to keep track of the removals. It can perform range queries and updates, allowing for the efficient calculation of elements removed between positions.

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

Trade-offs: Segment trees generally have a larger constant factor than BITs and can be more complex to implement, but they also offer greater flexibility in terms of queries that can be performed.

✨

Key Insights

1Insight 1: The core of the problem lies in efficiently tracking the positions of the elements as they are being removed or shifted within the array.
2Insight 2: Using sorting to determine the order in which elements are removed is crucial. However, directly simulating array rotations is inefficient.
3Insight 3: A Binary Indexed Tree (BIT), also known as a Fenwick Tree, is an ideal data structure to track removed elements and efficiently query the number of elements removed before a given position.

📚

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!

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: Input array is already sorted in ascending order. In this case, each element is removed from the front, so the answer will be equal to the number of elements in the array.
Edge case 2: Input array is sorted in descending order. In this case, we have to rotate the array multiple times until the smallest element comes to the front. The BIT helps in efficiently calculating how many rotations are necessary.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Directly simulating the array operations can lead to a time complexity of O(n^2), which will result in a 'Time Limit Exceeded' error for larger input sizes.

✗Common mistake 2: Forgetting to update the Binary Indexed Tree after removing an element will result in incorrect move counts and ultimately a wrong answer.

✗Common mistake 3: Not handling the case where `curr_pos` is smaller than `last_pos` correctly, leading to an incorrect calculation of moves across the boundary of the array.

✅

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: Task scheduling where tasks need to be executed in a specific order based on priority, and the system must efficiently track completed tasks to schedule the next ones.

🏦Real-world scenario 2: Data stream processing where you need to efficiently determine the rank of an element or the number of elements less than or equal to a given value within a sliding window.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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 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

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

@TechLeadJul 14

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

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 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.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Make Array Empty"?

For this hard-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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