What is the time complexity of Distribute Elements Into Two Arrays II?

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

What data structures are used to solve Distribute Elements Into Two Arrays II?

This problem uses Array, Binary Indexed Tree, Segment Tree, Simulation. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Distribute Elements Into Two Arrays II?

Insight 1: Using Binary Indexed Tree (BIT) or Segment Tree is crucial for efficiently calculating the `greaterCount` in each array during each iteration. A brute-force approach of iterating each array for every element would be too slow, leading to TLE. Insight 2: Rank mapping can efficiently map the input values to a smaller range (1 to m, where m is the number of unique elements). This allows us to use BIT which operates on index and not the direct values, which can be very large (up to 10^9)

Which companies ask LeetCode 3072. Distribute Elements Into Two Arrays II in interviews?

This problem is frequently asked at Autodesk.

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Distribute Elements Into Two Arrays II - LeetCode 3072 Solution

Distribute Elements Into Two Arrays II - Complete Solution Guide

Distribute Elements Into Two Arrays II is LeetCode problem 3072, 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 a 1-indexed array of integers nums of length n . We define a function greaterCount such that greaterCount(arr, val) returns the number of elements in arr that are strictly greater than val . You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1 . In the second operation, append nums[2] to arr2 . Afterwards, in the i th operation: If greaterCount(arr1, nums[i]) > greaterCount(arr2, nums[i])

Detailed Explanation

The problem requires us to distribute elements from an input array `nums` into two arrays, `arr1` and `arr2`, based on the count of elements greater than the current element in each array. The distribution process involves comparing `greaterCount(arr1, nums[i])` and `greaterCount(arr2, nums[i])` and appending `nums[i]` to either `arr1` or `arr2` according to the given rules. If the counts are equal, the element is appended to the array with fewer elements. In case of a tie in both counts and lengths, the element goes to `arr1`. Finally, the `result` array is the concatenation of `arr1` and `arr2`.

Solution Approach

The solution uses Binary Indexed Trees (BITs) to efficiently compute `greaterCount`. It first maps the input numbers to their ranks among the unique numbers. Two BITs, `bit1` and `bit2`, are used to maintain the cumulative counts of elements in `arr1` and `arr2`, respectively. For each number in `nums`, the rank is obtained, and the counts of numbers greater than that number in `arr1` and `arr2` are computed using the BITs. The current number is then appended to the appropriate array based on the comparison of these counts and the lengths of the arrays, and the corresponding BIT is updated.

Step-by-Step Algorithm

Step 1: Extract unique elements from `nums`, sort them, and create a rank map to map each element to its rank (index + 1) in the sorted list.
Step 2: Initialize two Binary Indexed Trees (BITs), `bit1` and `bit2`, with size m+1 where 'm' is number of unique elements, to keep track of the presence of numbers in `arr1` and `arr2` respectively.
Step 3: Initialize `arr1` with `nums[0]` and update `bit1` with the rank of `nums[0]`.
Step 4: Initialize `arr2` with `nums[1]` and update `bit2` with the rank of `nums[1]`.
Step 5: Iterate through the remaining elements of `nums` starting from index 2.
Step 6: For each element `nums[i]`, get its rank from the rank map.
Step 7: Calculate `count1` = number of elements in `arr1` greater than `nums[i]` using `bit1`. This can be done by querying number of elements less than or equals to `nums[i]` and subtracting it from the `arr1.size()`.
Step 8: Calculate `count2` = number of elements in `arr2` greater than `nums[i]` using `bit2`.
Step 9: Compare `count1` and `count2`: if `count1 > count2`, append `nums[i]` to `arr1` and update `bit1`; if `count2 > count1`, append `nums[i]` to `arr2` and update `bit2`; otherwise, compare the lengths of `arr1` and `arr2` and append to the smaller array or `arr1` if lengths are equal and update the appropriate BIT.
Step 10: After the loop, concatenate `arr1` and `arr2` to form the `result` array.

Key Insights

Insight 1: Using Binary Indexed Tree (BIT) or Segment Tree is crucial for efficiently calculating the `greaterCount` in each array during each iteration. A brute-force approach of iterating each array for every element would be too slow, leading to TLE.
Insight 2: Rank mapping can efficiently map the input values to a smaller range (1 to m, where m is the number of unique elements). This allows us to use BIT which operates on index and not the direct values, which can be very large (up to 10^9)
Insight 3: Careful consideration of all the tie-breaking conditions is crucial to get the correct output. Pay close attention to length comparisons and initial array population.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Indexed Tree, Segment Tree, Simulation.

Companies

Asked at: Autodesk.

Distribute Elements Into Two Arrays II | LeetCode 3072

Hard

Array Binary Indexed Tree Segment Tree Simulation

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] resultArray(int[] nums) {
        Set<Integer> uniqueNumsSet = new HashSet<>();
        for (int num : nums) {
            uniqueNumsSet.add(num);
        }

        List<Integer> sortedUniqueNums = new ArrayList<>(uniqueNumsSet);
        Collections.sort(sortedUniqueNums);

        int m = sortedUniqueNums.size();
        Map<Integer, Integer> rankMap = new HashMap<>();
        for (int i = 0; i < m; i++) {
            rankMap.put(sortedUniqueNums.get(i), i + 1);
        }

        int[] bit1 = new int[m + 1];
        int[] bit2 = new int[m + 1];

        List<Integer> arr1 = new ArrayList<>();
        List<Integer> arr2 = new ArrayList<>();

        arr1.add(nums[0]);
        update(bit1, rankMap.get(nums[0]), m);

        arr2.add(nums[1]);
        update(bit2, rankMap.get(nums[1]), m);

        for (int i = 2; i < nums.length; i++) {
            int val = nums[i];
            int rank = rankMap.get(val);

            int count_le_1 = query(bit1, rank);
            int count_le_2 = query(bit2, rank);

            int count1 = arr1.size() - count_le_1;
            int count2 = arr2.size() - count_le_2;

            if (count1 > count2) {
                arr1.add(val);
                update(bit1, rank, m);
            } else if (count2 > count1) {
                arr2.add(val);
                update(bit2, rank, m);
            } else {
                if (arr1.size() <= arr2.size()) {
                    arr1.add(val);
                    update(bit1, rank, m);
                } else {
                    arr2.add(val);
                    update(bit2, rank, m);
                }
            }
        }

        int[] result = new int[arr1.size() + arr2.size()];
        int index = 0;
        for (int num : arr1) {
            result[index++] = num;
        }
        for (int num : arr2) {
            result[index++] = num;
        }

        return result;
    }

    private void update(int[] tree, int index, int m) {
        while (index <= m) {
            tree[index] += 1;
            index += index & -index;
        }
    }

    private int query(int[] tree, int index) {
        int s = 0;
        while (index > 0) {
            s += tree[index];
            index -= index & -index;
        }
        return s;
    }
}

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Problem Info

Description

Problem 3072: Distribute Elements Into Two Arrays II - Detailed explanation and solution approaches.

Topics

Array Binary Indexed Tree Segment Tree Simulation

Companies

Autodesk

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3072

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: Extract unique elements from `nums`, sort them, and create a rank map to map each element to its rank (index + 1) in the sorted list.
2
Step 2: Initialize two Binary Indexed Trees (BITs), `bit1` and `bit2`, with size m+1 where 'm' is number of unique elements, to keep track of the presence of numbers in `arr1` and `arr2` respectively.
3
Step 3: Initialize `arr1` with `nums[0]` and update `bit1` with the rank of `nums[0]`.
4
Step 4: Initialize `arr2` with `nums[1]` and update `bit2` with the rank of `nums[1]`.
5
Step 5: Iterate through the remaining elements of `nums` starting from index 2.
6
Step 6: For each element `nums[i]`, get its rank from the rank map.
7
Step 7: Calculate `count1` = number of elements in `arr1` greater than `nums[i]` using `bit1`. This can be done by querying number of elements less than or equals to `nums[i]` and subtracting it from the `arr1.size()`.
8
Step 8: Calculate `count2` = number of elements in `arr2` greater than `nums[i]` using `bit2`.
9
Step 9: Compare `count1` and `count2`: if `count1 > count2`, append `nums[i]` to `arr1` and update `bit1`; if `count2 > count1`, append `nums[i]` to `arr2` and update `bit2`; otherwise, compare the lengths of `arr1` and `arr2` and append to the smaller array or `arr1` if lengths are equal and update the appropriate BIT.
10
Step 10: After the loop, concatenate `arr1` and `arr2` to form the `result` 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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the loop that iterates through the `nums` array (O(n)). Inside the loop, we perform BIT queries and updates, which take O(log m) time, where m is the number of unique elements in `nums`. Since m <= n, the complexity is O(n log n). Creating the sorted unique list initially takes O(n log n) time. The space complexity is O(n) because we store the sorted unique numbers, rank map, arr1, arr2 and the two BITs, all of which can grow up to the size of n in worst case scenarios.

📖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 could be used instead of a Binary Indexed Tree to store and query the counts of elements in the arrays. A segment tree supports more complex range queries.

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

Trade-offs: Segment Trees generally have a larger constant factor than BITs and may use a bit more memory. BITs are preferred here for their concise implementation and speed, given the problem's requirements for cumulative frequency calculations only.

Brute Force

For each element nums[i] greaterCount for both the array can be calculated by looping and counting greater elements

Time: O(n^2)Space: O(n)

Trade-offs: Will cause TLE on larger test cases, hence using BIT improves performance

✨

Key Insights

1Insight 1: Using Binary Indexed Tree (BIT) or Segment Tree is crucial for efficiently calculating the `greaterCount` in each array during each iteration. A brute-force approach of iterating each array for every element would be too slow, leading to TLE.
2Insight 2: Rank mapping can efficiently map the input values to a smaller range (1 to m, where m is the number of unique elements). This allows us to use BIT which operates on index and not the direct values, which can be very large (up to 10^9)
3Insight 3: Careful consideration of all the tie-breaking conditions is crucial to get the correct output. Pay close attention to length comparisons and initial array population.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Duplicate numbers in `nums`. The solution correctly handles this by using `set` to extract unique numbers, rank mapping to convert the elements to ranks, and using BIT to store the cumulative counts. The `greaterCount` is calculated correctly even with duplicate numbers.
Edge case 2: All numbers are the same in `nums`. The solution handles this case by properly comparing the lengths of the arrays when `greaterCount` are the same. The distribution will eventually converge to nearly equal sized arrays.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to use a BIT or Segment Tree and attempting to iterate through `arr1` and `arr2` for each element of `nums` will lead to a time complexity of O(n^2), resulting in a Time Limit Exceeded (TLE) error.

✗Common mistake 2: Not considering the case when `greaterCount` is equal in both arrays and failing to handle the tie-breaking conditions based on the array lengths can lead to incorrect results. Also, it's important to initialize arr1 and arr2 with nums[0] and nums[1] respectively.

✅

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: This algorithm can be used in load balancing scenarios where tasks (represented by `nums`) need to be distributed among two servers (`arr1` and `arr2`) based on their current load (approximated by `greaterCount`). The goal is to balance the load and avoid overloading a single server.

🏦Real-world scenario 2: In database management, it can be used for distributing data across multiple shards or partitions based on the distribution of certain key values. The `greaterCount` can represent the distribution of data with keys greater than a certain value in a given shard.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: June 13, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 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 approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Distribute Elements Into Two Arrays II"?

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

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