What is the time complexity of Partition Array Into Two Arrays to Minimize Sum Difference?

The optimal solution has a time complexity of O(n^2 * 2^n).

What is the space complexity of Partition Array Into Two Arrays to Minimize Sum Difference?

The space complexity is O(n * 2^n).

What data structures are used to solve Partition Array Into Two Arrays to Minimize Sum Difference?

This problem uses Array, Two Pointers, Binary Search, Dynamic Programming, Bit Manipulation, Ordered Set, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Partition Array Into Two Arrays to Minimize Sum Difference?

Insight 1: Splitting the array into two halves and generating all possible subset sums for each half is crucial for efficiently exploring possible partitions. Insight 2: Using binary search to find the closest sum in one half that complements a sum in the other half allows for a more efficient search than checking all possible combinations.

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Partition Array Into Two Arrays to Minimize Sum Difference - LeetCode 2035 Solution

Q: Which companies ask LeetCode 2035. Partition Array Into Two Arrays to Minimize Sum Difference in interviews?

This problem is frequently asked at Arcesium, Millennium, PhonePe, Salesforce, Samsung and 1 more companies.

Partition Array Into Two Arrays to Minimize Sum Difference - Complete Solution Guide

Partition Array Into Two Arrays to Minimize Sum Difference is LeetCode problem 2035, 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 of 2 * n integers. You need to partition nums into two arrays of length n to minimize the absolute difference of the sums of the arrays. To partition nums , put each element of nums into one of the two arrays. Return the minimum possible absolute difference . Example 1: Input: nums = [3,9,7,3] Output: 2 Explanation: One optimal partition is: [3,9] and [7,3]. The absolute difference between the sums of the arrays is abs((3 + 9) - (7 + 3)) = 2. Example 2: Input:

Detailed Explanation

The problem requires partitioning an array `nums` of even length (2*n) into two subarrays, each of length `n`. The goal is to minimize the absolute difference between the sums of these two subarrays. In essence, you want to find a subset of size 'n' whose sum is as close as possible to half of the total sum of all elements in `nums`.

Solution Approach

The provided solution uses a divide-and-conquer approach combined with binary search. First, the input array is split into two halves. Then, it generates all possible subset sums for each half, storing these sums based on the number of elements included in each subset. Finally, it iterates through the subset sums of the first half, and for each sum, it uses binary search on the sorted subset sums of the second half to find the sum that, when combined, gets closest to half of the total sum of the original array. This allows to efficiently find the minimum absolute difference.

Step-by-Step Algorithm

Step 1: Divide the input array `nums` into two subarrays, `left_half` and `right_half`, each of size 'n'.
Step 2: Create a function `get_all_subset_sums` that takes a subarray (either `left_half` or `right_half`) as input and returns a list of sets. Each set in the list represents all possible subset sums for a given number of elements chosen from the input subarray. Specifically, `sums[i]` stores the set of all possible sums using `i` elements.
Step 3: Call `get_all_subset_sums` on both `left_half` and `right_half` to generate `left_sums` and `right_sums`, respectively.
Step 4: Calculate the total sum of all elements in the original array `nums`.
Step 5: Iterate through `left_sums` using nested loops. The outer loop iterates through the number of elements 'k' (from 0 to n), representing the number of elements chosen from the left half. The inner loop iterates through all possible subset sums `sum_l` in `left_sums[k]`.
Step 6: For each `sum_l`, calculate the target sum `target_r` that the corresponding subset from the `right_half` should have to make the total sum of the two partitions as close as possible to half of the total sum: `target_r = (total_sum / 2) - sum_l`.
Step 7: Use binary search on the sorted list of subset sums from `right_sums[n - k]` (subset sums using `n-k` elements from the right half) to find the sum closest to `target_r`. Since a set is not sortable, you need to create a new list from the set and sort this list.
Step 8: Consider the closest sums found via binary search (the sum at the index returned by binary search and the sum at the index before it) and calculate the absolute difference between the total sum and twice the sum of the current partition (sum_l + sum_r). Update `min_diff` with the minimum absolute difference found so far.
Step 9: Return the final `min_diff` as an integer.

Key Insights

Insight 1: Splitting the array into two halves and generating all possible subset sums for each half is crucial for efficiently exploring possible partitions.
Insight 2: Using binary search to find the closest sum in one half that complements a sum in the other half allows for a more efficient search than checking all possible combinations.
Insight 3: The optimal partition will have the sum of one partition as close as possible to the total sum divided by 2. This is used as the target in binary search.

Complexity Analysis

Time Complexity: O(n^2 * 2^n)

Space Complexity: O(n * 2^n)

Topics

This problem involves: Array, Two Pointers, Binary Search, Dynamic Programming, Bit Manipulation, Ordered Set, Bitmask.

Companies

Asked at: Arcesium, Millennium, PhonePe, Salesforce, Samsung, Texas Instruments.

Partition Array Into Two Arrays to Minimize Sum Difference | LeetCode 2035

Hard

Array Two Pointers Binary Search Dynamic Programming Bit Manipulation Ordered Set Bitmask

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public int minimumDifference(int[] nums) {
        int n = nums.length / 2;

        List<Set<Integer>> left_sums = get_all_subset_sums(java.util.Arrays.copyOfRange(nums, 0, n), n);
        List<Set<Integer>> right_sums = get_all_subset_sums(java.util.Arrays.copyOfRange(nums, n, nums.length), n);

        int total_sum = 0;
        for (int num : nums) {
            total_sum += num;
        }

        int min_diff = Integer.MAX_VALUE;

        for (int i = 0; i <= n; i++) {
            List<Integer> sortedRightSums = new ArrayList<>(right_sums.get(i));
            Collections.sort(sortedRightSums);
            right_sums.set(i, new HashSet<>(sortedRightSums));
        }


        for (int k = 0; k <= n; k++) {
            for (int sum_l : left_sums.get(k)) {
                List<Integer> r_list = new ArrayList<>(right_sums.get(n - k));
                Collections.sort(r_list);

                if (r_list.isEmpty()) {
                    continue;
                }
                double target_r = ((double) total_sum / 2) - sum_l;
                
                int idx = binarySearch(r_list, target_r);

                if (idx < r_list.size()) {
                    int sum_r = r_list.get(idx);
                    int current_sum_partition1 = sum_l + sum_r;
                    min_diff = Math.min(min_diff, Math.abs(total_sum - 2 * current_sum_partition1));
                }

                if (idx > 0) {
                    int sum_r = r_list.get(idx - 1);
                    int current_sum_partition1 = sum_l + sum_r;
                    min_diff = Math.min(min_diff, Math.abs(total_sum - 2 * current_sum_partition1));
                }
            }
        }

        return min_diff;
    }

    private List<Set<Integer>> get_all_subset_sums(int[] half_nums, int n) {
        List<Set<Integer>> sums = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            sums.add(new HashSet<>());
        }
        sums.get(0).add(0);

        for (int num : half_nums) {
            for (int i = n; i > 0; i--) {
                for (int s : new HashSet<>(sums.get(i - 1))) {
                    sums.get(i).add(s + num);
                }
            }
        }
        return sums;
    }

    private int binarySearch(List<Integer> list, double target) {
        int low = 0;
        int high = list.size() - 1;
        int ans = list.size();

        while (low <= high) {
            int mid = (low + high) / 2;
            if (list.get(mid) >= target) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }

        return ans;
    }
}

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Problem Info

Description

Problem 2035: Partition Array Into Two Arrays to Minimize Sum Difference - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Dynamic Programming Bit Manipulation Ordered Set Bitmask

Companies

ArcesiumMillenniumPhonePeSalesforceSamsungTexas Instruments

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:2035

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

9 steps

1
Step 1: Divide the input array `nums` into two subarrays, `left_half` and `right_half`, each of size 'n'.
2
Step 2: Create a function `get_all_subset_sums` that takes a subarray (either `left_half` or `right_half`) as input and returns a list of sets. Each set in the list represents all possible subset sums for a given number of elements chosen from the input subarray. Specifically, `sums[i]` stores the set of all possible sums using `i` elements.
3
Step 3: Call `get_all_subset_sums` on both `left_half` and `right_half` to generate `left_sums` and `right_sums`, respectively.
4
Step 4: Calculate the total sum of all elements in the original array `nums`.
5
Step 5: Iterate through `left_sums` using nested loops. The outer loop iterates through the number of elements 'k' (from 0 to n), representing the number of elements chosen from the left half. The inner loop iterates through all possible subset sums `sum_l` in `left_sums[k]`.
6
Step 6: For each `sum_l`, calculate the target sum `target_r` that the corresponding subset from the `right_half` should have to make the total sum of the two partitions as close as possible to half of the total sum: `target_r = (total_sum / 2) - sum_l`.
7
Step 7: Use binary search on the sorted list of subset sums from `right_sums[n - k]` (subset sums using `n-k` elements from the right half) to find the sum closest to `target_r`. Since a set is not sortable, you need to create a new list from the set and sort this list.
8
Step 8: Consider the closest sums found via binary search (the sum at the index returned by binary search and the sum at the index before it) and calculate the absolute difference between the total sum and twice the sum of the current partition (sum_l + sum_r). Update `min_diff` with the minimum absolute difference found so far.
9
Step 9: Return the final `min_diff` as an integer.

💡

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^2 * 2^n)

💾

Space Complexity

O(n * 2^n)

📝

Why?

Generating all subset sums for each half takes O(n * 2^n) because for each of the 'n' elements, we are potentially doubling the number of sums for each subset size from 1 to n. Sorting subsets within the loop `i=0 to n` and then binary search requires O(n * 2^n log(2^n)) which simplifies to O(n^2 * 2^n) since log(2^n) is n. The space complexity arises from storing all subset sums for both halves, leading to O(n * 2^n), since we store up to 2^n sums for each number of elements ranging from 0 to n.

📖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

Dynamic Programming with Memoization

Instead of generating all subset sums upfront, we can use dynamic programming with memoization to calculate the minimum difference recursively. This approach can potentially reduce redundant calculations.

Time: O(n * sum)Space: O(n * sum)

Trade-offs: The time and space complexity depends on sum of the numbers. Not recommended if the numbers can be large. This approach is harder to implement for this specific problem given the constraint of needing to split into two equal halves.

Brute Force

Generate all possible partitions of the array and calculate the absolute difference of sums for each partition. This is highly inefficient.

Time: O(2^(2n))Space: O(1)

Trade-offs: This approach is only viable for extremely small input sizes due to the exponential time complexity. It's simple to understand but completely impractical for the given constraints.

✨

Key Insights

1Insight 1: Splitting the array into two halves and generating all possible subset sums for each half is crucial for efficiently exploring possible partitions.
2Insight 2: Using binary search to find the closest sum in one half that complements a sum in the other half allows for a more efficient search than checking all possible combinations.
3Insight 3: The optimal partition will have the sum of one partition as close as possible to the total sum divided by 2. This is used as the target in binary search.

📚

Key Terms Explained

For Students

Array

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

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!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array (or when n = 0): The code implicitly handles this case because the loops won't execute, and the initial value of min_diff will be returned if properly initialized.
Edge case 2: Array with only positive or only negative numbers: The solution works correctly in this case, as it explores all possible partitions and finds the minimum absolute difference.
Edge case 3: Large input values that cause integer overflow. The problem statement indicates -10^7 <= nums[i] <= 10^7. Summing 'n' (max 15) such numbers will not exceed the integer limit. However, one should be careful with doubling and subtractions involving total_sum. The Java code uses `double` to prevent such issues.
Edge case 4: Empty r_list in one of the iterations of k. The code includes `if (r_list.isEmpty()) { continue; }` to handle this case gracefully. If no right sums exist for the current number of elements being considered, the loop continues to the next left sum.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering all possible subset sums for both halves. It's crucial to explore all combinations to find the global minimum difference.

✗Common mistake 2: Integer overflow when calculating the total sum or differences. Using appropriate data types (long or double if necessary) is important.

✗Common mistake 3: Incorrectly implementing the binary search, leading to inaccurate results.

✗Common mistake 4: Not sorting the right_sums lists before using binary search. Binary search requires a sorted list.

✅

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: Imagine you need to divide a set of tasks (represented by their workload) between two teams to balance their workload as much as possible.

🏦Real-world scenario 2: Load balancing: Distributing tasks across multiple servers to minimize the difference in server load, ensuring optimal performance.

🎮Real-world scenario 3: In finance, dividing a portfolio of assets into two sub-portfolios to minimize the difference in their overall risk profiles or expected returns. This can also be used for fair division of assets in legal cases (e.g. divorce settlements).

💡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.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 26, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

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.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

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 "Partition Array Into Two Arrays to Minimize Sum Difference"?

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

🤖 Need help with this problem?