What is the time complexity of Find Array Given Subset Sums?

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

What data structures are used to solve Find Array Given Subset Sums?

This problem uses Array, Divide and Conquer. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find Array Given Subset Sums?

Insight 1: Sorting the 'sums' array is crucial. The difference between the smallest two elements in the sorted array is a strong candidate for one of the elements (or its negation) in the original array. Insight 2: The presence of 0 in the subset sums helps differentiate between 'd' and '-d'. If the subset sums generated from the remaining elements after removing one number contain 0, the extracted element is 'd'. Otherwise, the extracted element is '-d'.

Which companies ask LeetCode 1982. Find Array Given Subset Sums in interviews?

This problem is frequently asked at Mindtickle.

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Find Array Given Subset Sums - LeetCode 1982 Solution

Q: What is the space complexity of Find Array Given Subset Sums?

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

Find Array Given Subset Sums - Complete Solution Guide

Find Array Given Subset Sums is LeetCode problem 1982, 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 n representing the length of an unknown array that you are trying to recover. You are also given an array sums containing the values of all 2 n subset sums of the unknown array (in no particular order). Return the array ans of length n representing the unknown array. If multiple answers exist, return any of them . An array sub is a subset of an array arr if sub can be obtained from arr by deleting some (possibly zero or all) elements of arr . The sum of the elements in s

Detailed Explanation

The problem requires you to reconstruct an unknown array of length 'n' given a list of all possible subset sums of that array. The 'sums' array contains 2^n elements, representing the sums of all subsets (including the empty set with a sum of 0). The challenge is to find the original array, knowing only these subset sums. Importantly, multiple valid solutions might exist, and the goal is to return any one of them.

Solution Approach

The solution employs a recursive divide-and-conquer strategy. It starts by sorting the 'sums' array. Then, it iteratively identifies and extracts one element of the unknown array. The difference between the smallest two numbers in the 'sums' array, 'd', is a strong candidate for either 'd' or '-d' being an element of the target array. The algorithm uses a Counter/HashMap to track the frequencies of each sum. By comparing the frequency of sums[i] with sums[i]+d we can find elements of A, which represent subsets without 'd'. Then by checking if A has a subset sum of 0, we can determine if the element added to the answer should be 'd' or '-d'. After extracting an element, it rebuilds the 'sums' array to reflect the subsets sums of the remaining elements, and repeat.

Step-by-Step Algorithm

Step 1: Sort the 'sums' array in ascending order.
Step 2: Iterate 'n' times (where 'n' is the length of the unknown array).
Step 3: Calculate 'd' as the difference between the second smallest and smallest element in the 'sums' array (sums[1] - sums[0]).
Step 4: Create a Counter (or HashMap) to store the frequencies of each sum in the 'sums' array.
Step 5: Create a new array 'A' to store subset sums that do *not* contain the element 'd'.
Step 6: Iterate through 'sums'. If a sum 's' is present in the Counter, decrement its count and check if 's + d' is also present. If 's + d' is present, decrement its count, and add 's' to array 'A'. This effectively splits the original subset sums into those containing 'd' and those not containing 'd'.
Step 7: Check if 'A' contains 0. If it does, it means the extracted element is 'd', and the next 'sums' array is 'A'. If 'A' does not contain 0, the extracted element is '-d', and the next 'sums' array is derived by adding 'd' to each element in 'A'.
Step 8: Add the extracted element (either 'd' or '-d') to the 'ans' array.
Step 9: Repeat steps 3-8 until 'n' elements have been extracted.
Step 10: Return the 'ans' array.

Key Insights

Insight 1: Sorting the 'sums' array is crucial. The difference between the smallest two elements in the sorted array is a strong candidate for one of the elements (or its negation) in the original array.
Insight 2: The presence of 0 in the subset sums helps differentiate between 'd' and '-d'. If the subset sums generated from the remaining elements after removing one number contain 0, the extracted element is 'd'. Otherwise, the extracted element is '-d'.
Insight 3: Divide and conquer: The problem can be reduced recursively. By identifying one element of the array (either 'd' or '-d'), we can effectively reduce the size of the problem. We create new subsets without this found element and repeat.

Complexity Analysis

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

Space Complexity: O(2**n)

Topics

This problem involves: Array, Divide and Conquer.

Companies

Asked at: Mindtickle.

Find Array Given Subset Sums | LeetCode 1982

Hard

Array Divide and Conquer

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

Solution Code

Available in:

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

class Solution {
    public int[] recoverArray(int n, int[] sums) {
        Arrays.sort(sums);

        List<Integer> ans = new ArrayList<>();

        for (int i = n; i > 0; i--) {
            int d = sums[1] - sums[0];

            Map<Integer, Integer> count = new HashMap<>();
            for (int s : sums) {
                count.put(s, count.getOrDefault(s, 0) + 1);
            }
            List<Integer> A = new ArrayList<>();

            boolean has_zero_in_A = false;

            for (int s : sums) {
                if (count.get(s) != null && count.get(s) > 0) {
                    count.put(s, count.get(s) - 1);
                    if (count.containsKey(s + d) && count.get(s + d) != null && count.get(s + d) > 0) {
                        count.put(s + d, count.get(s + d) - 1);
                        A.add(s);
                        if (s == 0) {
                            has_zero_in_A = true;
                        }
                    }
                }
            }

            if (has_zero_in_A) {
                ans.add(d);
                sums = A.stream().mapToInt(Integer::intValue).toArray();
            } else {
                ans.add(-d);
                List<Integer> nextSums = new ArrayList<>();
                for (int s : A) {
                    nextSums.add(s + d);
                }
                sums = nextSums.stream().mapToInt(Integer::intValue).toArray();
            }
        }
        
        int[] result = new int[ans.size()];
        for (int i = 0; i < ans.size(); i++) {
            result[i] = ans.get(i);
        }
        return result;
    }
}

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Problem Info

Description

Problem 1982: Find Array Given Subset Sums - Detailed explanation and solution approaches.

Topics

Array Divide and Conquer

Companies

Mindtickle

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1982

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: Sort the 'sums' array in ascending order.
2
Step 2: Iterate 'n' times (where 'n' is the length of the unknown array).
3
Step 3: Calculate 'd' as the difference between the second smallest and smallest element in the 'sums' array (sums[1] - sums[0]).
4
Step 4: Create a Counter (or HashMap) to store the frequencies of each sum in the 'sums' array.
5
Step 5: Create a new array 'A' to store subset sums that do *not* contain the element 'd'.
6
Step 6: Iterate through 'sums'. If a sum 's' is present in the Counter, decrement its count and check if 's + d' is also present. If 's + d' is present, decrement its count, and add 's' to array 'A'. This effectively splits the original subset sums into those containing 'd' and those not containing 'd'.
7
Step 7: Check if 'A' contains 0. If it does, it means the extracted element is 'd', and the next 'sums' array is 'A'. If 'A' does not contain 0, the extracted element is '-d', and the next 'sums' array is derived by adding 'd' to each element in 'A'.
8
Step 8: Add the extracted element (either 'd' or '-d') to the 'ans' array.
9
Step 9: Repeat steps 3-8 until 'n' elements have been extracted.
10
Step 10: Return the 'ans' 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*2**n)

💾

Space Complexity

O(2**n)

📝

Why?

The outer loop runs 'n' times. Inside the loop, creating and iterating through the 'sums' array takes O(2^n) time due to the size of 'sums' being 2^n. Sorting at the start only runs once, is O(2^n log 2^n), and gets dwarfed by the other operations. The space complexity is determined by the size of the 'sums' array, the counter/HashMap, and the intermediate array 'A', all of which can grow up to 2^n in size. Therefore, the space complexity is O(2^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

Backtracking

A backtracking approach could be used to explore all possible combinations of elements that could form the original array. However, this is generally less efficient.

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

Trade-offs: Backtracking would likely have a much higher time complexity, making it impractical for larger values of 'n'. It might be simpler to understand conceptually but is less performant than the divide-and-conquer approach.

✨

Key Insights

1Insight 1: Sorting the 'sums' array is crucial. The difference between the smallest two elements in the sorted array is a strong candidate for one of the elements (or its negation) in the original array.
2Insight 2: The presence of 0 in the subset sums helps differentiate between 'd' and '-d'. If the subset sums generated from the remaining elements after removing one number contain 0, the extracted element is 'd'. Otherwise, the extracted element is '-d'.
3Insight 3: Divide and conquer: The problem can be reduced recursively. By identifying one element of the array (either 'd' or '-d'), we can effectively reduce the size of the problem. We create new subsets without this found element and repeat.

📚

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: 'n' is small (e.g., n = 1 or n = 2). The algorithm works correctly even for small values of 'n'. The base case is implicitly handled by the loop condition `i > 0`.
Edge case 2: 'sums' contains duplicate values. The Counter/HashMap correctly handles duplicate values, ensuring that the algorithm accurately identifies and extracts the correct elements.
Edge case 3: When d=0, a potential division by zero could be anticipated, but the code correctly handles this since `count[s+d]` becomes `count[s]`, so the right element is removed correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the 'sums' array. Sorting is essential for the algorithm to correctly identify the potential elements of the unknown array.

✗Common mistake 2: Incorrectly handling duplicate values in the 'sums' array. Using a Counter/HashMap is crucial for accurately tracking the frequencies of each sum.

✗Common mistake 3: Assuming 'd' is always a valid element. It's essential to check if 'd' or '-d' is the correct element by verifying the presence of 0 in the 'A' array or its transformed version.

✅

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: Error correction codes. The problem of recovering an array from its subset sums is analogous to certain error correction problems, where partial information needs to be used to reconstruct the original data.

🏦Real-world scenario 2: Signal processing. Subset sums can be used to represent the frequency components of a signal. Recovering the original array is akin to isolating the individual frequency components from the composite signal.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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 3, 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 14

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

@CodeNinjaJul 7

❓

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 "Find Array Given Subset Sums"?

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

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