What is the time complexity of Closest Subsequence Sum?

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

What is the space complexity of Closest Subsequence Sum?

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

What data structures are used to solve Closest Subsequence Sum?

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

What are the key insights for solving Closest Subsequence Sum?

Insight 1: Since the array size can be up to 40, generating all possible subsequences (2^40) will result in Time Limit Exceeded (TLE). Therefore, a divide and conquer approach is needed. Insight 2: We can split the array into two halves and generate all possible subsequence sums for each half. This reduces the complexity to O(2^(n/2)) for each half.

Which companies ask LeetCode 1755. Closest Subsequence Sum in interviews?

This problem is frequently asked at LTI, Sprinklr.

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Closest Subsequence Sum - LeetCode 1755 Solution

Closest Subsequence Sum - Complete Solution Guide

Closest Subsequence Sum is LeetCode problem 1755, 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 and an integer goal . You want to choose a subsequence of nums such that the sum of its elements is the closest possible to goal . That is, if the sum of the subsequence's elements is sum , then you want to minimize the absolute difference abs(sum - goal) . Return the minimum possible value of abs(sum - goal) . Note that a subsequence of an array is an array formed by removing some elements (possibly all or none) of the original array. Example 1: Input: nums =

Detailed Explanation

The problem asks us to find a subsequence of a given integer array `nums` such that the sum of the elements in the subsequence is as close as possible to a given integer `goal`. We need to return the minimum absolute difference between any possible subsequence sum and the `goal`. A subsequence is formed by removing some (possibly none or all) elements from the original array. For example, given `nums = [5, -7, 3, 5]` and `goal = 6`, the subsequence consisting of all elements of `nums` has a sum of 6, which is equal to the goal. Thus, the minimum absolute difference is 0.

Solution Approach

The solution uses a divide-and-conquer strategy coupled with a two-pointer technique. The `nums` array is split into two halves. For each half, all possible subsequence sums are generated and stored in sorted lists (`sums1` and `sums2`). Then, two pointers are initialized, one at the beginning of `sums1` and the other at the end of `sums2`. The pointers are moved towards each other, and at each step, the absolute difference between the sum of the pointed values and the `goal` is calculated and compared with the current minimum difference. The two-pointer movement is based on whether the current sum is less than or greater than the `goal`.

Step-by-Step Algorithm

Step 1: Split the input array `nums` into two halves.
Step 2: Generate all possible subsequence sums for the first half of the array (`sums1`). Store the sums in a set to avoid duplicates, and then convert it to a sorted list.
Step 3: Generate all possible subsequence sums for the second half of the array (`sums2`). Store the sums in a set to avoid duplicates, and then convert it to a sorted list.
Step 4: Sort both `sums1` and `sums2` in ascending order.
Step 5: Initialize two pointers, `p1` at the beginning of `sums1` and `p2` at the end of `sums2`.
Step 6: Iterate while `p1` is within the bounds of `sums1` and `p2` is within the bounds of `sums2`.
Step 7: Calculate the current sum: `current_sum = sums1[p1] + sums2[p2]`.
Step 8: Update the minimum difference: `min_diff = min(min_diff, abs(current_sum - goal))`.
Step 9: If `current_sum` is less than `goal`, increment `p1` to consider larger sums from `sums1`.
Step 10: If `current_sum` is greater than `goal`, decrement `p2` to consider smaller sums from `sums2`.
Step 11: If `current_sum` is equal to `goal`, return 0 immediately since the minimum difference is found.
Step 12: Return the final `min_diff`.

Key Insights

Insight 1: Since the array size can be up to 40, generating all possible subsequences (2^40) will result in Time Limit Exceeded (TLE). Therefore, a divide and conquer approach is needed.
Insight 2: We can split the array into two halves and generate all possible subsequence sums for each half. This reduces the complexity to O(2^(n/2)) for each half.
Insight 3: After generating the sums for each half, we can use a two-pointer approach on the sorted lists of sums to efficiently find the pair of sums (one from each half) that minimizes the absolute difference with the target `goal`.

Complexity Analysis

Time Complexity: O(2^(n/2)log(2^(n/2)))

Space Complexity: O(2^(n/2))

Topics

This problem involves: Array, Two Pointers, Dynamic Programming, Bit Manipulation, Sorting, Bitmask.

Companies

Asked at: LTI, Sprinklr.

Closest Subsequence Sum | LeetCode 1755

Hard

Array Two Pointers Dynamic Programming Bit Manipulation Sorting Bitmask

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

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 minAbsDifference(int[] nums, int goal) {
        
        List<Integer> sums1 = generateSums(nums, 0, nums.length / 2);
        List<Integer> sums2 = generateSums(nums, nums.length / 2, nums.length);

        Collections.sort(sums1);
        Collections.sort(sums2);

        int minDiff = Integer.MAX_VALUE;

        int p1 = 0;
        int p2 = sums2.size() - 1;

        while (p1 < sums1.size() && p2 >= 0) {
            int currentSum = sums1.get(p1) + sums2.get(p2);

            minDiff = Math.min(minDiff, Math.abs(currentSum - goal));

            if (currentSum < goal) {
                p1++;
            } else if (currentSum > goal) {
                p2--;
            } else {
                return 0;
            }
        }

        return minDiff;
    }

    private List<Integer> generateSums(int[] arr, int start, int end) {
        Set<Integer> sumsSet = new HashSet<>();
        sumsSet.add(0);

        for (int i = start; i < end; i++) {
            Set<Integer> newSums = new HashSet<>();
            for (int sum : sumsSet) {
                newSums.add(sum + arr[i]);
            }
            sumsSet.addAll(newSums);
        }

        return new ArrayList<>(sumsSet);
    }
}

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Problem Info

Description

Problem 1755: Closest Subsequence Sum - Detailed explanation and solution approaches.

Topics

Array Two Pointers Dynamic Programming Bit Manipulation Sorting Bitmask

Companies

LTISprinklr

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1755

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

12 steps

1
Step 1: Split the input array `nums` into two halves.
2
Step 2: Generate all possible subsequence sums for the first half of the array (`sums1`). Store the sums in a set to avoid duplicates, and then convert it to a sorted list.
3
Step 3: Generate all possible subsequence sums for the second half of the array (`sums2`). Store the sums in a set to avoid duplicates, and then convert it to a sorted list.
4
Step 4: Sort both `sums1` and `sums2` in ascending order.
5
Step 5: Initialize two pointers, `p1` at the beginning of `sums1` and `p2` at the end of `sums2`.
6
Step 6: Iterate while `p1` is within the bounds of `sums1` and `p2` is within the bounds of `sums2`.
7
Step 7: Calculate the current sum: `current_sum = sums1[p1] + sums2[p2]`.
8
Step 8: Update the minimum difference: `min_diff = min(min_diff, abs(current_sum - goal))`.
9
Step 9: If `current_sum` is less than `goal`, increment `p1` to consider larger sums from `sums1`.
10
Step 10: If `current_sum` is greater than `goal`, decrement `p2` to consider smaller sums from `sums2`.
11
Step 11: If `current_sum` is equal to `goal`, return 0 immediately since the minimum difference is found.
12
Step 12: Return the final `min_diff`.

💡

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(2^(n/2)log(2^(n/2)))

💾

Space Complexity

O(2^(n/2))

📝

Why?

Generating all subsets for half of the array (n/2) takes O(2^(n/2)) time. Sorting the generated sums takes O(2^(n/2)log(2^(n/2))) time. The two-pointer approach takes O(2^(n/2)) time in the worst case. The space complexity is dominated by storing the sums in two lists, each of size O(2^(n/2)).

📖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

A dynamic programming approach could be attempted to build up subsequence sums, but the state space would be large, potentially leading to memory issues or TLE.

Time: Potentially O(n*sum_range) where sum_range is the range of the possible sums. This is pseudo-polynomial.Space: O(n*sum_range)

Trade-offs: The DP approach is unlikely to be more efficient than the current solution and can easily lead to memory issues given the constraint ranges for the input numbers and the goal.

✨

Key Insights

1Insight 1: Since the array size can be up to 40, generating all possible subsequences (2^40) will result in Time Limit Exceeded (TLE). Therefore, a divide and conquer approach is needed.
2Insight 2: We can split the array into two halves and generate all possible subsequence sums for each half. This reduces the complexity to O(2^(n/2)) for each half.
3Insight 3: After generating the sums for each half, we can use a two-pointer approach on the sorted lists of sums to efficiently find the pair of sums (one from each half) that minimizes the absolute difference with the target `goal`.

📚

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.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty array or array with one element. The solution handles these implicitly because the generation of sums will still work correctly, producing at least the sum 0.
Edge case 2: The `goal` is very large or very small. The algorithm handles this by iterating through all possible sums and updating the minimum difference accordingly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not splitting the array. Trying to generate all possible subsequence sums for the entire array leads to exponential time complexity and a Time Limit Exceeded error.

✗Common mistake 2: Not sorting the sum lists. The two-pointer approach requires the lists to be sorted to efficiently find the closest sum to the target.

✗Common mistake 3: Using a naive search for the closest sum after generating the sums, rather than using the two-pointer technique.

✅

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 have a set of tasks each with a certain cost, and you have a limited budget. You want to find a combination of tasks whose total cost is as close as possible to your budget without exceeding it. This problem is similar to finding a subsequence with a sum closest to a target value.

🏦Real-world scenario 2: Portfolio optimization: You have a set of investment options each with a different return. You want to choose a portfolio of investments that achieves a target return with minimal risk. Finding a subset of investments whose return is closest to the target is a related problem.

💡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.

👨‍💻 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: October 14, 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 "Closest Subsequence Sum"?

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?