What is the time complexity of Length of the Longest Subsequence That Sums to Target?

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

What is the space complexity of Length of the Longest Subsequence That Sums to Target?

The space complexity is O(target).

What data structures are used to solve Length of the Longest Subsequence That Sums to Target?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Length of the Longest Subsequence That Sums to Target?

Insight 1: This problem is a variation of the classic knapsack problem, specifically the unbounded knapsack problem where each number can be used multiple times (though in this instance we don't want to). But by reusing elements we can build up longer subsequences. Insight 2: Dynamic programming is a suitable technique to solve this problem efficiently. We can use a 1D DP array to store the maximum length of a subsequence that sums to a particular value.

Which companies ask LeetCode 2915. Length of the Longest Subsequence That Sums to Target in interviews?

This problem is frequently asked at Intuit.

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Length of the Longest Subsequence That Sums to Target - LeetCode 2915 Solution

Length of the Longest Subsequence That Sums to Target - Complete Solution Guide

Length of the Longest Subsequence That Sums to Target is LeetCode problem 2915, a Medium 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 0-indexed array of integers nums , and an integer target . Return the length of the longest subsequence of nums that sums up to target . If no such subsequence exists, return -1 . A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements. Example 1: Input: nums = [1,2,3,4,5], target = 9 Output: 3 Explanation: There are 3 subsequences with a sum equal to 9: [4,5], [1,3,5], and [2,3,4]. The

Detailed Explanation

The problem asks us to find the length of the longest subsequence within a given array `nums` whose elements sum up to a specified `target` value. A subsequence is formed by selecting elements from the original array without changing their order, but potentially skipping some elements. If no subsequence sums to the target, we should return -1.

Solution Approach

The provided solution uses dynamic programming to solve the problem. It builds a DP array `dp` of size `target + 1`, where `dp[i]` represents the length of the longest subsequence that sums to `i`. The array is initialized with -1, except for `dp[0]` which is initialized to 0 (representing an empty subsequence summing to 0). The algorithm then iterates through the input array `nums`. For each number, it iterates backwards through the `dp` array from `target` down to `num`. If `dp[j - num]` is not -1 (meaning there's a subsequence that sums to `j - num`), then `dp[j]` can be updated to `max(dp[j], dp[j - num] + 1)`. This means that if we can form a subsequence that sums to `j - num`, then we can add `num` to that subsequence to form a subsequence that sums to `j`, and the length of this new subsequence will be one greater than the length of the subsequence that sums to `j - num`. Finally, the value of `dp[target]` will be the length of the longest subsequence that sums to the target, or -1 if no such subsequence exists.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `target + 1` with -1. `dp[i]` will store the length of the longest subsequence summing to `i`.
Step 2: Set `dp[0] = 0` because an empty subsequence has a sum of 0 and a length of 0.
Step 3: Iterate through each number `num` in the `nums` array.
Step 4: Iterate through the `dp` array from `target` down to `num`.
Step 5: Inside the inner loop, check if `dp[j - num]` is not -1. This condition checks if it's possible to form a sum of `j - num` using a subsequence.
Step 6: If `dp[j - num]` is not -1, update `dp[j]` to `max(dp[j], dp[j - num] + 1)`. This means we consider the possibility of including `num` in the subsequence to reach the sum `j`.
Step 7: After iterating through all numbers in `nums`, return `dp[target]`, which will contain the length of the longest subsequence that sums to `target`, or -1 if no such subsequence exists.

Key Insights

Insight 1: This problem is a variation of the classic knapsack problem, specifically the unbounded knapsack problem where each number can be used multiple times (though in this instance we don't want to). But by reusing elements we can build up longer subsequences.
Insight 2: Dynamic programming is a suitable technique to solve this problem efficiently. We can use a 1D DP array to store the maximum length of a subsequence that sums to a particular value.
Insight 3: We can iterate through the numbers in `nums` and update our DP table, effectively deciding at each step whether or not to include the current number in our subsequence.

Complexity Analysis

Time Complexity: O(n*target)

Space Complexity: O(target)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Intuit.

Length of the Longest Subsequence That Sums to Target | LeetCode 2915

Medium

Array Dynamic Programming

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

Solution Code

Available in:

import java.util.List;

class Solution {
    public int lengthOfLongestSubsequence(List<Integer> nums, int target) {
        int[] dp = new int[target + 1];
        for (int i = 0; i <= target; i++) {
            dp[i] = -1;
        }
        dp[0] = 0;

        for (int num : nums) {
            for (int j = target; j >= num; j--) {
                if (dp[j - num] != -1) {
                    dp[j] = Math.max(dp[j], dp[j - num] + 1);
                }
            }
        }

        return dp[target];
    }
}

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Problem Info

Description

Problem 2915: Length of the Longest Subsequence That Sums to Target - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Intuit

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2915

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

7 steps

1
Step 1: Initialize a DP array `dp` of size `target + 1` with -1. `dp[i]` will store the length of the longest subsequence summing to `i`.
2
Step 2: Set `dp[0] = 0` because an empty subsequence has a sum of 0 and a length of 0.
3
Step 3: Iterate through each number `num` in the `nums` array.
4
Step 4: Iterate through the `dp` array from `target` down to `num`.
5
Step 5: Inside the inner loop, check if `dp[j - num]` is not -1. This condition checks if it's possible to form a sum of `j - num` using a subsequence.
6
Step 6: If `dp[j - num]` is not -1, update `dp[j]` to `max(dp[j], dp[j - num] + 1)`. This means we consider the possibility of including `num` in the subsequence to reach the sum `j`.
7
Step 7: After iterating through all numbers in `nums`, return `dp[target]`, which will contain the length of the longest subsequence that sums to `target`, or -1 if no such subsequence exists.

💡

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*target)

💾

Space Complexity

O(target)

📝

Why?

The time complexity arises from the nested loops. The space complexity comes from the DP array which is proportional to target. No additional significant space is used.

📖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

Recursive approach with memoization

A recursive approach can be used where we try including or excluding each element in `nums`. Memoization can be used to avoid redundant computations.

Time: O(n*target)Space: O(n*target) due to the memoization table and recursion stack.

Trade-offs: Recursive approach can be more intuitive to understand initially, but can lead to stack overflow for larger inputs if memoization isn't carefully implemented, or if the recursion depth is too high. The iterative DP solution usually performs better due to less overhead and controlled memory usage.

✨

Key Insights

1Insight 1: This problem is a variation of the classic knapsack problem, specifically the unbounded knapsack problem where each number can be used multiple times (though in this instance we don't want to). But by reusing elements we can build up longer subsequences.
2Insight 2: Dynamic programming is a suitable technique to solve this problem efficiently. We can use a 1D DP array to store the maximum length of a subsequence that sums to a particular value.
3Insight 3: We can iterate through the numbers in `nums` and update our DP table, effectively deciding at each step whether or not to include the current number in our subsequence.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: If the input `nums` is empty. The outer loop won't execute, and dp[target] will remain -1 if target != 0 or 0 if target == 0, which is the correct behavior.
Edge case 2: If the target is zero. dp[0] is initialized to 0, which correctly reflects the empty subsequence summing to 0. If there are any zeros in the input array, they will allow forming a subsequence with zero sum and the result will be at least 1.
Edge case 3: If no subsequence sums to target. The `dp[target]` will remain -1, indicating that no subsequence was found.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize `dp[0]` to 0. This will result in incorrect results because the algorithm relies on the fact that an empty subsequence has a sum of 0 and a length of 0.

✗Common mistake 2: Iterating through the `dp` array in the wrong direction. Iterating from `num` to `target` (instead of `target` down to `num`) would mean that you would be reusing the *same* `num` multiple times within the inner loop's calculation for each `j`. Iterating backwards prevents this unintended multiple use within the inner loop and properly calculates subsequences.

✗Common mistake 3: Not handling the case where no subsequence sums to the target. Forgetting to return -1 in this case leads to incorrect results.

✅

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: Inventory management. Determining the optimal combination of items to include in a shipment to meet a specific weight or value target, while maximizing the number of items.

🏦Real-world scenario 2: Financial planning. Finding the optimal combination of investments to reach a financial goal, while considering the returns and risks of each investment. The target could be a specific return or portfolio value.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 6, 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 14

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

@AlgoMasterJul 7

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

@InterviewProJun 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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Length of the Longest Subsequence That Sums to Target"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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