What is the time complexity of Constrained Subsequence Sum?

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

What data structures are used to solve Constrained Subsequence Sum?

This problem uses Array, Dynamic Programming, Queue, Sliding Window, Heap (Priority Queue), Monotonic Queue. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Constrained Subsequence Sum?

Insight 1: Dynamic Programming (DP) is suitable because the maximum sum at index `i` depends on the maximum sums at previous indices within the range `[i-k, i-1]`. Insight 2: A Monotonic Queue (Deque) is crucial for efficiently maintaining the maximum sum within the sliding window of size `k`. It allows us to quickly access the largest sum from the valid previous elements.

Which companies ask LeetCode 1425. Constrained Subsequence Sum in interviews?

This problem is frequently asked at Akuna Capital.

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Constrained Subsequence Sum - LeetCode 1425 Solution

Q: How to solve Constrained Subsequence Sum?

The provided solutions employ dynamic programming with a monotonic queue to solve the problem efficiently. The algorithm iterates through the `nums` array, updating each element `nums[i]` with the maximum subsequence sum ending at that index. The monotonic queue stores indices of potentially optimal previous elements. Specifically, it ensures that the queue only contains indices `j` such that `nums[j]` contributes to the largest subsequence sum leading to `nums[i]`. The front of the queue always holds the index of the element within the valid `k` range with the highest value seen so far. The elements of the `nums` array are effectively used as dp table, storing the maximum sum ending at that index at any point of time.

Constrained Subsequence Sum - Complete Solution Guide

Constrained Subsequence Sum is LeetCode problem 1425, 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

Given an integer array nums and an integer k , return the maximum sum of a non-empty subsequence of that array such that for every two consecutive integers in the subsequence, nums[i] and nums[j] , where i < j , the condition j - i <= k is satisfied. A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order. Example 1: Input: nums = [10,2,-10,5,20], k = 2 Output: 37 Explanation: The subsequence i

Detailed Explanation

The problem asks us to find the maximum sum of a non-empty subsequence from a given array `nums`, subject to the constraint that for any two consecutive elements `nums[i]` and `nums[j]` in the subsequence (where `i < j`), their indices must satisfy `j - i <= k`. Essentially, we want to find the best subsequence where the indices of consecutive elements are no more than `k` apart. The subsequence must contain at least one element.

Solution Approach

The provided solutions employ dynamic programming with a monotonic queue to solve the problem efficiently. The algorithm iterates through the `nums` array, updating each element `nums[i]` with the maximum subsequence sum ending at that index. The monotonic queue stores indices of potentially optimal previous elements. Specifically, it ensures that the queue only contains indices `j` such that `nums[j]` contributes to the largest subsequence sum leading to `nums[i]`. The front of the queue always holds the index of the element within the valid `k` range with the highest value seen so far. The elements of the `nums` array are effectively used as dp table, storing the maximum sum ending at that index at any point of time.

Step-by-Step Algorithm

Step 1: Initialize a deque (double-ended queue) `q` to store indices of elements within the valid window of size `k`.
Step 2: Iterate through the `nums` array from left to right (index `i`).
Step 3: Remove elements from the front of the queue that are outside the current window (i.e., `q.front() < i - k`).
Step 4: If the queue is not empty, update `nums[i]` with the maximum subsequence sum ending at index `i`. This involves adding `nums[q.front()]` to `nums[i]` if `nums[q.front()]` is positive.
Step 5: Maintain the monotonic property of the queue. Remove elements from the back of the queue that are smaller than or equal to `nums[i]` (i.e., `nums[q.back()] <= nums[i]`). This ensures that the queue always stores indices of elements in decreasing order of their values.
Step 6: Add the current index `i` to the back of the queue.
Step 7: After iterating through the entire array, find the maximum value in the modified `nums` array. This maximum value is the maximum constrained subsequence sum.

Key Insights

Insight 1: Dynamic Programming (DP) is suitable because the maximum sum at index `i` depends on the maximum sums at previous indices within the range `[i-k, i-1]`.
Insight 2: A Monotonic Queue (Deque) is crucial for efficiently maintaining the maximum sum within the sliding window of size `k`. It allows us to quickly access the largest sum from the valid previous elements.
Insight 3: If all numbers are negative, the subsequence consisting of only the largest element in the array should be returned. The solution incorporates this by comparing nums[i] with the accumulated maximum sum.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: Array, Dynamic Programming, Queue, Sliding Window, Heap (Priority Queue), Monotonic Queue.

Companies

Asked at: Akuna Capital.

Constrained Subsequence Sum | LeetCode 1425

Hard

Array Dynamic Programming Queue Sliding Window Heap (Priority Queue)Monotonic Queue

Time: O(n)Space: O(k)

Solution Code

Available in:

import java.util.Deque;
import java.util.LinkedList;

class Solution {
    public int constrainedSubsetSum(int[] nums, int k) {
        Deque<Integer> q = new LinkedList<>();
        for (int i = 0; i < nums.length; i++) {
            if (!q.isEmpty() && q.peekFirst() < i - k) {
                q.pollFirst();
            }
            
            if (!q.isEmpty()) {
                if (nums[q.peekFirst()] > 0) {
                    nums[i] += nums[q.peekFirst()];
                }
            }
            
            while (!q.isEmpty() && nums[q.peekLast()] <= nums[i]) {
                q.pollLast();
            }
            
            q.offer(i);
            
        }
        
        int max = Integer.MIN_VALUE;
        for (int num : nums) {
            max = Math.max(max, num);
        }
        return max;
    }
}

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Problem Info

Description

Problem 1425: Constrained Subsequence Sum - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Queue Sliding Window Heap (Priority Queue)Monotonic Queue

Companies

Akuna Capital

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1425

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 deque (double-ended queue) `q` to store indices of elements within the valid window of size `k`.
2
Step 2: Iterate through the `nums` array from left to right (index `i`).
3
Step 3: Remove elements from the front of the queue that are outside the current window (i.e., `q.front() < i - k`).
4
Step 4: If the queue is not empty, update `nums[i]` with the maximum subsequence sum ending at index `i`. This involves adding `nums[q.front()]` to `nums[i]` if `nums[q.front()]` is positive.
5
Step 5: Maintain the monotonic property of the queue. Remove elements from the back of the queue that are smaller than or equal to `nums[i]` (i.e., `nums[q.back()] <= nums[i]`). This ensures that the queue always stores indices of elements in decreasing order of their values.
6
Step 6: Add the current index `i` to the back of the queue.
7
Step 7: After iterating through the entire array, find the maximum value in the modified `nums` array. This maximum value is the maximum constrained subsequence sum.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(k)

📝

Why?

The algorithm iterates through the input array once. The deque stores at most k elements since it keeps track of the valid indices within the constraint j - i <= k. Thus the operations inside the loop is O(1) and space complexity is O(k).

📖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 without Monotonic Queue

A straightforward DP approach without a monotonic queue would involve iterating through the previous `k` elements for each element in `nums` to find the maximum subsequence sum. This is less efficient.

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

Trade-offs: Simpler to implement but less efficient if `k` is large, making it unsuitable when k is of the same order as n. It is better to use the monotonic queue approach as it offers better time complexity.

✨

Key Insights

1Insight 1: Dynamic Programming (DP) is suitable because the maximum sum at index `i` depends on the maximum sums at previous indices within the range `[i-k, i-1]`.
2Insight 2: A Monotonic Queue (Deque) is crucial for efficiently maintaining the maximum sum within the sliding window of size `k`. It allows us to quickly access the largest sum from the valid previous elements.
3Insight 3: If all numbers are negative, the subsequence consisting of only the largest element in the array should be returned. The solution incorporates this by comparing nums[i] with the accumulated maximum sum.

📚

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.

Queue

FIFO structure - First In, First Out. Like a line at a store!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: All numbers are negative. The solution correctly handles this by considering each number individually and choosing the largest negative number. This is handled because if the current queue is empty (which it will be), the value of nums[i] is not updated.
Edge case 2: `k` is greater than or equal to the length of `nums`. The algorithm still works correctly because the queue maintains a window of at most `k` elements. If k is larger than the length of nums, the condition `q.front() < i - k` never becomes true before i reaches the end of the array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where all numbers are negative. This will result in an incorrect result. Must return the largest negative number in that case.

✗Common mistake 2: Incorrectly implementing the monotonic queue, leading to incorrect results or increased time complexity. Pay close attention to when elements are added and removed from the queue and ensure that the queue maintains the monotonic property.

✅

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: Stock trading analysis. Identifying profitable trading subsequences within a certain time window (k) subject to constraints.

🏦Real-world scenario 2: Signal processing. Finding the strongest signal within a limited time frame by analyzing a series of signal strength measurements.

💡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: November 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

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.

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.

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 "Constrained 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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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