What is the time complexity of Find the Sum of Subsequence Powers?

The optimal solution has a time complexity of O(n^3k + n^2log(n)).

What is the space complexity of Find the Sum of Subsequence Powers?

The space complexity is O(n^2 + nk).

What data structures are used to solve Find the Sum of Subsequence Powers?

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

What are the key insights for solving Find the Sum of Subsequence Powers?

Insight 1: Sorting the input array `nums` is crucial because the minimum difference in a subsequence is easily found when the elements are sorted. This allows us to systematically consider possible differences. Insight 2: Dynamic Programming (DP) is essential for efficiently counting the number of subsequences of length `k` with a minimum difference of at least `d`. The DP table `dp[i][j]` stores the number of subsequences of length `j` ending at index `i` that satisfy the minimum difference constraint.

Which companies ask LeetCode 3098. Find the Sum of Subsequence Powers in interviews?

This problem is frequently asked at Rubrik.

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Find the Sum of Subsequence Powers - LeetCode 3098 Solution

Find the Sum of Subsequence Powers - Complete Solution Guide

Find the Sum of Subsequence Powers is LeetCode problem 3098, 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 length n , and a positive integer k . The power of a subsequence is defined as the minimum absolute difference between any two elements in the subsequence. Return the sum of powers of all subsequences of nums which have length equal to k . Since the answer may be large, return it modulo 10 9 + 7 . Example 1: Input: nums = [1,2,3,4], k = 3 Output: 4 Explanation: There are 4 subsequences in nums which have length 3: [1,2,3] , [1,3,4] , [1,2,4] , and [2,3,4] .

Detailed Explanation

The problem requires us to find the sum of 'powers' of all subsequences of a given array `nums` that have a length equal to `k`. The 'power' of a subsequence is defined as the minimum absolute difference between any two elements within that subsequence. The final result should be returned modulo 10^9 + 7 to prevent integer overflow. For example, if nums = [1, 2, 3, 4] and k = 3, the subsequences of length 3 are [1, 2, 3], [1, 2, 4], [1, 3, 4], and [2, 3, 4]. Their corresponding powers (minimum absolute differences) are |2-1| = 1, |2-1| = 1, |3-1| = 2, and |3-2| = 1. The sum of powers is 1 + 1 + 2 + 1 = 5.

Solution Approach

The provided solution utilizes a combination of sorting, dynamic programming, and memoization to efficiently compute the sum of subsequence powers. First, the input array `nums` is sorted. Then, all possible minimum differences between elements in the sorted array are identified and sorted. The algorithm iterates through these sorted differences and uses dynamic programming to count the number of subsequences of length `k` with a minimum difference of at least `d`. Memoization is employed to store the results of the DP calculations for each `d`, avoiding redundant computations. Finally, the sum of powers is calculated by multiplying each difference increase (d - prev_d) by the count of subsequences having at least that minimum difference and accumulating it into the total sum.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order.
Step 2: Identify all unique possible minimum differences between elements in `nums` and store them in a sorted list called `sorted_diffs`.
Step 3: Define a helper function `count_subsequences_with_min_diff(d)` that uses dynamic programming and memoization to count the number of subsequences of length `k` with a minimum difference of at least `d`.
Step 4: Inside `count_subsequences_with_min_diff(d)`: Create a DP table `dp[i][j]` where `dp[i][j]` represents the number of subsequences of length `j` ending at index `i` with a minimum difference of at least `d`. Initialize the first column (j=1) to 1, since any element can form a subsequence of length 1.
Step 5: Iterate through the DP table, filling it using the following recurrence relation: `dp[i][j] = sum(dp[p][j-1]) for all p < i such that nums[i] - nums[p] >= d`. This means that the number of subsequences of length `j` ending at index `i` is the sum of subsequences of length `j-1` ending at all indices `p` before `i` such that the difference between `nums[i]` and `nums[p]` is at least `d`.
Step 6: Store the result of `count_subsequences_with_min_diff(d)` in the `memo` to avoid recalculating it later.
Step 7: Iterate through the `sorted_diffs` list. For each difference `d`, calculate the number of subsequences with at least that minimum difference using `count_subsequences_with_min_diff(d)`.
Step 8: Calculate the contribution to the total sum of powers: `term = (d - prev_d) * count`, where `prev_d` is the previous minimum difference.
Step 9: Add `term` to the `total_sum_of_powers`, taking the modulo 10^9 + 7 at each step to prevent integer overflow.
Step 10: Update `prev_d = d` for the next iteration.
Step 11: Return the `total_sum_of_powers`.

Key Insights

Insight 1: Sorting the input array `nums` is crucial because the minimum difference in a subsequence is easily found when the elements are sorted. This allows us to systematically consider possible differences.
Insight 2: Dynamic Programming (DP) is essential for efficiently counting the number of subsequences of length `k` with a minimum difference of at least `d`. The DP table `dp[i][j]` stores the number of subsequences of length `j` ending at index `i` that satisfy the minimum difference constraint.
Insight 3: Memoization is used to avoid recomputing the number of subsequences with a given minimum difference `d`. This significantly improves the performance by storing and reusing previously computed results.

Complexity Analysis

Time Complexity: O(n^3k + n^2log(n))

Space Complexity: O(n^2 + nk)

Topics

This problem involves: Array, Dynamic Programming, Sorting.

Companies

Asked at: Rubrik.

Find the Sum of Subsequence Powers | LeetCode 3098

Hard

Array Dynamic Programming Sorting

Time: O(n^3k + n^2log(n))Space: O(n^2 + nk)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.ArrayList;
import java.util.Collections;

class Solution {
    private final int MOD = 1_000_000_007;

    public int sumOfPowers(int[] nums, int k) {
        int n = nums.length;

        Arrays.sort(nums);

        Map<Integer, Integer> memo = new HashMap<>();

        
        class Helper {
            public int count_subsequences_with_min_diff(int d) {
                if (memo.containsKey(d)) {
                    return memo.get(d);
                }

                int[][] dp = new int[n][k + 1];

                for (int i = 0; i < n; i++) {
                    dp[i][1] = 1;
                }

                for (int j = 2; j <= k; j++) {
                    int p = 0;
                    int current_sum = 0;
                    for (int i = 0; i < n; i++) {
                        while (p < i && nums[i] - nums[p] >= d) {
                            current_sum = (current_sum + dp[p][j - 1]) % MOD;
                            p += 1;
                        }
                        dp[i][j] = current_sum;
                    }
                }

                int total_count = 0;
                for (int i = 0; i < n; i++) {
                    total_count = (total_count + dp[i][k]) % MOD;
                }

                memo.put(d, total_count);
                return total_count;
            }
        }
        
        Helper helper = new Helper();


        Set<Integer> diffsSet = new HashSet<>();
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                diffsSet.add(nums[j] - nums[i]);
            }
        }

        List<Integer> diffsList = new ArrayList<>(diffsSet);
        Collections.sort(diffsList);

        int total_sum_of_powers = 0;
        int prev_d = 0;

        for (int d : diffsList) {
            int count = helper.count_subsequences_with_min_diff(d);
            long term = (long) (d - prev_d) * count % MOD;
            total_sum_of_powers = (int) (((long) total_sum_of_powers + term) % MOD);
            prev_d = d;
        }

        return total_sum_of_powers;
    }
}

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Problem Info

Description

Problem 3098: Find the Sum of Subsequence Powers - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Sorting

Companies

Rubrik

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3098

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

11 steps

1
Step 1: Sort the input array `nums` in ascending order.
2
Step 2: Identify all unique possible minimum differences between elements in `nums` and store them in a sorted list called `sorted_diffs`.
3
Step 3: Define a helper function `count_subsequences_with_min_diff(d)` that uses dynamic programming and memoization to count the number of subsequences of length `k` with a minimum difference of at least `d`.
4
Step 4: Inside `count_subsequences_with_min_diff(d)`: Create a DP table `dp[i][j]` where `dp[i][j]` represents the number of subsequences of length `j` ending at index `i` with a minimum difference of at least `d`. Initialize the first column (j=1) to 1, since any element can form a subsequence of length 1.
5
Step 5: Iterate through the DP table, filling it using the following recurrence relation: `dp[i][j] = sum(dp[p][j-1]) for all p < i such that nums[i] - nums[p] >= d`. This means that the number of subsequences of length `j` ending at index `i` is the sum of subsequences of length `j-1` ending at all indices `p` before `i` such that the difference between `nums[i]` and `nums[p]` is at least `d`.
6
Step 6: Store the result of `count_subsequences_with_min_diff(d)` in the `memo` to avoid recalculating it later.
7
Step 7: Iterate through the `sorted_diffs` list. For each difference `d`, calculate the number of subsequences with at least that minimum difference using `count_subsequences_with_min_diff(d)`.
8
Step 8: Calculate the contribution to the total sum of powers: `term = (d - prev_d) * count`, where `prev_d` is the previous minimum difference.
9
Step 9: Add `term` to the `total_sum_of_powers`, taking the modulo 10^9 + 7 at each step to prevent integer overflow.
10
Step 10: Update `prev_d = d` for the next iteration.
11
Step 11: Return the `total_sum_of_powers`.

💡

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^3k + n^2log(n))

💾

Space Complexity

O(n^2 + nk)

📝

Why?

The time complexity is dominated by the nested loops in the dynamic programming and the process of calculating all possible differences. The space complexity is primarily determined by the memoization map and the DP table.

📖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

Brute-force

Generate all possible subsequences of length `k`, calculate the minimum difference for each subsequence, and sum these differences. This approach is highly inefficient.

Time: O(C(n, k) * k^2), where C(n, k) is the binomial coefficient (n choose k). This is approximately O(n^k) for a fixed kSpace: O(k)

Trade-offs: The brute-force approach is simple to understand but completely impractical for larger values of `n` and `k`. The given problem constraints restrict `n` to be up to 50 and `k` to be up to `n`, so the brute force approach is extremely slow.

✨

Key Insights

1Insight 1: Sorting the input array `nums` is crucial because the minimum difference in a subsequence is easily found when the elements are sorted. This allows us to systematically consider possible differences.
2Insight 2: Dynamic Programming (DP) is essential for efficiently counting the number of subsequences of length `k` with a minimum difference of at least `d`. The DP table `dp[i][j]` stores the number of subsequences of length `j` ending at index `i` that satisfy the minimum difference constraint.
3Insight 3: Memoization is used to avoid recomputing the number of subsequences with a given minimum difference `d`. This significantly improves the performance by storing and reusing previously computed results.

📚

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.

Sorting

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

Edge Cases

Edge case 1: When k = n, the subsequence is simply the entire array. The power is then the minimum difference between any two elements in the sorted array. The algorithm handles this correctly as it calculates all unique minimum differences and uses DP to count the number of subsequences.
Edge case 2: When all elements in `nums` are the same, the minimum difference is 0. The algorithm correctly identifies this and calculates the number of subsequences with a minimum difference of at least 0, which is simply the number of ways to choose k elements from n, and it correctly accounts for this when computing the sum of powers.
Edge case 3: When n = k = 2 and `nums` = [a, b] where a = b, the result should be 0. The code handles this correctly as it identifies the difference to be 0 and calculates the sum of powers accordingly.
Edge case 4: The modulo operation is essential to prevent integer overflow, especially when `n` and `k` are large. The code correctly applies the modulo operation at each step of the summation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the input array `nums`. Sorting is crucial for efficiently finding the minimum difference between elements in a subsequence.

✗Common mistake 2: Not using memoization, which leads to significant performance degradation due to redundant calculations. The dynamic programming calculations are repeated for each difference considered, resulting in a much higher time complexity without memoization.

✗Common mistake 3: Integer overflow. The sum of powers can be very large, so it is essential to use modulo arithmetic (modulo 10^9 + 7) to prevent integer overflow.

✗Common mistake 4: Incorrectly calculating the minimum difference within the subsequence. The minimum difference needs to be calculated correctly in each subsequence.

✅

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: Network optimization: Consider a network where nodes have different processing speeds. The power can represent the minimum difference in processing speed among a subset of nodes selected for a specific task. The algorithm can help to determine optimal node combinations for tasks with specific speed requirements.

🏦Real-world scenario 2: Financial analysis: Consider a portfolio of assets with different risk levels. The minimum difference can represent the minimum difference in risk levels among a selection of assets. This algorithm can find investment strategies where we minimize the possible lowest risk gap by picking k assets.

💡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: July 9, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 the Sum of Subsequence Powers"?

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.

🤖 Need help with this problem?