What is the time complexity of Find the Sum of the Power of All Subsequences?

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

What data structures are used to solve Find the Sum of the Power of All Subsequences?

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

What are the key insights for solving Find the Sum of the Power of All Subsequences?

Insight 1: The problem can be solved using dynamic programming. We can maintain a `dp` array to store the number of subsequences with a specific sum up to a certain index in the `nums` array. Insight 2: The key is to efficiently count the number of subsequences with a sum equal to 'k' for each subsequence of 'nums'. Because we have to consider all subsequences of `nums`, we multiply `dp[k]` by 2^(n) * (1/2)^|T|, which is the same as 2^(n - |T|), to consider all combinations of including/excluding the remaining elements of `nums` after we have found a subsequence of some length |T| that sums to `k`.

Which companies ask LeetCode 3082. Find the Sum of the Power of All Subsequences in interviews?

This problem is frequently asked at DE Shaw.

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Find the Sum of the Power of All Subsequences - LeetCode 3082 Solution

Find the Sum of the Power of All Subsequences - Complete Solution Guide

Find the Sum of the Power of All Subsequences is LeetCode problem 3082, 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 an array of integers is defined as the number of subsequences with their sum equal to k . Return the sum of power of all subsequences of nums . Since the answer may be very large, return it modulo 10 9 + 7 . Example 1: Input: nums = [1,2,3], k = 3 Output: 6 Explanation: There are 5 subsequences of nums with non-zero power: The subsequence [ 1 , 2 , 3 ] has 2 subsequences with sum == 3 : [1,2, 3 ] and [ 1 , 2

Detailed Explanation

The problem asks us to find the sum of the 'power' of all subsequences of a given integer array `nums`. The 'power' of a subsequence is defined as the number of its subsequences whose sum equals a given target value `k`. The final result needs to be returned modulo 10^9 + 7 to prevent integer overflow. The problem essentially requires us to iterate through all possible subsequences of `nums`, calculate the number of subsequences of each subsequence with a sum equal to `k`, and then sum up these counts. This sum must then be taken modulo 10^9 + 7.

Solution Approach

The provided solution uses dynamic programming to efficiently compute the sum of the power of all subsequences. It initializes a `dp` array of size `k+1` where `dp[j]` stores the sum of (1/2)^|T| where the sum of subsequence T is equal to j. It iterates through each number in `nums` and updates the `dp` array based on whether the current number is included or excluded from the subsequence. Specifically, it iterates backwards through the dp table. After processing all numbers, `dp[k]` represents the sum of (1/2)^|T| for all subsequences T with a sum of k. Finally, the result is obtained by multiplying dp[k] by 2^n (modulo MOD), which effectively accounts for all possible subsequences of `nums`. The modular inverse of 2 is calculated to incorporate elements into current subsequences.

Step-by-Step Algorithm

Step 1: Initialize a `dp` array of size `k+1` with all elements set to 0. Set `dp[0]` to 1, representing the empty subsequence with a sum of 0.
Step 2: Calculate the modular inverse of 2 (i.e., 2^(MOD-2) % MOD). This value is needed for dividing by 2 in modular arithmetic.
Step 3: Iterate through the `nums` array. For each number `num`, iterate through the `dp` array from `k` down to `num`.
Step 4: In the inner loop, update `dp[j]` as `dp[j] = (dp[j] + dp[j - num] * inv2) % MOD`. This step considers whether or not the current number 'num' is included into the current subsequence.
Step 5: After iterating through all numbers in `nums`, calculate 2^n % MOD where n is the length of `nums`. This value represents the total number of subsequences of `nums` if you include/exclude each element in nums.
Step 6: Calculate the final result as `(dp[k] * power_of_2_n) % MOD`. This is the sum of the power of all subsequences of `nums`.
Step 7: Return the final result.

Key Insights

Insight 1: The problem can be solved using dynamic programming. We can maintain a `dp` array to store the number of subsequences with a specific sum up to a certain index in the `nums` array.
Insight 2: The key is to efficiently count the number of subsequences with a sum equal to 'k' for each subsequence of 'nums'. Because we have to consider all subsequences of `nums`, we multiply `dp[k]` by 2^(n) * (1/2)^|T|, which is the same as 2^(n - |T|), to consider all combinations of including/excluding the remaining elements of `nums` after we have found a subsequence of some length |T| that sums to `k`.
Insight 3: Using modular arithmetic is crucial to prevent integer overflow, especially when calculating powers. We utilize the modular inverse of 2, which is 2^(MOD-2) % MOD, when including elements into the current subsequence.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(k)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: DE Shaw.

Find the Sum of the Power of All Subsequences | LeetCode 3082

Hard

Array Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int sumOfPower(int[] nums, int k) {
        int MOD = (int) 1e9 + 7;
        int n = nums.length;

        // dp[j] will store sum_{T s.t. sum(T)=j} (1/2)^|T| over subsequences T
        // of the numbers processed so far.
        long[] dp = new long[k + 1];
        dp[0] = 1;  // For the empty subsequence: sum=0, length=0, (1/2)^0=1

        // Modular inverse of 2
        long inv2 = power(2, MOD - 2, MOD);

        for (int num : nums) {
            // We update dp in place. Iterate j from k down to num to use
            // dp values from before processing the current num.
            for (int j = k; j >= num; j--) {
                // dp_new[j] = dp_old[j] (don't include num) +
                //             dp_old[j - num] * 1/2 (include num)
                dp[j] = (dp[j] + dp[j - num] * inv2) % MOD;
            }
        }

        // The total sum of powers is sum_{T, sum(T)=k} 2^(n-|T|)
        // = 2^n * sum_{T, sum(T)=k} (1/2)^|T|
        // = 2^n * dp[k]

        long power_of_2_n = power(2, n, MOD);

        long result = (dp[k] * power_of_2_n) % MOD;

        return (int) result;
    }

    // Function to calculate (base^exp) % mod efficiently using binary exponentiation
    private long power(long base, long exp, long mod) {
        long res = 1;
        base %= mod;
        while (exp > 0) {
            if ((exp & 1) == 1) res = (res * base) % mod;
            base = (base * base) % mod;
            exp >>= 1;
        }
        return res;
    }
}

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Problem Info

Description

Problem 3082: Find the Sum of the Power of All Subsequences - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

DE Shaw

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3082

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 of size `k+1` with all elements set to 0. Set `dp[0]` to 1, representing the empty subsequence with a sum of 0.
2
Step 2: Calculate the modular inverse of 2 (i.e., 2^(MOD-2) % MOD). This value is needed for dividing by 2 in modular arithmetic.
3
Step 3: Iterate through the `nums` array. For each number `num`, iterate through the `dp` array from `k` down to `num`.
4
Step 4: In the inner loop, update `dp[j]` as `dp[j] = (dp[j] + dp[j - num] * inv2) % MOD`. This step considers whether or not the current number 'num' is included into the current subsequence.
5
Step 5: After iterating through all numbers in `nums`, calculate 2^n % MOD where n is the length of `nums`. This value represents the total number of subsequences of `nums` if you include/exclude each element in nums.
6
Step 6: Calculate the final result as `(dp[k] * power_of_2_n) % MOD`. This is the sum of the power of all subsequences of `nums`.
7
Step 7: Return the final result.

💡

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*k)

💾

Space Complexity

O(k)

📝

Why?

The nested loops contribute to the O(n*k) time complexity. The `dp` array's size determines the O(k) space complexity. The binary exponentiation function `power` runs in O(log n) time, where n is the exponent, but it's only called a few times, so its impact is negligible compared to the O(n*k) from the nested loops.

📖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 to explore all possible subsequences. Memoization can be applied to store intermediate results and avoid redundant computations.

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

Trade-offs: The recursive approach with memoization can be more intuitive but may use more memory due to the recursive call stack and the memoization table. The iterative DP approach is generally more efficient in terms of memory.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming. We can maintain a `dp` array to store the number of subsequences with a specific sum up to a certain index in the `nums` array.
2Insight 2: The key is to efficiently count the number of subsequences with a sum equal to 'k' for each subsequence of 'nums'. Because we have to consider all subsequences of `nums`, we multiply `dp[k]` by 2^(n) * (1/2)^|T|, which is the same as 2^(n - |T|), to consider all combinations of including/excluding the remaining elements of `nums` after we have found a subsequence of some length |T| that sums to `k`.
3Insight 3: Using modular arithmetic is crucial to prevent integer overflow, especially when calculating powers. We utilize the modular inverse of 2, which is 2^(MOD-2) % MOD, when including elements into the current 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: `k = 0`. The solution correctly handles this case because `dp[0]` is initialized to 1, representing the empty subsequence with a sum of 0. When `k` is 0, `dp[k]` will eventually hold the number of ways to reach a sum of 0 using subsequences.
Edge case 2: `nums` is empty. The loop won't execute, `power_of_2_n` will be pow(2, 0, MOD) which is 1, and the result will be dp[k]*1. If k is zero dp[k] is also initialized to 1 and hence the answer is 1. If k is non-zero then dp[k] is initialised to 0 and the answer will be 0.
Edge case 3: All elements in `nums` are greater than `k`. In this situation, unless k=0, no subsequences can sum up to k and the result will be 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to use modular arithmetic. The intermediate results can become very large, leading to integer overflow and incorrect results.

✗Common mistake 2: Not understanding the modular inverse. When dividing in modular arithmetic, it's important to multiply by the modular inverse instead of directly dividing.

✗Common mistake 3: Updating the DP table in the wrong order. The DP table must be updated in reverse order to avoid overwriting values that are needed for later calculations.

✅

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: Financial modeling where calculating the probabilities of different investment scenarios summing to a specific target portfolio value might need this dynamic programming approach.

🏦Real-world scenario 2: Combinatorial optimization problems where we are searching for the number of ways to form a specific target value, where limited resources or values must be used.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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 23, 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 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 the Power of All Subsequences"?

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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