#3082

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Find the Sum of the Power of All Subsequences

Find the sum of the power of all subsequences of an integer array where their sum equals a given number.

Problem Statement

You are given an integer array nums of length n and a positive integer k. The task is to find the sum of the power of all subsequences of nums, where the power is defined as the number of subsequences whose sum equals k.

To solve the problem, you must count subsequences of nums whose elements add up to k, with each subsequence contributing 2^(n - j) to the answer, where j is the length of the subsequence. Your goal is to return the total sum of powers for all such subsequences.

Examples

Example 1

Input: nums = [1,2,3], k = 3

Output: 6

There are 5 subsequences of nums with non-zero power: Hence the answer is 2 + 1 + 1 + 1 + 1 = 6 .

Example 2

Input: nums = [2,3,3], k = 5

Output: 4

There are 3 subsequences of nums with non-zero power: Hence the answer is 2 + 1 + 1 = 4 .

Example 3

Input: nums = [1,2,3], k = 7

Output: 0

There exists no subsequence with sum 7 . Hence all subsequences of nums have power = 0 .

Constraints

1 <= n <= 100
1 <= nums[i] <= 104
1 <= k <= 100

Solution Approach

Dynamic Programming State Transition

The problem can be solved efficiently using dynamic programming. Maintain a DP array where each index represents the number of subsequences that sum up to a specific value. For every number in the array, update the DP table to account for new subsequences that include this number.

Subset Sum Calculation

Calculate the number of valid subsequences for each possible sum up to k. As you iterate over the array, check if adding a number can form a subsequence that contributes to the sum k. The contribution is calculated by adding powers of two depending on the subsequence length.

Efficient Power Calculation

To compute the power of each subsequence efficiently, use the relation 2^(n - j), where n is the total length of the array and j is the length of the subsequence. This allows you to calculate the contribution of each subsequence without explicitly generating them.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the approach used. A naive approach may have a time complexity of O(2^n), but using dynamic programming to manage state transitions brings it down to O(n*k), where n is the length of the array and k is the target sum. Space complexity is O(k), as we only need to store intermediate results for sums up to k.

What Interviewers Usually Probe

The problem tests knowledge of dynamic programming and subset sum problems.
Candidates should demonstrate understanding of state transitions in dynamic programming.
Pay attention to how the solution scales with n and k, as efficiency is key for larger inputs.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the contribution of subsequences to the power sum.
Failing to update the DP table correctly during iteration.
Not handling edge cases such as when no subsequence sums to k, resulting in zero power.

Follow-up variants

Consider modifying the problem by changing the target sum k or adding constraints like limiting subsequence length.
What if the array contains duplicates or the integers in nums have different ranges?
Extend the problem to handle multiple target sums, requiring the calculation of power for each.

FAQ

How can dynamic programming be applied to the "Find the Sum of the Power of All Subsequences" problem?

Dynamic programming can be used to track subsequences and their sums efficiently. By maintaining a DP table that tracks the number of subsequences for each sum, we can compute the power contribution for each subsequence.

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

The time complexity is O(n*k) using dynamic programming, where n is the length of the array and k is the target sum.

What is the significance of the power 2^(n - j) in this problem?

The power 2^(n - j) is used to calculate the contribution of each subsequence, where n is the total length of the array and j is the length of the subsequence.

Are there any edge cases I should consider for the "Find the Sum of the Power of All Subsequences" problem?

Yes, you should handle cases where no subsequences sum to k, which would result in a power of zero for all subsequences.

How does the state transition work in dynamic programming for this problem?

The state transition involves updating the DP table to account for new subsequences formed by including the current element in the array. This allows for efficient calculation of subsequences and their sums.

terminal

Solution

Solution 1: Dynamic Programming

The problem requires us to find all subsequences $\textit{S}$ in the given array $\textit{nums}$, and then calculate the number of ways for each subsequence $\textit{T}$ such that the sum of $\textit{T}$ equals $\textit{k}$.

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class Solution:
    def sumOfPower(self, nums: List[int], k: int) -> int:
        mod = 10**9 + 7
        n = len(nums)
        f = [[0] * (k + 1) for _ in range(n + 1)]
        f[0][0] = 1
        for i, x in enumerate(nums, 1):
            for j in range(k + 1):
                f[i][j] = f[i - 1][j] * 2 % mod
                if j >= x:
                    f[i][j] = (f[i][j] + f[i - 1][j - x]) % mod
        return f[n][k]

Solution 2: Dynamic Programming (Optimization)

In the state transition equation from Solution 1, the value of $f[i][j]$ only depends on $f[i-1][j]$ and $f[i-1][j-x]$. Therefore, we can optimize the first dimension of the space, reducing the space complexity to $O(k)$.

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class Solution:
    def sumOfPower(self, nums: List[int], k: int) -> int:
        mod = 10**9 + 7
        n = len(nums)
        f = [[0] * (k + 1) for _ in range(n + 1)]
        f[0][0] = 1
        for i, x in enumerate(nums, 1):
            for j in range(k + 1):
                f[i][j] = f[i - 1][j] * 2 % mod
                if j >= x:
                    f[i][j] = (f[i][j] + f[i - 1][j - x]) % mod
        return f[n][k]

Continue Practicing

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