Introduction

In this blog post, we will explore how to solve the LeetCode problem "3082" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 109 + 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 [1,2,3]. The subsequence [1,2,3] has 1 subsequence with sum == 3: [1,2,3]. The subsequence [1,2,3] has 1 subsequence with sum == 3: [1,2,3]. The subsequence [1,2,3] has 1 subsequence with sum == 3: [1,2,3]. The subsequence [1,2,3] has 1 subsequence with sum == 3: [1,2,3]. Hence the answer is 2 + 1 + 1 + 1 + 1 = 6. Example 2: Input: nums = [2,3,3], k = 5 Output: 4 Explanation: There are 3 subsequences of nums with non-zero power: The subsequence [2,3,3] has 2 subsequences with sum == 5: [2,3,3] and [2,3,3]. The subsequence [2,3,3] has 1 subsequence with sum == 5: [2,3,3]. The subsequence [2,3,3] has 1 subsequence with sum == 5: [2,3,3]. Hence the answer is 2 + 1 + 1 = 4. Example 3: Input: nums = [1,2,3], k = 7 Output: 0 Explanation: 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

Explanation

Here's the solution to the problem:

• High-Level Approach:

  • Use dynamic programming to count the number of subsequences that sum up to k for each possible subsequence of nums.
  • Iterate through all possible subsequences of nums, and for each subsequence, compute its power using DP. Accumulate the power across all subsequences.
  • Utilize memoization within the dynamic programming to optimize performance.

• Complexity:

  • Runtime Complexity: O(n 2ⁿ k), where n is the length of nums and k is the target sum. The O(2ⁿ) comes from the nested loop that goes through all subsequences, and the O(n*k) is for computing the sum of each subsequences.
  • Storage Complexity: O(n * k) due to the memoization table used in the DP.

Code

    def solve():
    def subsequence_sum_count(subsequence, target):
        n = len(subsequence)
        dp = {}

        def count_ways(index, current_sum):
            if (index, current_sum) in dp:
                return dp[(index, current_sum)]

            if current_sum == target:
                return 1

            if index == n or current_sum > target:
                return 0

            include = count_ways(index + 1, current_sum + subsequence[index])
            exclude = count_ways(index + 1, current_sum)

            dp[(index, current_sum)] = include + exclude
            return dp[(index, current_sum)]

        return count_ways(0, 0)

    def sum_power_of_subsequences(nums, k):
        n = len(nums)
        total_power = 0
        MOD = 10**9 + 7

        for i in range(1 << n):
            subsequence = []
            for j in range(n):
                if (i >> j) & 1:
                    subsequence.append(nums[j])

            power = subsequence_sum_count(subsequence, k)
            total_power = (total_power + power) % MOD

        return total_power

    n = int(input())
    nums = list(map(int, input().split()))
    k = int(input())
    print(sum_power_of_subsequences(nums, k))

solve()