How do you solve Find Sum of Array Product of Magical Sequences on LeetCode?

Find Sum of Array Product of Magical Sequences (LeetCode #3539) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * k * n) time complexity and O(m * k) space complexity. Key insight: Understanding bit manipulation is crucial for counting set bits.

What is the time complexity of Find Sum of Array Product of Magical Sequences?

The optimal solution for Find Sum of Array Product of Magical Sequences runs in O(m * k * n) time and uses O(m * k) space. The complexity is driven by the nested loops over m, k, and n, leading to a polynomial time complexity.

What is the brute force solution for Find Sum of Array Product of Magical Sequences?

The brute force approach for Find Sum of Array Product of Magical Sequences is Brute Force, with O(n^m) time complexity and O(1) space complexity. Generate all possible sequences of length m from indices of nums. Check each sequence for the number of set bits in its binary representation and calculate the product if valid. The optimal Optimal Solution approach improves this to O(m * k * n).

What are the interview tips for Find Sum of Array Product of Magical Sequences?

Clarify the problem constraints and requirements. Discuss your thought process and approach before coding. Test your solution with edge cases.

What are common mistakes when solving Find Sum of Array Product of Magical Sequences?

Overlooking the modulo operation in calculations. Not properly counting set bits in binary representation.

#3539

Find Sum of Array Product of Magical Sequences

Hard

ArrayMathDynamic ProgrammingBit ManipulationCombinatoricsBitmaskDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^m)	O(m * k * n)
Space	O(1)	O(m * k)

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Intuition

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

Use dynamic programming to build up solutions for sequences of increasing length while tracking the number of set bits.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table dp[i][j][mask] to store the sum of products for sequences of length i with j set bits and the last used mask.
2Step 2: Iterate through each number and update the DP table based on previous states, ensuring the set bits count matches k.
3Step 3: Sum all valid products from the DP table where the sequence length is m and the set bits count is k.

solution.py12 lines

1def magical_sum(m, k, nums):
2    MOD = 10**9 + 7
3    dp = [[0] * (k + 1) for _ in range(m + 1)]
4    dp[0][0] = 1
5    for num in nums:
6        for i in range(m, 0, -1):
7            for j in range(k, -1, -1):
8                if dp[i-1][j] > 0:
9                    new_set_bits = bin(num).count('1')
10                    if j + new_set_bits <= k:
11                        dp[i][j + new_set_bits] = (dp[i][j + new_set_bits] + dp[i-1][j] * num) % MOD
12    return sum(dp[m][k]) % MOD

ℹ

Complexity note: The complexity is driven by the nested loops over m, k, and n, leading to a polynomial time complexity.

1Understanding bit manipulation is crucial for counting set bits.
2Dynamic programming can significantly reduce the number of computations needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.