#3539

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Find Sum of Array Product of Magical Sequences

Use state transition dynamic programming to count magical index sequences and accumulate weighted products without enumerating exponential orderings.

array math dynamic programming bit manipulation combinatorics

Problem Statement

You are given integers m and k and an array nums. Build sequences of length m using indices of nums, and define the value of a sequence as the product of the selected entries in order. A sequence is magical when the binary carry behavior of its chosen indices makes the final power-sum representation contain exactly k set bits, so repeated picks can merge upward instead of behaving like simple distinct selections.

Return the sum of sequence products over every magical sequence. This is why the examples split the way they do: when m = 2 and k = 2, any ordered pair with different indices stays as two set bits, while when m = 5 and k = 5 with five usable positions, only permutations with no repeated index avoid carry merges and keep five set bits.

Examples

Example 1

Input: m = 5, k = 5, nums = [1,10,100,10000,1000000]

Output: 991600007

All permutations of [0, 1, 2, 3, 4] are magical sequences, each with an array product of 10 13 .

Example 2

Input: m = 2, k = 2, nums = [5,4,3,2,1]

Output: 170

The magical sequences are [0, 1] , [0, 2] , [0, 3] , [0, 4] , [1, 0] , [1, 2] , [1, 3] , [1, 4] , [2, 0] , [2, 1] , [2, 3] , [2, 4] , [3, 0] , [3, 1] , [3, 2] , [3, 4] , [4, 0] , [4, 1] , [4, 2] , and [4, 3] .

Example 3

Input: m = 1, k = 1, nums = [28]

Output: 28

The only magical sequence is [0] .

Constraints

1 <= k <= m <= 30
1 <= nums.length <= 50
1 <= nums[i] <= 108

Solution Approach

Model the hidden condition as binary carry transitions

The key rewrite is to view a sequence by the multiset of chosen indices, not by the raw product first. Each pick adds one copy of 2^i, and equal indices generate carries just like binary addition. After all m picks, the sequence is magical only if the final carried form has popcount k. That turns the problem into counting ways to distribute picks across indices while preserving product contribution.

Run state transition DP over index positions and carry state

Process nums positions from low index to high index. For each position, decide how many times that index is used, multiply the current contribution by nums[i]^count, and push the resulting carry into the next position. The DP state stores how many picks have been used so far and the carry entering the current bit. When you advance, the local bit parity determines whether this position contributes a set bit to the final popcount, which lets you accumulate how many ones have appeared.

Aggregate product sums instead of separate counting and scoring

Do not first count magical sequences and then try to recover products. The transition itself should add weighted sums, because every choice of count at index i contributes both combinatorial placements and a multiplicative factor from nums[i]. Precompute combinations for distributing positions in the ordered sequence and powers of nums[i] for repeated use. The final answer is the DP total whose used length is m and whose finished carried representation has exactly k set bits.

Complexity Analysis

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

The complexity depends on how tightly you compress the state, but the intended solution is polynomial in nums.length, m, k, and the bounded carry range, which is feasible because m <= 30 keeps the carry dimension small. Space is the number of reachable DP states after rolling over index positions. The main trade-off is between a cleaner DP with more dimensions and a compressed DP that reduces memory but needs more careful transitions.

What Interviewers Usually Probe

They mention Dynamic Programming because the real obstacle is carry propagation, not plain subset enumeration.
If they ask why permutations appear in the first example, they want you to connect distinct picks with a no-carry popcount of k.
If they push on brute force, they expect you to explain that product accumulation must be folded into the state transitions.

Common Pitfalls or Variants

Common pitfalls

Treating magical sequences as simple distinct-index selections fails when repeated indices can still become valid after carries.
Counting valid sequences first and multiplying later loses the exact weighted sum structure from nums powers.
Forgetting ordered placements undercounts badly because the same multiset of indices can form many different sequences.

Follow-up variants

Change the target from exactly k set bits to at most k, which only alters the final acceptance rule on the carried form.
Ask for the number of magical sequences instead of the product sum, which removes nums powers but keeps the same carry DP skeleton.
Restrict each index usage count, which adds another transition filter on how many copies of each position can be chosen.

FAQ

What is the main insight for Find Sum of Array Product of Magical Sequences?

The main insight is to reinterpret repeated index selections as binary additions of powers of two. Once you track carries across indices, the magical condition becomes a popcount target, and the problem turns into state transition dynamic programming.

Why is brute force impossible here?

A raw search must inspect an enormous number of length-m ordered sequences, and each one also needs a validity check. With m up to 30, the useful structure is in how counts per index collapse through carries, so exhaustive generation is the wrong axis.

Why does the solution need combinatorics as well as DP?

When an index is chosen multiple times, you need the number of ways to place those choices among the m sequence positions. Those multinomial-style placement counts are what convert a frequency decision inside the DP into the correct number of ordered sequences.

What does the state transition dynamic programming state usually track?

A practical state tracks the current index position, how many picks have been used, the carry entering this bit, and how many set bits have already appeared in the normalized carried form. The transition chooses a usage count for the current index and updates both carry and popcount contribution.

What is the easiest bug to make on this problem?

The easiest bug is assuming that repeated indices are automatically invalid. In this problem, repeats can merge upward through carries, so validity depends on the final normalized bit count, not on whether the original sequence had duplicates.

terminal

Solution

Solution 1: Combinatorics + Memoized Search

We design a function $\text{dfs}(i, j, k, st)$, which represents the number of ways when we are currently processing the $i$-th element of array $\textit{nums}$, still need to select numbers from the remaining $j$ positions to fill into the magical sequence, still need to satisfy having $k$ set bits in binary form, and the current carry from the previous bit is $st$. Then the answer is $\text{dfs}(0, m, k, 0)$.

mx = 30
mod = 10**9 + 7
f = [1] + [0] * mx
g = [1] + [0] * mx

for i in range(1, mx + 1):
    f[i] = f[i - 1] * i % mod
    g[i] = pow(f[i], mod - 2, mod)


def comb(m: int, n: int) -> int:
    return f[m] * g[n] * g[m - n] % mod


class Solution:
    def magicalSum(self, m: int, k: int, nums: List[int]) -> int:
        @cache
        def dfs(i: int, j: int, k: int, st: int) -> int:
            if k < 0 or (i == len(nums) and j > 0):
                return 0
            if i == len(nums):
                while st:
                    k -= st & 1
                    st >>= 1
                return int(k == 0)
            res = 0
            for t in range(j + 1):
                nt = t + st
                p = pow(nums[i], t, mod)
                nk = k - (nt & 1)
                res += comb(j, t) * p * dfs(i + 1, j - t, nk, nt >> 1)
                res %= mod
            return res

        ans = dfs(0, m, k, 0)
        dfs.cache_clear()
        return ans

Continue Practicing

#3444 minimum increments for target multiples in an array #2572 count the number of square free subsets #3533 concatenated divisibility