3539. Find Sum of Product of Magical Sequences Solution

The problem

You are given two integers, m and k, and an integer array nums.

A sequence of integers seq is called magical if:

seq has a size of m.
0 <= seq[i] < nums.length
The binary representation of 2^seq[0] + 2^seq[1] + ... + 2^{seq[m - 1]} has k set bits.

The array product of this sequence is defined as prod(seq) = (nums[seq[0]] * nums[seq[1]] * ... * nums[seq[m - 1]]).

Return the sum of the array products for all valid magical sequences.

Since the answer may be large, return it modulo 10⁹ + 7.

A set bit refers to a bit in the binary representation of a number that has a value of 1.

Example 1:

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

Output: 991600007

Explanation:

All permutations of [0, 1, 2, 3, 4] are magical sequences, each with an array product of 10¹³.

Example 2:

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

Output: 170

Explanation:

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

Explanation:

The only magical sequence is [0].

Constraints:

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

Approach overview

Problem Overview: You are given constraints that define “magical sequences” and must compute the total sum of products of arrays derived from those sequences. Direct enumeration quickly explodes because the number of valid sequences grows exponentially, so the challenge is counting contributions without constructing every sequence explicitly.

Approach 1: Brute Force Enumeration (Exponential Time, O(k^n) time, O(n) space)

The most direct idea is to generate every valid magical sequence, build the corresponding array, compute its product, and accumulate the sum. This requires recursive backtracking that tries all allowed values at each position while validating the sequence constraints. Although simple to reason about, the number of candidate sequences grows exponentially with the sequence length. Even moderate input sizes make this infeasible. The brute force approach is mainly useful for understanding the structure of the problem or validating optimized solutions on small inputs.

Approach 2: Dynamic Programming with Bitmask State (O(n · 2^m) time, O(2^m) space)

A better strategy is to represent the state of a partially built sequence using a bitmask. Each bit encodes which elements or constraints have already been used. Instead of recomputing the same partial configurations repeatedly, store intermediate results in a DP table keyed by the mask and position. For each transition, update the cumulative product contribution and propagate it to the next valid state. This technique converts repeated exponential exploration into a manageable state-space traversal. Problems involving subsets and constraints often benefit from bitmask based DP combined with memoization.

Approach 3: Combinatorics + Memoized Search (O(states) time, O(states) space)

The optimal approach observes that many sequences contribute identical multiplicative patterns. Instead of enumerating each sequence individually, count how many sequences produce the same structural configuration and multiply that count by the resulting product contribution. Use combinatorial formulas to determine how many ways a configuration can be formed, then perform a memoized DFS over valid states. Each state stores the number of ways and the accumulated product contribution. Memoization avoids recomputation, while combinatorics collapses large groups of equivalent sequences into a single calculation. This hybrid technique blends combinatorics, dynamic programming, and bit-level state encoding to scale to the problem’s limits.

Recommended for interviews: Start by explaining the brute force enumeration to show you understand the definition of magical sequences. Then move quickly to the combinatorics + memoized search approach. Interviewers expect you to recognize the exponential explosion and reduce it using counting techniques and cached subproblems. Demonstrating the transition from naive enumeration to state compression with combinatorics shows strong problem-solving maturity.

Solution

We design a function dfs(i, j, k, st), which represents the number of ways when we are currently processing the i-th element of array 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 dfs(0, m, k, 0).

The execution process of function dfs(i, j, k, st) is as follows:

If k < 0 or i = n and j > 0, it means the current solution is not feasible, return 0.

If i = n, it means we have finished processing array nums. We need to check if there are still set bits in the current carry st, and if so, we need to decrease k. If k = 0 at this point, it means the current solution is feasible, return 1, otherwise return 0.

Otherwise, we enumerate selecting t numbers at position i to fill into the magical sequence (0 leq t leq j). The number of ways to fill t numbers into the magical sequence is \binom{j}{t}, the array product is nums[i]^t, the updated carry is (t + st) >> 1, the updated number of required set bits is k - ((t + st) \& 1), and we recursively call dfs(i + 1, j - t, k - ((t + st) \& 1), (t + st) >> 1). The sum of all t solutions is dfs(i, j, k, st).

To efficiently compute the binomial coefficient \binom{m}{n}, we preprocess the factorial array f and the inverse factorial array g, where f[i] = i! \mod (10^9 + 7) and g[i] = (i!)^{-1} \mod (10^9 + 7). Then \binom{m}{n} = f[m] cdot g[n] cdot g[m - n] \mod (10^9 + 7).

The time complexity is O(n cdot m^3 cdot k) and the space complexity is O(n cdot m^2 cdot k), where n is the length of array nums, and m and k are the parameters in the problem.

Python

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

Java

class Solution {
    static final int N = 31;
    static final long MOD = 1_000_000_007L;
    private static final long[] f = new long[N];
    private static final long[] g = new long[N];
    private Long[][][][] dp;

    static {
        f[0] = 1;
        g[0] = 1;
        for (int i = 1; i < N; ++i) {
            f[i] = f[i - 1] * i % MOD;
            g[i] = qpow(f[i], MOD - 2);
        }
    }

    public static long qpow(long a, long k) {
        long res = 1;
        while (k != 0) {
            if ((k & 1) == 1) {
                res = res * a % MOD;
            }
            a = a * a % MOD;
            k >>= 1;
        }
        return res;
    }

    public static long comb(int m, int n) {
        return f[m] * g[n] % MOD * g[m - n] % MOD;
    }

    public int magicalSum(int m, int k, int[] nums) {
        int n = nums.length;
        dp = new Long[n + 1][m + 1][k + 1][N];
        long ans = dfs(0, m, k, 0, nums);
        return (int) ans;
    }

    private long dfs(int i, int j, int k, int st, int[] nums) {
        if (k < 0 || (i == nums.length && j > 0)) {
            return 0;
        }
        if (i == nums.length) {
            while (st > 0) {
                k -= (st & 1);
                st >>= 1;
            }
            return k == 0 ? 1 : 0;
        }

        if (dp[i][j][k][st] != null) {
            return dp[i][j][k][st];
        }

        long res = 0;
        for (int t = 0; t <= j; t++) {
            int nt = t + st;
            int nk = k - (nt & 1);
            long p = qpow(nums[i], t);
            long tmp = comb(j, t) * p % MOD * dfs(i + 1, j - t, nk, nt >> 1, nums) % MOD;
            res = (res + tmp) % MOD;
        }

        return dp[i][j][k][st] = res;
    }
}

C++

const int N = 31;
const long long MOD = 1'000'000'007;

long long f[N], g[N];

long long qpow(long long a, long long k) {
    long long res = 1;
    while (k) {
        if (k & 1) res = res * a % MOD;
        a = a * a % MOD;
        k >>= 1;
    }
    return res;
}

int init = []() {
    f[0] = g[0] = 1;
    for (int i = 1; i < N; ++i) {
        f[i] = f[i - 1] * i % MOD;
        g[i] = qpow(f[i], MOD - 2);
    }
    return 0;
}();

long long comb(int m, int n) {
    return f[m] * g[n] % MOD * g[m - n] % MOD;
}

class Solution {
    vector<vector<vector<vector<long long>>>> dp;

    long long dfs(int i, int j, int k, int st) {
        if (k < 0 || (i == nums.size() && j > 0)) {
            return 0;
        }
        if (i == nums.size()) {
            while (st > 0) {
                k -= (st & 1);
                st >>= 1;
            }
            return k == 0 ? 1 : 0;
        }

        long long& res = dp[i][j][k][st];
        if (res != -1) {
            return res;
        }

        res = 0;
        for (int t = 0; t <= j; ++t) {
            int nt = t + st;
            int nk = k - (nt & 1);
            long long p = qpow(nums[i], t);
            long long tmp = comb(j, t) * p % MOD * dfs(i + 1, j - t, nk, nt >> 1) % MOD;
            res = (res + tmp) % MOD;
        }
        return res;
    }

public:
    int magicalSum(int m, int k, vector<int>& nums) {
        int n = nums.size();
        this->nums = nums;
        dp.assign(n + 1, vector<vector<vector<long long>>>(m + 1, vector<vector<long long>>(k + 1, vector<long long>(N, -1))));
        return dfs(0, m, k, 0);
    }

private:
    vector<int> nums;
};

const N = 31
const MOD = 1_000_000_007

var f [N]int64
var g [N]int64

func init() {
	f[0], g[0] = 1, 1
	for i := 1; i < N; i++ {
		f[i] = f[i-1] * int64(i) % MOD
		g[i] = qpow(f[i], MOD-2)
	}
}

func qpow(a, k int64) int64 {
	res := int64(1)
	for k > 0 {
		if k&1 == 1 {
			res = res * a % MOD
		}
		a = a * a % MOD
		k >>= 1
	}
	return res
}

func comb(m, n int) int64 {
	if n < 0 || n > m {
		return 0
	}
	return f[m] * g[n] % MOD * g[m-n] % MOD
}

func magicalSum(m int, k int, nums []int) int {
	n := len(nums)
	dp := make([][][][]int64, n+1)
	for i := 0; i <= n; i++ {
		dp[i] = make([][][]int64, m+1)
		for j := 0; j <= m; j++ {
			dp[i][j] = make([][]int64, k+1)
			for l := 0; l <= k; l++ {
				dp[i][j][l] = make([]int64, N)
				for s := 0; s < N; s++ {
					dp[i][j][l][s] = -1
				}
			}
		}
	}

	var dfs func(i, j, k, st int) int64
	dfs = func(i, j, k, st int) int64 {
		if k < 0 || (i == n && j > 0) {
			return 0
		}
		if i == n {
			for st > 0 {
				k -= st & 1
				st >>= 1
			}
			if k == 0 {
				return 1
			}
			return 0
		}
		if dp[i][j][k][st] != -1 {
			return dp[i][j][k][st]
		}
		res := int64(0)
		for t := 0; t <= j; t++ {
			nt := t + st
			nk := k - (nt & 1)
			p := qpow(int64(nums[i]), int64(t))
			tmp := comb(j, t) * p % MOD * dfs(i+1, j-t, nk, nt>>1) % MOD
			res = (res + tmp) % MOD
		}
		dp[i][j][k][st] = res
		return res
	}

	return int(dfs(0, m, k, 0))
}

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Complexity analysis

Approach	Time	Space	When to Use
Brute Force Enumeration	O(k^n)	O(n)	Small inputs or for validating correctness of optimized approaches
Bitmask Dynamic Programming	O(n · 2^m)	O(2^m)	When sequence constraints map naturally to subset states
Combinatorics + Memoized Search	O(states)	O(states)	Best general solution for large constraints; avoids enumerating every sequence

Frequently asked questions

Is Find Sum of Array Product of Magical Sequences easy or hard?

The problem is classified as Hard because it combines multiple advanced techniques: combinatorics, dynamic programming, and bitmask state compression. Solving it efficiently requires recognizing repeated substructures and replacing brute force enumeration with counted contributions.

Find Sum of Array Product of Magical Sequences Python/Java solution

Implement the solution using a memoized DFS that tracks the current state (often encoded as a bitmask or index combination). Use a dictionary or map for memoization and apply combinatorial counting to aggregate the number of ways each configuration appears. The same logic translates cleanly across Python, Java, C++, and Go.

How to solve Find Sum of Array Product of Magical Sequences in O(n)?

An O(n) solution is generally not achievable because the problem requires exploring combinations of sequence states or subsets. The practical optimization is reducing exponential enumeration using memoized search and combinatorial counting. This avoids generating every sequence while still accounting for all valid contributions.

What is the best approach for Find Sum of Array Product of Magical Sequences?

The most efficient strategy combines combinatorics with memoized search. Instead of enumerating every magical sequence, group equivalent configurations and count how many sequences produce the same product contribution. Memoization caches intermediate states so each configuration is computed once, reducing the exponential search space to a manageable number of states.

Is Find Sum of Array Product of Magical Sequences asked at Google/Amazon/Meta?

Problems combining combinatorics, bitmask states, and dynamic programming frequently appear in interviews at companies like Google, Amazon, and Meta. Variants involving counting sequences, subset states, or combinatorial contributions are especially common in senior-level algorithm interviews.

What data structure is used in Find Sum of Array Product of Magical Sequences?

Typical implementations rely on hash maps or arrays for memoization, bitmasks to represent subset states, and recursion or DFS for exploring configurations. These structures allow efficient caching of intermediate results while computing contributions from each valid configuration.

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

The optimized combinatorics + memoization solution runs in O(states) time where states represent unique configuration or bitmask combinations explored by the search. Space complexity is also O(states) due to caching. A naive brute force approach would require exponential time O(k^n), which is not practical for typical constraints.

Find Sum of Array Product of Magical Sequences solution and explanation

The problem

Approach overview

Solution

Complexity analysis

Frequently asked questions

Comments

Find Sum of Array Product of Magical Sequences solution and explanation

The problem

Approach overview

Solution

Complexity analysis

Frequently asked questions

Comments

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