LeetCode #3539 — HARD

Find Sum of Array Product of Magical Sequences

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

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 2seq[0] + 2seq[1] + ... + 2seq[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
Patterns Used
📊Dynamic Programming→⚡Bit Manipulation→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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 2seq[0] + 2seq[1] + ... + 2seq[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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Dynamic Programming · Bit Manipulation

Example 1

5
5
[1,10,100,10000,1000000]

Example 2

2
2
[5,4,3,2,1]

Example 3

1
1
[28]

Related Problems

  • Product of Array Except Self (product-of-array-except-self)
  • Smallest Number With All Set Bits (smallest-number-with-all-set-bits)
Step 02

Core Insight

What unlocks the optimal approach

  • Use Dynamic Programming
  • Let <code>dp[i][j][mask]</code> be the state after choosing <code>i</code> numbers (indices)
  • The partial sum <code>S = 2^(seq[0]) + 2^(seq[1]) + ... + 2^(seq[i - 1])</code> has produced exactly <code>j</code> set bits once you’ve fully propagated any carries
  • The <code>mask</code> represents the "window" of lower-order bits from <code>S</code> that have not yet been fully processed (i.e. bits that might later create new set bits when additional terms are added)
  • Use combinatorics
  • How many ways are there to permute a sequence of entities where some are repetitive?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3539: Find Sum of Array Product of Magical Sequences
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;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n · m^3 · k)
Space
O(n · m^2 · k)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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