LeetCode 3539 - Find Sum of Array Product of Magical Sequences

This problem asks us to compute the sum of array products for all magical sequences of a given length. A magical sequence is defined as a sequence of indices into the input array nums of length m, such that when you sum 2 raised to each element of the sequence, the resulting…

leetcodehardarraymathdynamic-programmingbit-manipulationcombinatoricsbitmask

LeetCode Problem 3539

Difficulty: 🔴 Hard
Topics: Array, Math, Dynamic Programming, Bit Manipulation, Combinatorics, Bitmask

Solution

Problem Understanding

This problem asks us to compute the sum of array products for all magical sequences of a given length. A magical sequence is defined as a sequence of indices into the input array nums of length m, such that when you sum 2 raised to each element of the sequence, the resulting number in binary has exactly k set bits. The array product of a sequence is the product of the elements in nums corresponding to the indices in the sequence. Finally, the sum of these products across all magical sequences is returned modulo 10^9 + 7.

In other words, we are selecting sequences of indices from nums (indices range from 0 to len(nums)-1) with length m, and only keeping those sequences whose "binary exponent sum" has exactly k bits set. Then, for each valid sequence, we compute the product of the numbers at those indices, sum them up, and return the result modulo 10^9 + 7.

The constraints tell us that m and k can go up to 30, while the array length is at most 50. The values in nums can be large (up to 10^8). This means a naive enumeration of all sequences (len(nums)^m) is impossible because it would be astronomically large for m = 30.

Important edge cases include sequences of length 1, sequences where k equals m, and arrays with repeated or very large numbers. The problem guarantees k ≤ m, so no impossible counts need handling.

Approaches

The brute-force approach would generate all sequences of length m using indices from 0 to len(nums)-1. For each sequence, we would compute the sum of 2^index values, count the number of set bits in its binary representation, and if it matches k, multiply the corresponding numbers from nums to compute the product. This would give the correct answer, but the complexity is O(n^m) where n = len(nums), which is completely infeasible for large m or n.

The key insight is to use dynamic programming with bitmasking. Each number i contributes 2^i to a sum, and we care about the count of set bits in that sum. Instead of generating all sequences, we can track, for every possible bitmask, the sum of products for sequences that form that bitmask. This allows us to build solutions incrementally: for sequences of length l, we can extend them by one more index and update the corresponding mask. Using collections.Counter or a dictionary keyed by masks lets us avoid redundant computation.

Approach Time Complexity Space Complexity Notes
Brute Force O(n^m) O(1) Enumerates all sequences and filters by set bits
Optimal (DP + Bitmask) O(m * 2^n) O(2^n) Tracks sums by bitmask; feasible because n ≤ 50

Algorithm Walkthrough

  1. Initialize a dp dictionary to store intermediate results. The key is the bitmask representing the sum of 2^index for the current sequence, and the value is the sum of products that produce this bitmask.
  2. Start with sequences of length 0. This corresponds to a single entry {0: 1} in the dictionary (mask 0, product 1) because multiplying nothing is the multiplicative identity.
  3. For each sequence length from 1 to m, iterate over the current dp states. For each index i in nums, compute the new mask by performing a bitwise OR with 1 << i. Multiply the previous product sum by nums[i] and add it to the new mask in the next dp dictionary.
  4. Replace the old dp with the new dictionary after processing all indices for this sequence length.
  5. After reaching sequences of length m, sum the values in dp for masks that have exactly k set bits.
  6. Return the sum modulo 10^9 + 7.

Why it works: At each step, dp[mask] accurately represents the sum of array products for all sequences that produce that mask. By extending sequences iteratively and updating masks with bitwise OR, all sequences are counted exactly once. Filtering by bit counts at the end ensures we only include magical sequences.

Python Solution

from typing import List
from collections import defaultdict

class Solution:
    def magicalSum(self, m: int, k: int, nums: List[int]) -> int:
        MOD = 10**9 + 7
        n = len(nums)
        dp = defaultdict(int)
        dp[0] = 1
        
        for _ in range(m):
            next_dp = defaultdict(int)
            for mask, val in dp.items():
                for i in range(n):
                    new_mask = mask | (1 << i)
                    next_dp[new_mask] = (next_dp[new_mask] + val * nums[i]) % MOD
            dp = next_dp
        
        result = 0
        for mask, val in dp.items():
            if bin(mask).count('1') == k:
                result = (result + val) % MOD
        
        return result

The Python implementation uses defaultdict to automatically handle missing keys. The dp dictionary tracks sums of products keyed by masks, updated iteratively for each sequence length. The final sum filters masks with exactly k bits.

Go Solution

func magicalSum(m int, k int, nums []int) int {
    const MOD = 1_000_000_007
    n := len(nums)
    dp := map[int]int{0: 1}

    for step := 0; step < m; step++ {
        nextDP := make(map[int]int)
        for mask, val := range dp {
            for i := 0; i < n; i++ {
                newMask := mask | (1 << i)
                nextVal := (int64(val) * int64(nums[i])) % MOD
                nextDP[newMask] = (nextDP[newMask] + int(nextVal)) % MOD
            }
        }
        dp = nextDP
    }

    result := 0
    for mask, val := range dp {
        if bitsSet(mask) == k {
            result = (result + val) % MOD
        }
    }
    return result
}

func bitsSet(x int) int {
    count := 0
    for x > 0 {
        count += x & 1
        x >>= 1
    }
    return count
}

The Go version uses a map to track DP states, similar to Python. Integer overflow is avoided using int64 during multiplication. A helper function bitsSet counts set bits in a mask.

Worked Examples

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

At step 0, dp = {0:1}.

Step 1 extends sequences with one index:

mask=0 -> add indices 0-4
dp = {
  1: 1*1=1,
  2: 1*10=10,
  4: 1*100=100,
  8: 1*10000=10000,
  16: 1*1000000=1000000
}

Step 2 extends each mask again, multiplying the previous sums. Continue until length 5. At the end, select masks with exactly 5 set bits, sum their products modulo 10^9 + 7 → 991600007.

Example 2 and 3 follow similarly, updating dp iteratively and summing masks with the correct number of set bits.

Complexity Analysis

Measure Complexity Explanation
Time O(m * n * 2^n) For each sequence length, iterate all current masks (≤2^n) and each index (n)
Space O(2^n) Store sum of products for all possible masks

This complexity is feasible because n ≤ 50, but m ≤ 30 means the outer loop is limited. The number of masks grows exponentially, but is manageable for this constraint.

Test Cases

# Provided examples
assert Solution().magicalSum(5, 5, [1,10,100,10000,1000000]) == 991600007  # Example 1
assert Solution().magicalSum(2, 2, [5,4,3,2,1]) == 170  # Example 2
assert Solution().magicalSum(1, 1, [28]) == 28  # Example 3

# Edge cases
assert Solution().magicalSum(1, 1, [1]) == 1  # single element
assert Solution().magicalSum(2, 1, [1,1]) == 4  # repeated numbers
assert Solution().magicalSum(3, 2, [1,2,3]) >= 0  # larger m
Test Why
m=5, k=5, nums=[1,10,100,10000,1000000] checks large products and correct masking