Input: m =5, k =5, nums =[1,10,100,10000,1000000]Output: 991600007Explanation:
All permutations of `[0, 1, 2, 3, 4]` are magical sequences, each with an
array product of 1013.
The most straightforward way is to generate every ordered sequence seq of length m where each element is an index in [0..n-1] (where n = nums.length), compute the integer S = sum(2^{seq[i]}) for that sequence, check if popcount(S) == k, and if so add the product nums[seq[0]] * ... * nums[seq[m-1]] to the running total (modulo 1e9+7). This is conceptually simple but quickly becomes infeasible because there are n^m ordered sequences.
Recursively (or with m nested loops) enumerate every ordered sequence seq of length m. At each recursion depth you pick an index in 0..n-1.
Maintain the running product of chosen nums values and the running integer sum of powers-of-two (S). When you reach depth m, compute popcount(S). If it equals k, add the product to the answer modulo MOD.
Simple pruning: if at any point you detect that even with the best/worst possible remaining contributions you cannot reach k set bits, you could skip (but in the naive approach this pruning is weak).
Why this is insufficient in practice
The number of sequences is n^m. With constraints up to n = 50 and m = 30 the naive method is astronomically large and unusable.
Even for moderate values (e.g., n=10, m=10) it’s usually too slow.
Why memoization or combinatorics is needed (lead-in to Method 2)
There is a great deal of repeated work in the naive enumeration: different orderings that share the same counts of each index produce the same product up to permutation multiplicity, and the binary addition behavior depends only on counts (and carries), not on order.
Recasting the problem to iterate over counts of each index (a multiset view) and accounting for permutations with combinatorial coefficients reduces the search space dramatically. Likewise, memoizing states keyed by (remaining, odd_needed, idx, carry) (or equivalent) collapses repeated subproblems.
We’ll implement the memoized DP that uses these observations in Method 2, and then provide a bottom-up DP variant in Method 3.
⏰ Time complexity:O(n^m * m) – Enumerating every ordered sequence (n^m) and computing product/popcount for each sequence costs another O(m) per sequence; hence the naive upper bound is O(n^m * m).
🧺 Space complexity:O(m) – recursion stack and O(m) temporary workspace for product/bit-sum calculations.
We can avoid enumerating every ordered sequence by processing indices one-by-one and choosing how many times to include each index (0..remaining). The binary sum’s behavior depends only on counts and carries per binary digit, and permutations of identical choices can be accounted for with combinatorics (binomial coefficients). Memoizing states keyed by (remaining, odd_needed, idx, carry) collapses repeated subproblems.
Precompute combinatorics (either factorials/inverses or a Pascal table C) up to m, and optionally precompute pow(nums[i], t, MOD) for t in 0..m to avoid repeated exponentiation.
Define a memoized recursion dfs(remaining, odd_needed, idx, carry) that returns the total sum of array-products modulo MOD for selecting exactly remaining items from indices idx..n-1 with current carry and still needing odd_needed set bits overall.
For the current idx, iterate take from 0..remaining:
ways = C[remaining][take]
prod_contrib = nums[idx]^take % MOD
new_carry = carry + take
lsb = new_carry & 1 (this bit contributes to the current bitcount)
recurse on dfs(remaining - take, odd_needed - lsb, idx+1, new_carry >> 1) and multiply by ways * prod_contrib.
Sum these contributions modulo MOD and memoize the result.
dfs(remaining, odd_needed, idx, carry) returns the total modular sum of products for selecting remaining more elements from indices idx..n-1 with pending carry and odd_needed remaining set bits to form.
⏰ Time complexity:O(n * m * m * states) – each memo state iterates take up to m, and the number of distinct states is bounded by O(n * m * maxCarry) where maxCarry is O(m); overall polynomial in n and m versus exponential n^m.
🧺 Space complexity:O(n * m * maxCarry) – for memoization and recursion stack.
Notes: For production performance, pack the memo key into a fixed-size integer or use an iterative 4D DP table when memory allows (see Method 3 planned next).
We can convert the memoized recursion into an iterative DP that fills a 4D table: dp[pos][bits][carry][chosen] — the total sum of products (mod MOD) after processing the first pos indices, having accumulated bits set bits so far, with carry pending and having chosen chosen elements in total. Transitions enumerate how many copies cnt of the current index to take and update the next state’s bits/carry/chosen accordingly while multiplying by combinatorial placement counts and power contributions.
Precompute binomial coefficients C[a][b] for 0..m and precompute pw[i][t] = nums[i]^t % MOD for t in 0..m.
Initialize dp[0][0][0][0] = 1 and iterate pos from 0 to n-1.
For each reachable state (bits, carry, chosen) at pos, iterate cnt from 0..remaining (remaining = m - chosen) and update dp[pos+1][new_bits][new_carry][chosen+cnt] += dp[pos][bits][carry][chosen] * C[remaining][cnt] * pw[pos][cnt].
After filling dp[n], aggregate the answer by considering leftover carry values: for each carry, compute carry_bits = popcount(carry) and add dp[n][k - carry_bits][carry][m] if carry_bits <= k.
classSolution {
privatestaticfinalint MOD = 1_000_000_007;
publicintmagicalSum(int m, int k, int[] nums) {
int n = nums.length;
long[][] C =newlong[m+1][m+1];
for (int i = 0; i <= m; i++) {
C[i][0]= C[i][i]= 1;
for (int j = 1; j < i; j++) C[i][j]= (C[i-1][j-1]+ C[i-1][j]) % MOD;
}
long[][] pw =newlong[n][m+1];
for (int i = 0; i < n; i++) {
pw[i][0]= 1;
for (int t = 1; t <= m; t++) pw[i][t]= pw[i][t-1]* nums[i]% MOD;
}
long[][][][] dp =newlong[n+1][k+1][m+1][m+1];
dp[0][0][0][0]= 1;
for (int pos = 0; pos < n; pos++) {
for (int bits = 0; bits <= k; bits++) {
for (int carry = 0; carry <= m; carry++) {
for (int chosen = 0; chosen <= m; chosen++) {
long cur = dp[pos][bits][carry][chosen];
if (cur == 0) continue;
int remaining = m - chosen;
for (int cnt = 0; cnt <= remaining; cnt++) {
int total = carry + cnt;
int newBits = bits + (total & 1);
int newCarry = total >> 1;
if (newBits <= k && newCarry <= m) {
long add = cur * C[remaining][cnt]% MOD * pw[pos][cnt]% MOD;
dp[pos+1][newBits][newCarry][chosen+cnt]= (dp[pos+1][newBits][newCarry][chosen+cnt]+ add) % MOD;
}
}
}
}
}
}
long res = 0;
for (int carry = 0; carry <= m; carry++) {
int carryBits = Integer.bitCount(carry);
if (carryBits <= k) {
res = (res + dp[n][k - carryBits][carry][m]) % MOD;
}
}
return (int) res;
}
}
classSolution {
privateval MOD = 1_000_000_007L
funmagicalSum(m: Int, k: Int, nums: IntArray): Int {
val n = nums.size
val C = Array(m + 1) { LongArray(m + 1) }
for (i in0..m) {
C[i][0] = 1L C[i][i] = 1Lfor (j in1 until i) C[i][j] = (C[i-1][j-1] + C[i-1][j]) % MOD
}
val pw = Array(n) { LongArray(m + 1) }
for (i in0 until n) {
pw[i][0] = 1Lfor (t in1..m) pw[i][t] = pw[i][t-1] * nums[i] % MOD
}
// dp[pos][bits][carry][chosen]
val dp = Array(n+1) { Array(k+1) { Array(m+1) { LongArray(m+1) } } }
dp[0][0][0][0] = 1Lfor (pos in0 until n) {
for (bits in0..k) {
for (carry in0..m) {
for (chosen in0..m) {
val cur = dp[pos][bits][carry][chosen]
if (cur ==0L) continueval remaining = m - chosen
for (cnt in0..remaining) {
val total = carry + cnt
val newBits = bits + (total and 1)
val newCarry = total shr 1if (newBits <= k && newCarry <= m) {
val add = cur * C[remaining][cnt] % MOD * pw[pos][cnt] % MOD
dp[pos+1][newBits][newCarry][chosen+cnt] = (dp[pos+1][newBits][newCarry][chosen+cnt] + add) % MOD
}
}
}
}
}
}
var res = 0Lfor (carry in0..m) {
val carryBits = Integer.bitCount(carry)
if (carryBits <= k) res = (res + dp[n][k - carryBits][carry][m]) % MOD
}
return res.toInt()
}
}
from typing import List
classSolution:
MOD =10**9+7defmagicalSum(self, m: int, k: int, nums: List[int]) -> int:
n = len(nums)
# binomial table C = [[0] * (m +1) for _ in range(m +1)]
for i in range(m +1):
C[i][0] =1 C[i][i] =1for j in range(1, i):
C[i][j] = (C[i-1][j-1] + C[i-1][j]) % self.MOD
# pow table pw = [[1] * (m +1) for _ in range(n)]
for i in range(n):
for t in range(1, m +1):
pw[i][t] = pw[i][t-1] * nums[i] % self.MOD
# dp[pos][bits][carry][chosen] dp = [ [ [ [0] * (m +1) for _ in range(m +1)] for _ in range(k +1)] for _ in range(n +1) ]
dp[0][0][0][0] =1for pos in range(n):
for bits in range(k +1):
for carry in range(m +1):
for chosen in range(m +1):
cur = dp[pos][bits][carry][chosen]
if cur ==0:
continue remaining = m - chosen
for cnt in range(remaining +1):
total = carry + cnt
new_bits = bits + (total &1)
new_carry = total >>1if new_bits <= k and new_carry <= m:
add = cur * C[remaining][cnt] % self.MOD * pw[pos][cnt] % self.MOD
dp[pos+1][new_bits][new_carry][chosen+cnt] = (dp[pos+1][new_bits][new_carry][chosen+cnt] + add) % self.MOD
res =0for carry in range(m +1):
carry_bits = bin(carry).count('1')
if carry_bits <= k:
res = (res + dp[n][k - carry_bits][carry][m]) % self.MOD
return res
⏰ Time complexity:O(n * k * m^3) – we iterate over pos (n), bits (k), carry (O(m)), chosen (O(m)) and for each state try up to O(m) values of cnt, yielding O(n * k * m^3).
🧺 Space complexity:O(n * k * m^2) – the 4D DP table has size O(n * k * (m+1) * (m+1)).