Find Sum of Array Product of Magical Sequences

Find Sum of Array Product of Magical Sequences - Problem

Array Math Dynamic Programming Bit Manipulation Combinatorics Bitmask Hard

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 10^9 + 7.

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

Input & Output

Example 1 — Basic Case

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

› Output: 9

💡 Note: Valid magical sequences: [0,1] (2^0+2^1=3, 2 set bits, product=2*3=6) and [1,2] (2^1+2^2=6, 2 set bits, product=3*1=3). Sum = 6+3 = 9.

Example 2 — Single Element

$ Input: nums = [5], m = 1, k = 1

› Output: 5

💡 Note: Only sequence [0] is valid: 2^0=1 has 1 set bit, product=5.

Example 3 — No Valid Sequences

$ Input: nums = [1,2], m = 2, k = 1

› Output: 0

💡 Note: No sequence of length 2 can have exactly 1 set bit since we need 2^i + 2^j where i≠j, which always has at least 2 set bits.

Constraints

1 ≤ nums.length ≤ 20
1 ≤ nums[i] ≤ 10⁹
1 ≤ m ≤ nums.length
1 ≤ k ≤ m

Visualization

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Understanding the Visualization

Input

Array nums=[2,3,1], m=2, k=2

Process

Find sequences where power sum has exactly k set bits

Output

Sum of all valid sequence products

Key Takeaway

🎯 Key Insight: Use dynamic programming to efficiently count magical sequences by tracking length and set bit combinations

Asked in

G Google 25 M Microsoft 18 f Meta 15 a Amazon 12

This problem requires finding all magical sequences of length m where the sum of powers of 2 has exactly k set bits. The key insight is using dynamic programming to track sequences by length and set bit count. The optimal approach uses DP with state dp[length][setBits] and processes each array element to build valid sequences. Time: O(n×m×k), Space: O(m×k).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Enumeration	O(n^m × m)	O(m)	Generate all possible sequences of length m and check which ones are magical
Optimized DP with Bit Manipulation	O(n × m × k)	O(m × k)	Use efficient bit manipulation and optimized DP state transitions
Dynamic Programming with Memoization	O(n × m × 2^n)	O(n × m × 2^n)	Use memoization to avoid recalculating the same subproblems

Brute Force Enumeration — Algorithm Steps

Generate all sequences of length m from array indices
For each sequence, compute 2^seq[0] + 2^seq[1] + ... + 2^seq[m-1]
Count set bits in the sum
If exactly k set bits, calculate product and add to result

Visualization

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Step-by-Step Walkthrough

Generate

Create all sequences of length m

Check

Validate power sum has k set bits

Sum

Add valid products to result

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MOD 1000000007

int countSetBits(long long x) {
    int count = 0;
    while (x) {
        count += x & 1;
        x >>= 1;
    }
    return count;
}

long long backtrack(int* nums, int n, int* seq, int seqLen, int m, int k) {
    if (seqLen == m) {
        long long powerSum = 0;
        for (int i = 0; i < seqLen; i++) {
            powerSum += (1LL << seq[i]);
        }
        if (countSetBits(powerSum) == k) {
            long long product = 1;
            for (int i = 0; i < seqLen; i++) {
                product = (product * nums[seq[i]]) % MOD;
            }
            return product;
        }
        return 0;
    }
    
    long long result = 0;
    for (int i = 0; i < n; i++) {
        seq[seqLen] = i;
        result = (result + backtrack(nums, n, seq, seqLen + 1, m, k)) % MOD;
    }
    return result;
}

int solution(int* nums, int numsSize, int m, int k) {
    int* seq = (int*)malloc(m * sizeof(int));
    long long result = backtrack(nums, numsSize, seq, 0, m, k);
    free(seq);
    return (int)result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int m, k;
    scanf("%d", &m);
    scanf("%d", &k);
    
    printf("%d\n", solution(nums, numsSize, m, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n^m × m)

Generate n^m sequences, each taking O(m) time to validate

✓ Linear Growth

Space Complexity

O(m)

Space for recursion stack and current sequence

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Enumeration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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