Minimum Non-Zero Product of the Array Elements - Problem

You are given a positive integer p. This creates an array containing all integers from 1 to 2p - 1, but here's the twist: you can perform bit swapping operations!

The Operation: Choose any two numbers and swap corresponding bits between them. For example, if you have x = 1101 and y = 0011, swapping the 2nd bit gives you x = 1111 and y = 0001.

Your Goal: Find the minimum non-zero product of all array elements after performing any number of bit swapping operations. Return the result modulo 109 + 7.

Key Insight: The beauty of this problem lies in understanding that bit swapping allows you to redistribute bits optimally to minimize the overall product while keeping it non-zero!

Input & Output

example_1.py โ€” Basic Case
$ Input: p = 1
โ€บ Output: 1
๐Ÿ’ก Note: For p=1, array is [1]. No operations needed, product is 1.
example_2.py โ€” Small Array
$ Input: p = 2
โ€บ Output: 6
๐Ÿ’ก Note: For p=2, array is [1,2,3]. We can swap bits to get [1,1,6] or similar. Product is 6.
example_3.py โ€” Larger Case
$ Input: p = 3
โ€บ Output: 1512
๐Ÿ’ก Note: For p=3, array is [1,2,3,4,5,6,7]. Optimal redistribution gives us 14^3 = 2744, but with proper bit manipulation we get 1512.

Constraints

  • 1 โ‰ค p โ‰ค 60
  • Result must be returned modulo 109 + 7
  • Important: Array contains integers from 1 to 2p - 1

Visualization

Tap to expand
Binary Trading Market (p=3)1234567Initial: Product = 1ร—2ร—3ร—4ร—5ร—6ร—7 = 50401116111Optimal: Product = 1ร—1ร—1ร—6ร—1ร—1ร—1 = 6Minimum Product Achieved!
Understanding the Visualization
1
Initial Market
All traders have unique wealth from 1 to 2^p-1
2
Strategic Trading
Traders exchange bits to create many poor traders (value 1)
3
Optimal Distribution
Most traders become poor, few remain wealthy but balanced
Key Takeaway
๐ŸŽฏ Key Insight: By redistributing bits strategically, we can create many 1's (which don't increase the product) and minimize the values of remaining numbers, achieving the minimum possible non-zero product.
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