Problem Description

You start with an array nums containing all integers from 1 to 2^p - 1 (where p is a given positive integer), represented in binary form.

You can perform the following operation any number of times:

• Select any two elements x and y from the array
• Pick any bit position and swap the bits at that position between x and y

For example, if x = 1101 (13 in decimal) and y = 0011 (3 in decimal), swapping the 2nd bit from the right gives x = 1111 (15) and y = 0001 (1).

Your goal is to find the minimum possible non-zero product of all elements in the array after performing any number of these bit-swapping operations. Return this minimum product modulo 10^9 + 7.

The key insight is that bit-swapping operations preserve the total sum of all elements. When the sum is fixed, to minimize the product, you want to maximize the differences between elements. The optimal strategy is to create pairs where one element is as small as possible (1) and the other is as large as possible within the constraints.

Through careful bit manipulation, you can transform most elements into pairs of (1, 2^p - 2), while keeping one element at 2^p - 1. This leads to the minimum product formula: (2^p - 1) × (2^p - 2)^(2^(p-1) - 1).

Intuition

The first key observation is that swapping bits between two numbers doesn't change their sum. If we have x and y, and we swap a bit where x has 1 and y has 0, we're essentially moving a value of 2^i from x to y. The total sum remains constant.

Since the sum is fixed and we want to minimize the product, we should think about what arrangement of numbers gives the smallest product for a fixed sum. From the AM-GM inequality, we know that for a fixed sum, the product is minimized when the numbers are as far apart as possible - essentially, we want some numbers to be very small and others to be very large.

The largest possible number in our range is 2^p - 1 (all bits set to 1). This number is special because it's already at the maximum, so we can't make it any larger through bit swaps. We should keep this number as is.

For all other numbers from 1 to 2^p - 2, we can pair them up strategically. Notice that for any number x in this range, the number 2^p - 1 - x is also in the range. These two numbers together have exactly p bits set to 1 between them (since their sum equals 2^p - 1).

Through a series of bit swaps between x and 2^p - 1 - x, we can redistribute their bits to make one number as small as possible (1, with just one bit set) and the other as large as possible (2^p - 2, with all bits except one set to 1).

Since we have 2^p - 1 total numbers and one of them (2^p - 1) stays unchanged, we have 2^p - 2 numbers to pair up, giving us 2^(p-1) - 1 pairs. Each pair contributes 1 × (2^p - 2) to the product.

Therefore, the minimum product is: (2^p - 1) × (2^p - 2)^(2^(p-1) - 1)

Learn more about Greedy, Recursion and Math patterns.

Solution Approach

The implementation follows a greedy strategy combined with fast power modular exponentiation.

Based on our intuition, we want to transform the array into:

• One element with value 2^p - 1 (the maximum)
• 2^(p-1) - 1 elements with value 2^p - 2
• 2^(p-1) - 1 elements with value 1

This gives us the minimum product: (2^p - 1) × (2^p - 2)^(2^(p-1) - 1)

The implementation steps are:

1. Calculate the maximum value: 2^p - 1 represents the largest element that remains unchanged.

2. Calculate the second largest value: 2^p - 2 represents the value that each "large" element in our pairs will become.

3. Calculate the exponent: 2^(p-1) - 1 represents how many pairs we have (and thus how many times 2^p - 2 appears in the product).

4. Use fast power with modulo: Since the numbers can be extremely large, we use Python's built-in pow(base, exponent, mod) function which efficiently computes (base^exponent) % mod using binary exponentiation.

The code implementation:

class Solution:
    def minNonZeroProduct(self, p: int) -> int:
        mod = 10**9 + 7
        return (2**p - 1) * pow(2**p - 2, 2 ** (p - 1) - 1, mod) % mod

The key optimization here is using pow(base, exp, mod) instead of computing (base**exp) % mod directly, as the latter would cause overflow for large values of p. The built-in pow function with three arguments performs modular exponentiation efficiently using the square-and-multiply algorithm, keeping intermediate results within manageable bounds.

Example Walkthrough

Let's walk through a small example with p = 3 to illustrate the solution approach.

Initial Setup:

• We start with all integers from 1 to 2³ - 1 = 7
• In binary: [001, 010, 011, 100, 101, 110, 111]
• In decimal: [1, 2, 3, 4, 5, 6, 7]
• Initial product: 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040

Key Observations:

• Total sum = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 (this remains constant)
• The maximum value 7 (111 in binary) already has all bits set, so we keep it unchanged
• We need to pair up the remaining 6 numbers

Pairing Strategy: We can pair numbers that sum to 7:

• Pair 1 (001) with 6 (110): sum = 7
• Pair 2 (010) with 5 (101): sum = 7
• Pair 3 (011) with 4 (100): sum = 7

Bit Manipulation Process: For each pair, we redistribute bits to create (1, 6):

Pair 1: (001, 110)

• Already optimal: 1 and 6

Pair 2: (010, 101)

• Swap bit at position 1: 010 → 000, 101 → 111 (but 111 = 7 is reserved)
• Swap bit at position 0: 010 → 011, 101 → 100
• Continue swapping to get: 001 and 110 (which equals 1 and 6)

Pair 3: (011, 100)

• Swap bits to redistribute: transform to 001 and 110 (1 and 6)

Final Configuration: After all swaps: [1, 1, 1, 6, 6, 6, 7]

Verification Using Formula:

• p = 3
• Maximum value: 2³ - 1 = 7
• Second largest: 2³ - 2 = 6
• Number of pairs: 2^(3-1) - 1 = 2² - 1 = 3

Minimum product = 7 × 6³ = 7 × 216 = 1512

Modulo Calculation: Since 1512 < 10⁹ + 7, the result is 1512.

This confirms our formula: (2^p - 1) × (2^p - 2)^(2^(p-1) - 1) gives us the minimum possible product.

Solution Implementation

Time and Space Complexity

The time complexity is O(p), and the space complexity is O(1).

The time complexity analysis:

• Computing 2**p takes O(p) time using repeated multiplication
• The pow(2**p - 2, 2**(p-1) - 1, mod) function uses fast modular exponentiation, which has time complexity O(log(exponent)). The exponent is 2**(p-1) - 1, so log(2**(p-1) - 1) ≈ p - 1. Therefore, this operation takes O(p) time
• The overall time complexity is O(p) + O(p) = O(p)

The space complexity analysis:

• The algorithm only uses a constant amount of extra space for storing intermediate values like mod, the results of 2**p, and the modular exponentiation result
• No additional data structures that scale with input size are used
• Therefore, the space complexity is O(1)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow When Computing Powers

The most critical pitfall is attempting to compute large powers directly without modular arithmetic. For example, when p = 60, computing (2^60 - 2)^(2^59 - 1) directly would result in astronomical numbers that exceed Python's integer limits in most programming languages and cause overflow or performance issues.

Incorrect approach:

# This will be extremely slow or cause overflow in other languages
result = ((2**p - 1) * ((2**p - 2) ** (2**(p-1) - 1))) % MOD

Correct approach:

# Use modular exponentiation
result = (max_value * pow(second_max, num_pairs, MOD)) % MOD

2. Forgetting Final Modulo Operation

Even when using pow(base, exp, MOD) for the exponentiation, the final multiplication with max_value can still exceed the modulo range, requiring an additional modulo operation.

Incorrect approach:

# Missing final modulo
return max_value * pow(second_max, num_pairs, MOD)

Correct approach:

# Apply modulo to the final result
return (max_value * pow(second_max, num_pairs, MOD)) % MOD

3. Edge Case: p = 1

When p = 1, the array only contains the single element 1. The formula would give 1 * 0^0, which is undefined. The code needs to handle this special case.

Solution:

def minNonZeroProduct(self, p: int) -> int:
    if p == 1:
        return 1  # Special case: only element is 1
  
    MOD = 10**9 + 7
    max_value = (1 << p) - 1
    second_max = max_value - 1
    num_pairs = (1 << (p - 1)) - 1
    return (max_value * pow(second_max, num_pairs, MOD)) % MOD

4. Using Regular Power Instead of Bit Shifts

While 2**p works correctly, using bit shifts 1 << p is more efficient and clearer for powers of 2, especially for large values of p.

Less efficient:

max_value = 2**p - 1
num_pairs = 2**(p-1) - 1

More efficient:

max_value = (1 << p) - 1
num_pairs = (1 << (p - 1)) - 1