Introduction

In this blog post, we will explore how to solve the LeetCode problem "1969" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a positive integer p. Consider an array nums (1-indexed) that consists of the integers in the inclusive range [1, 2p - 1] in their binary representations. You are allowed to do the following operation any number of times: Choose two elements x and y from nums. Choose a bit in x and swap it with its corresponding bit in y. Corresponding bit refers to the bit that is in the same position in the other integer. For example, if x = 1101 and y = 0011, after swapping the 2nd bit from the right, we have x = 1111 and y = 0001. Find the minimum non-zero product of nums after performing the above operation any number of times. Return this product modulo 109 + 7. Note: The answer should be the minimum product before the modulo operation is done. Example 1: Input: p = 1 Output: 1 Explanation: nums = [1]. There is only one element, so the product equals that element. Example 2: Input: p = 2 Output: 6 Explanation: nums = [01, 10, 11]. Any swap would either make the product 0 or stay the same. Thus, the array product of 1 2 3 = 6 is already minimized. Example 3: Input: p = 3 Output: 1512 Explanation: nums = [001, 010, 011, 100, 101, 110, 111] - In the first operation we can swap the leftmost bit of the second and fifth elements. - The resulting array is [001, 110, 011, 100, 001, 110, 111]. - In the second operation we can swap the middle bit of the third and fourth elements. - The resulting array is [001, 110, 001, 110, 001, 110, 111]. The array product is 1 6 1 6 1 6 7 = 1512, which is the minimum possible product. Constraints: 1 <= p <= 60

Explanation

• Key Idea: The core idea is that to minimize the product, we want to make as many numbers as possible equal to 1. We can achieve this by concentrating all the set bits into a single number, making the rest 1s. After performing the allowed operations, the optimal array will have (2^p - 1) - 1 elements equal to 1, and one element equal to (2^p - 1).

• Modulo Arithmetic: Since the product can be very large, we need to perform the calculations modulo 10⁹ + 7.

• Efficient Calculation: We can efficiently calculate (2^p - 2)^{(2p - 1) - 1} by using the fast modular exponentiation technique.

• Complexity:

  • Runtime: O(log(2^p - 2)) which simplifies to O(p) due to the fast power algorithm.
  • Storage: O(1)

Code

    def minNonZeroProduct(p: int) -> int:
    """
    Calculates the minimum non-zero product of nums after performing the allowed operations any number of times.

    Args:
        p: A positive integer.

    Returns:
        The minimum product modulo 10^9 + 7.
    """

    MOD = 10**9 + 7
    max_val = (1 << p) - 1
    count = max_val - 1

    def fast_power(base, power, mod):
        result = 1
        while power > 0:
            if power % 2 == 1:
                result = (result * base) % mod
            base = (base * base) % mod
            power //= 2
        return result

    result = fast_power(max_val - 1, count, MOD)
    result = (result * max_val) % MOD
    return result