Introduction

In this blog post, we will explore how to solve the LeetCode problem "982" 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

Given an integer array nums, return the number of AND triples. An AND triple is a triple of indices (i, j, k) such that: 0 <= i < nums.length 0 <= j < nums.length 0 <= k < nums.length nums[i] & nums[j] & nums[k] == 0, where & represents the bitwise-AND operator. Example 1: Input: nums = [2,1,3] Output: 12 Explanation: We could choose the following i, j, k triples: (i=0, j=0, k=1) : 2 & 2 & 1 (i=0, j=1, k=0) : 2 & 1 & 2 (i=0, j=1, k=1) : 2 & 1 & 1 (i=0, j=1, k=2) : 2 & 1 & 3 (i=0, j=2, k=1) : 2 & 3 & 1 (i=1, j=0, k=0) : 1 & 2 & 2 (i=1, j=0, k=1) : 1 & 2 & 1 (i=1, j=0, k=2) : 1 & 2 & 3 (i=1, j=1, k=0) : 1 & 1 & 2 (i=1, j=2, k=0) : 1 & 3 & 2 (i=2, j=0, k=1) : 3 & 2 & 1 (i=2, j=1, k=0) : 3 & 1 & 2 Example 2: Input: nums = [0,0,0] Output: 27 Constraints: 1 <= nums.length <= 1000 0 <= nums[i] < 216

Explanation

Here's a breakdown of the approach, complexity, and the Python code for solving the AND triples problem:

• Key Idea: Instead of directly iterating through all possible triples (which would be O(n^3)), precompute the counts of all possible bitwise AND results of pairs of numbers in the input array. Then, iterate through the array once, and for each number, check which precomputed AND results, when ANDed with the current number, yield zero.

• Optimization: Since the numbers are limited to the range [0, 2¹⁶), the possible AND results are also within this range. This limits the size of the precomputed counts, allowing for efficient lookups. We use a frequency array/hashmap to store the counts of the results for better performance.

• Complexity:

  • Runtime: O(n²) to calculate and store the AND pairs. O(n * max_value) where max_value is maximum possible value after AND operation which is 2¹⁶, in the second pass. Thus, it is practically O(n). Therefore, overall, it will be O(n² + n) -> O(n²).
  • Space: O(2¹⁶), to store the counts of the AND pairs.

Code

    def count_and_triples(nums):
    """
    Counts the number of AND triples in the given integer array.

    Args:
        nums: A list of integers.

    Returns:
        The number of AND triples (i, j, k) such that nums[i] & nums[j] & nums[k] == 0.
    """

    count = 0
    and_counts = {}  # Dictionary to store counts of num[i] & num[j]

    # Precompute and count all possible AND results of pairs
    for i in range(len(nums)):
        for j in range(len(nums)):
            and_val = nums[i] & nums[j]
            and_counts[and_val] = and_counts.get(and_val, 0) + 1

    # Iterate through the array and count the triples
    for k in range(len(nums)):
        for and_val, and_count in and_counts.items():
            if (and_val & nums[k]) == 0:
                count += and_count

    return count