Introduction

In this blog post, we will explore how to solve the LeetCode problem "2518" 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 an array nums consisting of positive integers and an integer k. Partition the array into two ordered groups such that each element is in exactly one group. A partition is called great if the sum of elements of each group is greater than or equal to k. Return the number of distinct great partitions. Since the answer may be too large, return it modulo 109 + 7. Two partitions are considered distinct if some element nums[i] is in different groups in the two partitions. Example 1: Input: nums = [1,2,3,4], k = 4 Output: 6 Explanation: The great partitions are: ([1,2,3], [4]), ([1,3], [2,4]), ([1,4], [2,3]), ([2,3], [1,4]), ([2,4], [1,3]) and ([4], [1,2,3]). Example 2: Input: nums = [3,3,3], k = 4 Output: 0 Explanation: There are no great partitions for this array. Example 3: Input: nums = [6,6], k = 2 Output: 2 Explanation: We can either put nums[0] in the first partition or in the second partition. The great partitions will be ([6], [6]) and ([6], [6]). Constraints: 1 <= nums.length, k <= 1000 1 <= nums[i] <= 109

Explanation

Here's the approach to solve this problem:

• Dynamic Programming: The core idea is to use dynamic programming to determine the number of subsets of nums that sum to a specific value. We'll focus on finding subsets whose sum is less than k.
• Symmetry: If the total sum of nums is total_sum, we need both subset sums to be greater than or equal to k. So, if a subset has sum s < k, the other subset must have sum total_sum - s. If total_sum - s < k as well then we have an invalid partition. Therefore we use dynamic programming to find the number of subsets with sum less than k. Then calculate all possible arrangements - invalid arrangements to get the final answer.

• Combinatorics: The total number of possible partitions is 2ⁿ, where n is the length of nums. If we count the number of "bad" partitions where one of the subsets sums to less than k, we can subtract that from the total number of partitions. The total arrangements are calculated using pow(2, len(nums), MOD)

• Time & Space Complexity: O(n*k) time, O(k) space.

Code

    def count_great_partitions(nums, k):
    """
    Counts the number of distinct great partitions of nums such that the sum of elements of each group is >= k.

    Args:
        nums: A list of positive integers.
        k: An integer.

    Returns:
        The number of distinct great partitions modulo 10^9 + 7.
    """
    MOD = 10**9 + 7
    n = len(nums)
    total_sum = sum(nums)

    if total_sum < 2 * k:
        return 0

    dp = [0] * k
    dp[0] = 1

    for num in nums:
        for j in range(k - 1, num - 1, -1):
            dp[j] = (dp[j] + dp[j - num]) % MOD

    invalid_partitions = sum(dp) % MOD
    total_partitions = pow(2, n, MOD)

    return (total_partitions - invalid_partitions + MOD) % MOD