2518. Number of Great Partitions

Problem

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

Solution (Java)

class Solution {
    public int countPartitions(int[] A, int k) {
        long mod = (long)1e9 + 7, total = 0, res = 1, dp[] = new long[k];
        dp[0] = 1;
        for (int a : A) {
            for (int i = k - 1 - a; i >= 0; --i)
                dp[i + a] = (dp[i + a] + dp[i]) % mod;
            res = res * 2 % mod;
            total += a;
        }
        for (int i = 0; i < k; ++i) {
            res -= total - i < k ? dp[i] : dp[i] * 2;
        }
        return (int)((res % mod + mod) % mod);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).