2518. Number of Great Partitions

Problem Link

Description

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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

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Python3

class Solution:
    def countPartitions(self, nums: List[int], k: int) -> int:
        N = len(nums)
        M = 10 ** 9 + 7
 
        if sum(nums) < 2 * k: return 0
 
        @cache
        def dp(index, s):
            if index >= N: return 1
 
            skip = dp(index + 1, s)
            take = 0 if s + nums[index] >= k else dp(index + 1, s + nums[index])
 
            return (skip + take) % M
        
        return (pow(2, N, M) - 2 * dp(0, 0) + M) % M

C++

class Solution {
public:
    int N, K;
    int MOD = 1e9 + 7;
    int cache[1001][1001];
 
    int powmod (int a, int b, int k) {
        int result = 1 ;
        while (b--) {
            result *= a ;
            result %= k ;
        }
        return result ;
    }
 
    int dp(vector<int>& nums, int index, int s) {
        if (cache[index][s] != -1) {
            return cache[index][s];
        }
 
        if (s >= K) return 0;
        if (index >= N) return 1;
 
        int skip = dp(nums, index + 1, s);
        int take = s + nums[index] <= 1000 ? dp(nums, index + 1, s + nums[index]) : 0;
 
        return cache[index][s] = (skip + take) % MOD;
    }
 
    int countPartitions(vector<int>& nums, int k) {
        long long total = accumulate(nums.begin(), nums.end(), 0LL);
        if (total < 2 * k) return 0;
 
        memset(cache, -1, sizeof(cache));
        N = nums.size();
        K = k;
        
        int all = powmod(2, N, MOD);
        int invalid = 2 * dp(nums, 0, 0);
 
        return (all - invalid + MOD) % MOD;
    }
};