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2518. Number of Great Partitions

Description

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

Solutions

Solution 1

Thinking

Each element goes to exactly one of two groups, and both group sums must be at least \(k\). There are \(2^n\) assignments; \(n\) and \(k\) are up to \(10^3\), so listing them is impossible, but a group whose sum is already below \(k\) is exactly the forbidden case.

If the total sum is less than \(2k\), both sides cannot succeed and the answer is \(0\). Otherwise subtract the bad partitions from \(2^n\). A bad partition is a subset with sum \(<k\) used as one side — a \(0\)-\(1\) knapsack of capacity \(k-1\). Either side may be the small one, so the bad count is twice the number of such subsets.

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class Solution:
    def countPartitions(self, nums: List[int], k: int) -> int:
        if sum(nums) < k * 2:
            return 0
        mod = 10**9 + 7
        n = len(nums)
        f = [[0] * k for _ in range(n + 1)]
        f[0][0] = 1
        ans = 1
        for i in range(1, n + 1):
            ans = ans * 2 % mod
            for j in range(k):
                f[i][j] = f[i - 1][j]
                if j >= nums[i - 1]:
                    f[i][j] = (f[i][j] + f[i - 1][j - nums[i - 1]]) % mod
        return (ans - sum(f[-1]) * 2 + mod) % mod

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class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int countPartitions(int[] nums, int k) {
        long s = 0;
        for (int v : nums) {
            s += v;
        }
        if (s < k * 2) {
            return 0;
        }
        int n = nums.length;
        long[][] f = new long[n + 1][k];
        f[0][0] = 1;
        long ans = 1;
        for (int i = 1; i <= n; ++i) {
            int v = nums[i - 1];
            ans = ans * 2 % MOD;
            for (int j = 0; j < k; ++j) {
                f[i][j] = f[i - 1][j];
                if (j >= v) {
                    f[i][j] = (f[i][j] + f[i - 1][j - v]) % MOD;
                }
            }
        }
        for (int j = 0; j < k; ++j) {
            ans = (ans - f[n][j] * 2 % MOD + MOD) % MOD;
        }
        return (int) ans;
    }
}

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class Solution {
public:
    const int mod = 1e9 + 7;

    int countPartitions(vector<int>& nums, int k) {
        long s = accumulate(nums.begin(), nums.end(), 0l);
        if (s < k * 2) return 0;
        int n = nums.size();
        long f[n + 1][k];
        int ans = 1;
        memset(f, 0, sizeof f);
        f[0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            int v = nums[i - 1];
            ans = ans * 2 % mod;
            for (int j = 0; j < k; ++j) {
                f[i][j] = f[i - 1][j];
                if (j >= v) {
                    f[i][j] = (f[i][j] + f[i - 1][j - v]) % mod;
                }
            }
        }
        for (int j = 0; j < k; ++j) {
            ans = (ans - f[n][j] * 2 % mod + mod) % mod;
        }
        return ans;
    }
};

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func countPartitions(nums []int, k int) int {
    s := 0
    for _, v := range nums {
        s += v
    }
    if s < k*2 {
        return 0
    }
    const mod int = 1e9 + 7
    n := len(nums)
    f := make([][]int, n+1)
    for i := range f {
        f[i] = make([]int, k)
    }
    f[0][0] = 1
    ans := 1
    for i := 1; i <= n; i++ {
        v := nums[i-1]
        ans = ans * 2 % mod
        for j := 0; j < k; j++ {
            f[i][j] = f[i-1][j]
            if j >= v {
                f[i][j] = (f[i][j] + f[i-1][j-v]) % mod
            }
        }
    }
    for j := 0; j < k; j++ {
        ans = (ans - f[n][j]*2%mod + mod) % mod
    }
    return ans
}

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