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

Python3JavaC++Go

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

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;
    }
}

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;
    }
};

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
}

2518. Number of Great Partitions

Description

Solutions

Solution 1

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