Problem Description

In this problem, we are given an array called nums which contains positive integers and another integer k. Our task is to find out how many distinct ways (great partitions) there are to split the array into two groups such that the sum of the numbers in each group is at least k. A great partition means that neither of the two groups has a sum that is less than k.

Each element from the array nums must be put into one, and only one, of the two groups. The order of elements within the groups is not important, but distinct partitions are those where at least one element appears in a different group compared to another partition.

Since there can be many distinct great partitions, we are required to return this number modulo 10^9 + 7 to keep the number manageable and within the bounds of typical integer limits.

One key point to understand is that if the sum of all elements in the array is less than 2k, we cannot possibly form two groups each having a sum greater or equal to k, hence the answer is straightforwardly 0 in such cases.

Intuition

To find the number of distinct great partitions, we can start by considering a dynamic programming approach that helps us understand whether it is possible to reach a certain sum with a subset of the given elements. The primary intuition behind the solution is to build up the count of subsets that can make up sums from 0 to k-1 respectively since sums from k to 2k-1 will automatically form a partition where both groups have at least k.

To arrive at the solution, we initialize a matrix 'f' with dimensions (n+1) x k where n is the number of elements in nums. Each f[i][j] will represent the number of ways we can achieve a sum j using the first i numbers of the array nums.

At the start, we have only one way to achieve a sum of 0, which is by choosing no elements (f[0][0] = 1). As we iterate over the elements of nums, we keep updating the matrix based on whether we include the current element in the subset or not. This helps us accumulate counts of all possible subset sums up to k.

The total number of subsets of nums is 2^n. When building our dynamic programming matrix, we are actually counting the number of subsets that reach every possible sum less than k. These counts have to be subtracted from the total since combining them with their complements does not yield a great partition—because at least one of the groups would have a sum of less than k.

Once we calculate the subset counts, the final answer is adjusted to account for duplications and to ensure the result falls within the desired modulus of 10^9 + 7. Specifically, we:

• Multiply the number of all subsets (2^n) by 2 modulo 10^9 + 7.
• Subtract twice the sum of subsets that have a sum less than k from the total count.
• Add the modulus to ensure non-negativity, then take the result modulo 10^9 + 7 to get the final answer.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses a dynamic programming approach to solve the partition problem. Here's how the steps of the algorithm unfold:

1. First, we check if the total sum of the array is less than 2k. If it is, we immediately know it's impossible to partition the array into two groups each with a sum greater than or equal to k, so we return 0.

2. We define a two-dimensional array f with n+1 rows and k columns. n is the length of the input array nums. We initialize the first row, which corresponds to using 0 elements from the array, such that f[0][0] is 1 (one way to make a sum of 0 with 0 elements) and the rest are 0.

3. We then use nested loops to fill in the dynamic programming matrix f. The outer loop runs through elements of nums from 1 to n, inclusive. The inner loop runs through all possible sums j from 0 to k-1, inclusive.

4. For each element i and each sum j, f[i][j] is updated to be the sum of:

   • f[i - 1][j], which is the count of ways to reach the sum j without using the i-th element.
   • If j is greater than or equal to nums[i - 1] (the current element we're considering), we add f[i - 1][j - nums[i - 1]] to the count, which represents using the i-th element to reach the sum j.

5. All computations are done modulo 10^9 + 7 to ensure we don't encounter integer overflow and that all operations fit within the specified modulus. This is why we see % mod after every arithmetic operation.

6. As we calculate f[i][j], we also keep a running total of all possible subsets (represented by ans) by repeatedly doubling (ans * 2 % mod) since every element either can be included or excluded from a set.

7. After filling out the dynamic programming matrix, we calculate the final answer. We know ans is the count of all subsets, but we want to exclude the cases where subsets sum to less than k. We sum up all f[n][j] for j from 0 to k-1 (which are the counts of subsets that don't form a great partition) and subtract twice this sum from ans. We multiply by 2 because for each such subset that forms a sum less than k, both it and its complement would be counted in ans, but neither makes a great partition.

8. Finally, in case the subtraction makes the number negative, we add mod (which has no effect modulo mod) before taking the modulo to get the non-negative remainder within the desired range.

From a data structure perspective, we utilize a 2D array (list of lists in Python) to store the number of ways to achieve each possible sum with different subsets of the array. The dynamic programming matrix (f) is central to this solution, allowing us to build up the solution incrementally.

The code is a direct implementation of this dynamic programming approach, using the patterns of modularity arithmetic to ensure all operations stay within bounds of the problem's specifications.

Example Walkthrough

Let's walk through an example to illustrate the solution approach using a small array. Suppose we have an array nums = [1, 2, 3] and an integer k = 3.

1. The total sum of the array 1 + 2 + 3 is 6, which is greater than 2*k (6), so it's possible to have two groups each with a sum of at least k.

2. We define a dynamic programming matrix f with dimensions 4 x 3 (n+1 x k), as there are 3 elements in nums and we need to consider sums up to k-1 which is 2.

3. Initialize f to have the first row as [1, 0, 0] representing there is one way to reach a sum of 0 with 0 elements.

4. We fill in the matrix f. Iterating over nums and possible sums, the updated matrix would look like this at each step:

   • Include the first element (1):

     f = [
       [1, 0, 0],
       [1, 1, 0], // Include or exclude the 1
       [1, 0, 0],
       [1, 0, 0]
     ]

   • Include the second element (2):

     f = [
       [1, 0, 0],
       [1, 1, 0],
       [1, 1, 1], // Include or exclude the 2 (we can now reach sums 1 and 2 using 1 and/or 2)
       [1, 0, 0]
     ]

   • Include the third element (3):

     f = [
       [1, 0, 0],
       [1, 1, 0],
       [1, 1, 1],
       [1, 2, 2]  // Include or exclude the 3 (we can reach sums 1 and 2 using combinations of 1, 2, and 3)
     ]

5. All operations are done modulo 10^9 + 7.

6. The total number of subsets is 2^n = 2^3 = 8. We double this to account for all possible subsets and exclude invalid subsets later. Thus ans = 16 % (10^9 + 7).

7. From the matrix f, the subset counts for sums less than k are f[3][0], f[3][1], and f[3][2], which are 1, 2, and 2, respectively. Summing these up gives us 1 + 2 + 2 = 5.

8. Subtracting twice this sum from ans we get: 16 - 2*5 = 6. We add 10^9 + 7 to ensure non-negativity and take modulo 10^9 + 7 to appear within bounds, which in this case just confirms 6 as the final count of great partitions.

So, for the given nums = [1, 2, 3] and k = 3, the number of distinct great partitions is 6. These partitions are:

• {1, 2} & {3}
• {1, 3} & {2}
• {2, 3} & {1}
• {1} & {2, 3}
• {2} & {1, 3}
• {3} & {1, 2}

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code can be analyzed based on the nested loops it uses:

• There is an outer loop iterating over the array nums with n elements.
• Inside this outer loop, there is an inner loop that iterates k times.

Each operation inside the inner loop executes in constant time. Therefore, the total number of operations performed will be O(n * k).

Hence, the time complexity of the code is O(n * k).

Space Complexity

To analyze space complexity, we look at the memory allocation:

• The 2D list f of size (n+1) * k is the primary memory consumer, where n is the length of nums and k is the provided partition value.

No other significant memory usage is present that scales with the input size. Therefore, the space complexity is determined by the size of this 2D list.

Hence, the space complexity of the code is O(n * k).

Learn more about how to find time and space complexity quickly using problem constraints.