How do you solve Number of Great Partitions on LeetCode?

Number of Great Partitions (LeetCode #2518) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k) time complexity and O(k) space complexity. Key insight: If the total sum of the array is less than 2 * k, it's impossible to create great partitions.

What is the brute force solution for Number of Great Partitions?

The brute force approach for Number of Great Partitions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible partitions of the array and checks if each partition meets the criteria of being a great partition. This method is straightforward but inefficient due to the exponential number of partitions. The optimal Optimal Solution approach improves this to O(n * k).

What algorithm or data structure is used to solve Number of Great Partitions?

Number of Great Partitions uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Subset Sum.

What are the interview tips for Number of Great Partitions?

Always consider edge cases, such as small arrays or large values of k. Explain your thought process clearly while coding, especially when transitioning from brute force to optimal solutions. Practice dynamic programming problems to become familiar with common patterns.

What are common mistakes when solving Number of Great Partitions?

Forgetting to check if the total sum is less than 2 * k before proceeding. Not properly handling the modulo operation when counting partitions.

#2518

Number of Great Partitions

Hard

ArrayDynamic ProgrammingDynamic ProgrammingSubset Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k)
Space	O(1)	O(k)

💡

Intuition

Time O(n * k)Space O(k)

Instead of counting great partitions directly, we count the partitions that do not meet the criteria and subtract from the total possible partitions. This approach leverages dynamic programming to efficiently calculate valid partitions.

⚙️

Algorithm

4 steps

1Step 1: Calculate the total sum of the array. If total_sum < 2 * k, return 0 as it's impossible to partition.
2Step 2: Use dynamic programming to count subsets with sums less than k.
3Step 3: Calculate the total number of partitions as 2^n - 1 (excluding the empty set).
4Step 4: Subtract the count of invalid partitions from the total partitions to get the number of great partitions.

solution.py23 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def countGreatPartitions(nums, k):
5    total_sum = sum(nums)
6    n = len(nums)
7    if total_sum < 2 * k:
8        return 0
9    
10    # Count subsets with sum < k
11    dp = [0] * (k)
12    dp[0] = 1
13    
14    for num in nums:
15        for j in range(k - 1, num - 1, -1):
16            dp[j] = (dp[j] + dp[j - num]) % MOD
17    
18    invalid_count = sum(dp) % MOD
19    total_partitions = pow(2, n, MOD) - 1
20    return (total_partitions - invalid_count + MOD) % MOD
21
22# Example usage
23print(countGreatPartitions([1, 2, 3, 4], 4))  # Output: 6

ℹ

Complexity note: The time complexity is O(n * k) because we iterate through each number in the array and update the dp array of size k. This is significantly more efficient than the brute force approach.

1If the total sum of the array is less than 2 * k, it's impossible to create great partitions.
2Counting invalid partitions can be more efficient than counting valid ones directly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.