How do you solve Partition to K Equal Sum Subsets on LeetCode?

Partition to K Equal Sum Subsets (LeetCode #698) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(n) space complexity. Key insight: The total sum must be divisible by k to form equal subsets.

What is the brute force solution for Partition to K Equal Sum Subsets?

The brute force approach for Partition to K Equal Sum Subsets is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsets of the array and checking if we can form k subsets with equal sums. This is straightforward but inefficient, as it explores many combinations. The optimal Optimal Solution approach improves this to O(n * 2^n).

What are the interview tips for Partition to K Equal Sum Subsets?

Explain your thought process clearly when discussing backtracking. Consider edge cases and constraints before diving into the code. Practice explaining your code as you write it, as if teaching someone else.

What are common mistakes when solving Partition to K Equal Sum Subsets?

Not checking if the total sum is divisible by k. Failing to consider edge cases with larger numbers or fewer subsets.

#698

Partition to K Equal Sum Subsets

Medium

ArrayDynamic ProgrammingBacktrackingBit ManipulationMemoizationBitmaskBacktrackingMemoizationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^n)
Space	O(1)	O(n)

💡

Intuition

Time O(n * 2^n)Space O(n)

The optimal solution uses backtracking with memoization to efficiently explore valid combinations of subsets. This significantly reduces the number of recursive calls by avoiding repeated calculations.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total sum and check if it can be divided by k.
2Step 2: Sort the array in descending order to optimize the backtracking process.
3Step 3: Use a recursive function with memoization to try to fill each subset while keeping track of the current index and the current subset sums.

solution.py25 lines

1# Full working Python code
2from functools import lru_cache
3
4def canPartitionKSubsets(nums, k):
5    total_sum = sum(nums)
6    if total_sum % k != 0:
7        return False
8    target = total_sum // k
9    nums.sort(reverse=True)
10
11    @lru_cache(None)
12    def backtrack(index, subset_sums):
13        if index == len(nums):
14            return all(s == target for s in subset_sums)
15        for i in range(k):
16            if subset_sums[i] + nums[index] <= target:
17                subset_sums[i] += nums[index]
18                if backtrack(index + 1, tuple(subset_sums)):
19                    return True
20                subset_sums[i] -= nums[index]
21            if subset_sums[i] == 0:
22                break
23        return False
24
25    return backtrack(0, tuple([0] * k))

ℹ

Complexity note: The time complexity is O(n * 2^n) due to the recursive nature of the backtracking with memoization. The space complexity is O(n) for the recursion stack and the memoization storage.

1The total sum must be divisible by k to form equal subsets.
2Sorting the array helps in optimizing the backtracking process.

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