Problem Description

You are given an integer array nums and an integer k. Your task is to determine whether it's possible to divide the array into exactly k non-empty subsets where each subset has the same sum.

For example, if nums = [4, 3, 2, 3, 5, 2, 1] and k = 4, you can divide it into 4 subsets: [5], [1, 4], [2, 3], and [2, 3], where each subset sums to 5. In this case, you would return true.

The key requirements are:

• You must use all elements from the array
• Each element can only belong to one subset
• You need exactly k subsets (no more, no less)
• All k subsets must have equal sums

The solution uses a depth-first search approach with backtracking. First, it checks if the total sum of the array is divisible by k - if not, it's impossible to create equal subsets. If divisible, the target sum for each subset is total_sum / k. The algorithm then tries to place each element into one of the k buckets (subsets), backtracking when a placement doesn't lead to a valid solution. The array is sorted in descending order to optimize the search by trying larger elements first, which helps identify invalid paths earlier.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem involves partitioning an array into subsets, not traversing nodes and edges in a graph structure.

Need to solve for kth smallest/largest?

• No: We're not looking for the kth smallest or largest element. We need to determine if a valid partition exists.

Involves Linked Lists?

• No: The problem works with an integer array, not linked list operations.

Does the problem have small constraints?

• Yes: Looking at typical constraints for this problem, we have:
  • Array length is usually limited (often ≤ 16)
  • k is typically small (often ≤ 4)
  • These small constraints make exhaustive search feasible

Brute force / Backtracking?

• Yes: Given the small constraints and the need to explore different ways of partitioning elements into k subsets, backtracking is appropriate. We need to:
  • Try placing each element in different subsets
  • Backtrack when a placement doesn't lead to a valid solution
  • Explore all possible assignments until we find a valid partition or exhaust all options

Conclusion: The flowchart correctly leads us to use a Backtracking approach. This aligns perfectly with the solution, which uses DFS with backtracking to try different element placements in k subsets, undoing choices that don't work and exploring alternative paths.

Intuition

The key insight is recognizing this as a partitioning problem where we need to make decisions about where each element belongs. Think of it like having k buckets and n items - we need to distribute all items such that each bucket has the same weight.

First, we need to check if equal partitioning is even possible. If the total sum isn't divisible by k, we can immediately return false. If it is divisible, each subset must have sum equal to total_sum / k.

Now, how do we assign elements to subsets? We can think of this as a decision tree where at each level, we decide which subset to place the current element in. For each element, we have up to k choices (which subset to add it to). This naturally leads to a backtracking approach - we try placing an element in subset 1, then recursively try to place the remaining elements. If that doesn't work, we backtrack and try subset 2, and so on.

To optimize our search:

1. Sort in descending order: Larger elements are more restrictive in where they can be placed. By trying them first, we can fail fast on impossible configurations.
2. Skip duplicate states: If two subsets currently have the same sum, trying to add the current element to either will lead to equivalent sub-problems. We only need to try one.
3. Prune invalid branches: If adding an element to a subset would exceed the target sum s, we don't explore that branch.

The beauty of this approach is that we're essentially building all k subsets simultaneously, making one placement decision at a time and verifying constraints as we go. When we successfully place all elements (reach the base case), we know we've found a valid partition.

Learn more about Memoization, Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

Let's walk through the implementation step by step, following the DFS + Pruning approach outlined in the reference solution.

Step 1: Check Feasibility

s, mod = divmod(sum(nums), k)
if mod:
    return False

First, we calculate the total sum and check if it's divisible by k. If there's a remainder (mod != 0), it's impossible to create equal subsets, so we return false immediately. The variable s stores the target sum for each subset.

Step 2: Initialize Data Structures

cur = [0] * k
nums.sort(reverse=True)

• cur: An array of size k tracking the current sum of each subset. Initially all zeros.
• We sort nums in descending order to try larger elements first, which helps prune the search tree earlier.

Step 3: DFS with Backtracking

def dfs(i: int) -> bool:
    if i == len(nums):
        return True

The recursive function dfs(i) tries to place element at index i into one of the k subsets. Base case: if we've successfully placed all elements (i == len(nums)), we've found a valid partition.

Step 4: Try Each Subset

for j in range(k):
    if j and cur[j] == cur[j - 1]:
        continue

For the current element nums[i], we try placing it in each subset j. The optimization if j and cur[j] == cur[j - 1] skips redundant attempts - if subset j has the same sum as subset j-1, trying to add the current element to either produces equivalent sub-problems.

Step 5: Backtrack Logic

cur[j] += nums[i]
if cur[j] <= s and dfs(i + 1):
    return True
cur[j] -= nums[i]

• Add nums[i] to subset j
• Only continue if the subset sum doesn't exceed target s
• Recursively try to place the next element
• If successful, propagate True up the call stack
• Otherwise, backtrack by removing nums[i] from subset j and try the next subset

Step 6: Return Result

return False

If we've tried all subsets for the current element and none worked, return false. The main function returns the result of dfs(0), starting from the first element.

The algorithm essentially builds a decision tree where each level represents choosing a subset for an element, with pruning based on sum constraints and duplicate states.

Example Walkthrough

Let's walk through a small example with nums = [4, 3, 2, 1] and k = 2.

Step 1: Check Feasibility

• Total sum = 4 + 3 + 2 + 1 = 10
• 10 ÷ 2 = 5 with remainder 0 ✓
• Target sum per subset: s = 5

Step 2: Initialize

• Sort array descending: nums = [4, 3, 2, 1]
• Create buckets: cur = [0, 0] (two empty subsets)

Step 3: DFS Process

Starting with dfs(0) - placing element nums[0] = 4:

• Try subset 0: cur[0] = 0 + 4 = 4 (≤ 5, valid)
• cur = [4, 0]
• Recursively call dfs(1)

At dfs(1) - placing element nums[1] = 3:

• Try subset 0: cur[0] = 4 + 3 = 7 (> 5, invalid - skip)
• Try subset 1: cur[1] = 0 + 3 = 3 (≤ 5, valid)
• cur = [4, 3]
• Recursively call dfs(2)

At dfs(2) - placing element nums[2] = 2:

• Try subset 0: cur[0] = 4 + 2 = 6 (> 5, invalid - skip)
• Try subset 1: cur[1] = 3 + 2 = 5 (= 5, valid)
• cur = [4, 5]
• Recursively call dfs(3)

At dfs(3) - placing element nums[3] = 1:

• Try subset 0: cur[0] = 4 + 1 = 5 (= 5, valid)
• cur = [5, 5]
• Recursively call dfs(4)

At dfs(4):

• i == len(nums) - all elements placed successfully!
• Return True

The algorithm successfully partitions the array into:

• Subset 0: {4, 1} with sum = 5
• Subset 1: {3, 2} with sum = 5

Each recursive call returns True, propagating the success back to the main function, which returns True.

Solution Implementation

Time and Space Complexity

Time Complexity: O(k^n)

The algorithm uses a depth-first search approach where at each level of recursion (corresponding to placing each number), we try to place the current number in one of k buckets. In the worst case:

• We have n numbers to place (depth of recursion tree is n)
• At each level, we try up to k different buckets
• This gives us a recursion tree with branching factor k and depth n
• Total nodes in the recursion tree: k^n

The pruning optimization if j and cur[j] == cur[j - 1]: continue helps skip redundant states when consecutive buckets have the same sum, but this doesn't change the worst-case complexity.

Space Complexity: O(n)

The space complexity consists of:

• Recursion call stack: O(n) - maximum depth is n when we process all numbers
• The cur array: O(k) - stores current sum for each subset
• Since k ≤ n (we need at least k elements to form k non-empty subsets), and the recursion stack dominates, the overall space complexity is O(n)

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

Common Pitfalls

Pitfall 1: Forgetting to Check if Any Single Element Exceeds Target Sum

The Problem: A common mistake is not checking whether any single element in the array is greater than the target sum before starting the backtracking process. If nums = [10, 5, 5, 2] and k = 3, the target sum would be 22/3 = 7.33.... While the division check catches this case, consider nums = [10, 5, 5] and k = 2 where target sum is 10. The element 10 alone equals the target, but without proper validation, the algorithm might waste time trying combinations.

Solution: Add an early termination check after sorting:

# After sorting in descending order
nums.sort(reverse=True)

# Check if largest element exceeds target
if nums[0] > target_sum:
    return False

Pitfall 2: Not Understanding Why Sorting in Descending Order Matters

The Problem: Developers often skip sorting or sort in ascending order, thinking it doesn't matter. However, sorting in descending order is crucial for performance. When we place larger elements first, we can identify invalid paths much earlier. For example, if a large element doesn't fit in any bucket, we know immediately rather than after trying many smaller elements first.

Solution: Always sort in descending order and understand that this optimization can reduce time complexity from timeout to acceptable:

# This is critical for performance
nums.sort(reverse=True)

Pitfall 3: Incorrect Duplicate State Detection

The Problem: The line if bucket_idx > 0 and bucket_sums[bucket_idx] == bucket_sums[bucket_idx - 1]: is often misunderstood or implemented incorrectly. Some developers might try to use a set to track seen sums or implement more complex duplicate detection, which either doesn't work correctly or adds unnecessary overhead.

Solution: The simple consecutive comparison works because:

• We're trying buckets in order (0, 1, 2, ...)
• If bucket j has the same sum as bucket j-1, placing the current element in either creates identical subproblems
• This only works because we process buckets sequentially in the loop

Pitfall 4: Not Handling Edge Cases

The Problem: Missing edge cases like:

• k = 1 (should return True for any valid array)
• k > len(nums) (impossible to create k non-empty subsets)
• Empty array or single element cases

Solution: Add edge case handling at the beginning:

def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
    # Edge cases
    if k == 1:
        return True
    if k > len(nums):
        return False
    if len(nums) == 0:
        return False
  
    # Continue with main logic...

Pitfall 5: Attempting Optimization with Memoization Incorrectly

The Problem: Some developers try to add memoization using the bucket sums as state, like @cache or using a dictionary with tuple(bucket_sums) as key. This often fails because:

1. The bucket sums array changes through backtracking
2. The order of buckets matters during the recursive process
3. The state space can still be very large

Solution: Instead of memoization, rely on the existing optimizations:

• Sorting in descending order
• Skipping duplicate bucket states
• Early termination when bucket sum exceeds target

If memoization is needed, use a bitmask to represent which elements have been used:

@cache
def dfs(used_mask: int, completed_subsets: int) -> bool:
    # Different approach using bitmask for used elements
    pass