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Problem solution · Python

Partition to K Equal Sum Subsets

Partition to K Equal Sum Subsets: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
29 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Partition to K Equal Sum Subsets, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 29 lines of Python from the credited upstream file 698.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codePartition to K Equal Sum Subsets · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def canPartitionKSubsets(self, nums: list[int], k: int) -> bool:    summ = sum(nums)    if summ % k != 0:      return False     target = summ // k  # the target sum of each subset    if any(num > target for num in nums):      return False     def dfs(s: int, remainingGroups: int, currSum: int, used: int) -> bool:      if remainingGroups == 0:        return True      if currSum > target:        return False      if currSum == target:  # Find a valid group, so fresh start.        return dfs(0, remainingGroups - 1, 0, used)       for i in range(s, len(nums)):        if used >> i & 1:          continue        if dfs(i + 1, remainingGroups, currSum + nums[i], used | 1 << i):          return True       return False     nums.sort(reverse=True)    return dfs(0, k, 0, 0)

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