CPMagical CP Library

Problem solution · C++

Partition to K Equal Sum Subsets

Partition to K Equal Sum Subsets: a C++ 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
37 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

  • 37 lines of C++ from the credited upstream file 698.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  bool canPartitionKSubsets(vector<int>& nums, int k) {    const int sum = accumulate(nums.begin(), nums.end(), 0);    if (sum % k != 0)      return false;     const int target = sum / k;  // the target sum of each subset    if (ranges::any_of(nums, [target](const int num) { return num > target; }))      return false;     ranges::sort(nums, greater<>());    return dfs(nums, 0, k, /*currSum=*/0, target, /*used=*/0);  }  private:  bool dfs(const vector<int>& nums, int s, int remainingGroups, int currSum,           const int subsetTargetSum, int used) {    if (remainingGroups == 0)      return true;    if (currSum > subsetTargetSum)      return false;    if (currSum == subsetTargetSum)  // Find a valid group, so fresh start.      return dfs(nums, 0, remainingGroups - 1, 0, subsetTargetSum, used);     for (int i = s; i < nums.size(); ++i) {      if (used >> i & 1)        continue;      if (dfs(nums, i + 1, remainingGroups, currSum + nums[i], subsetTargetSum,              used | 1 << i))        return true;    }     return false;  }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗