This approach involves using backtracking to attempt to build k subsets, each with a sum equal to totalSum / k. The decision tree can be pruned by sorting the numbers in decreasing order and tracking the sums of the current subset.
Time Complexity: O(kn), where n is the number of elements, due to exploring all partition possibilities.
Space Complexity: O(n + k) due to recursive call stack and subset sums array.
1#include <vector>
2#include <algorithm>
3
4class Solution {
5public:
6 bool canPartitionKSubsets(std::vector<int>& nums, int k) {
7 int totalSum = 0;
8 for (int num : nums) totalSum += num;
9 if (totalSum % k != 0) return false;
10 int targetSubsetSum = totalSum / k;
11 std::sort(nums.begin(), nums.end(), std::greater<int>());
12 std::vector<int> subsetSums(k, 0);
13 return canPartition(nums, subsetSums, targetSubsetSum, 0);
14 }
15
16private:
17 bool canPartition(std::vector<int>& nums, std::vector<int>& subsetSums, int targetSubsetSum, int index) {
18 if (index == nums.size()) return true;
19 for (int i = 0; i < subsetSums.size(); i++) {
20 if (subsetSums[i] + nums[index] <= targetSubsetSum) {
21 subsetSums[i] += nums[index];
22 if (canPartition(nums, subsetSums, targetSubsetSum, index + 1)) return true;
23 subsetSums[i] -= nums[index];
24 }
25 if (subsetSums[i] == 0) break;
26 }
27 return false;
28 }
29};The solution leverages a depth-first search (DFS) with backtracking. It uses a vector to maintain current subset sums and tries to distribute each number in the nums array into these subsets. Sorting the numbers high to low optimizes branching.
A less intuitive but potentially powerful technique involves using bitmasking to represent subsets and top-down dynamic programming to evaluate the feasibility of partitioning. This uses state compression where each state represents a set of elements already partitioned; 1 means used, 0 means unused.
Time Complexity: O(n * 2n), determined by the states and transitions across the full bitmask.
Space Complexity: O(2n), which is used to store DP states for all subset configurations.
1import java.util.Arrays;
2
3public class Solution {
4 public boolean canPartitionKSubsets(int[] nums, int k) {
5 int totalSum = Arrays.stream(nums).sum();
6 if (totalSum % k != 0) return false;
7 int targetSubsetSum = totalSum / k;
8
9 int n = nums.length;
10 int[] dp = new int[1 << n];
11 Arrays.fill(dp, -1);
12 dp[0] = 0;
13
14 for (int mask = 0; mask < (1 << n); ++mask) {
15 if (dp[mask] == -1) continue;
16 for (int i = 0; i < n; ++i) {
17 if ((mask & (1 << i)) == 0 && dp[mask] + nums[i] <= targetSubsetSum) {
18 int newMask = mask | (1 << i);
19 dp[newMask] = (dp[mask] + nums[i]) % targetSubsetSum;
20 }
21 }
22 }
23
24 return dp[(1 << n) - 1] == 0;
25 }
26
27 public static void main(String[] args) {
28 Solution sol = new Solution();
29 System.out.println(sol.canPartitionKSubsets(new int[]{4, 3, 2, 3, 5, 2, 1}, 4));
30 }
31}In this Java DP with bitmasking solution, bit shifts represent which numbers are in or out of a subset configuration. This method ensures each subset sum is proper modulo targetSubsetSum by using DP states. The state-mask expresses the current subset configuration as elements included or excluded.