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.
1using System;
2
3public class Solution {
4 public bool CanPartitionKSubsets(int[] nums, int k) {
5 int totalSum = 0;
6 foreach (var num in nums) totalSum += num;
7 if (totalSum % k != 0) return false;
8 int targetSubsetSum = totalSum / k;
9 Array.Sort(nums, (a, b) => b.CompareTo(a));
10 int[] subsetSums = new int[k];
11 return CanPartition(nums, subsetSums, targetSubsetSum, 0);
12 }
13
14 private bool CanPartition(int[] nums, int[] subsetSums, int targetSubsetSum, int index) {
15 if (index == nums.Length) return true;
16 for (int i = 0; i < subsetSums.Length; i++) {
17 if (subsetSums[i] + nums[index] <= targetSubsetSum) {
18 subsetSums[i] += nums[index];
19 if (CanPartition(nums, subsetSums, targetSubsetSum, index + 1)) return true;
20 subsetSums[i] -= nums[index];
21 }
22 if (subsetSums[i] == 0) break;
23 }
24 return false;
25 }
26
27 public static void Main() {
28 Solution sol = new Solution();
29 Console.WriteLine(sol.CanPartitionKSubsets(new int[] { 4, 3, 2, 3, 5, 2, 1 }, 4));
30 }
31}This C# solution wraps the backtracking method within the class structure of C#. Sorting in descending order and maintaining variable names consistent with numeric identifiers helps navigate potential unsolvable branches sooner.
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.