# 698 Partition to K Equal Sum Subsets

### Problem

Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal.

Example 1:

``````Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4
Output: True
Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.
``````

Note:

1 <= k <= len(nums) <= 16. 0 < nums[i] < 10000.

### Solutions

``````class Solution {
public boolean canPartitionKSubsets(int[] nums, int k) {
int sum = 0;
for (int i = 0; i < nums.length; i ++) {
sum += nums[i];
}
if (sum % k != 0) {
return false;
}
Arrays.sort(nums);
boolean[] visited = new boolean[nums.length];
return process(0, nums, visited, sum/k, sum/k, k);
}
private boolean process(int start, int[] nums, boolean[] visited, int sum, int left, int togo) {
if (togo == 1) {
return true;
}
if (left == 0) {
return process(0, nums, visited, sum, sum, togo - 1);
}
for (int i = start; i < nums.length; i ++) {
if (nums[i] > left) {
return false;
}
if (visited[i] == true) {
continue;
}
visited[i] = true;
if (process(i + 1, nums, visited, sum, left - nums[i], togo)) {
return true;
}
visited[i] = false;
}
return false;
}

}
``````

greedy:

``````class Solution {
public boolean canPartitionKSubsets(int[] nums, int k) {
int sum = 0;
for (int i = 0; i < nums.length; i ++) {
sum += nums[i];
}
if (sum % k != 0) {
return false;
}
Arrays.sort(nums);
int[] bucket = new int[k];
return process(bucket, nums, sum/k, nums.length - 1);
}
private boolean process(int[] bucket, int[] nums, int target, int index) {
if (index == -1) {
for (int i = 0; i < bucket.length; i ++) {
if (bucket[i] != target) {
return false;
}
}
return true;
}
int num = nums[index];
for (int i = 0; i < bucket.length; i ++) {
if (bucket[i] + num > target) {
continue;
}
bucket[i] += num;
if (process(bucket, nums, target, index - 1)) {
return true;
}
bucket[i] -= num;
}
return false;
}

}
``````