# 0698. Partition to K Equal Sum Subsets

<https://leetcode.com/problems/partition-to-k-equal-sum-subsets>

## Description

Given an integer array `nums` and an integer `k`, return `true` if it is 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.
```

**Example 2:**

```
**Input:** nums = [1,2,3,4], k = 3
**Output:** false
```

**Constraints:**

* `1 <= k <= nums.length <= 16`
* `1 <= nums[i] <= 104`
* The frequency of each element is in the range `[1, 4]`.

## ac1: DFS

```java
class Solution {
    public boolean canPartitionKSubsets(int[] nums, int k) {
        // edge cases
        if (nums == null || nums.length == 0 || k <= 0 || nums.length < k) return false;

        int sum = 0;
        for (int n : nums) sum += n;
        if (sum % k != 0) return false;

        return helper(nums, 0, new boolean[nums.length], 0, sum/k, k-1);
    }

    private boolean helper(int[] nums, int startIndex, boolean[] visited, int currSum, int targetSum, int remainSubset) {
        // exit
        if (currSum > targetSum) {  // this element not good, find next
            return false;
        } else if (currSum == targetSum) {  // find one subset
            if (remainSubset == 0) return true; // last subset, since every subset meet target sum, and it has k subset, so return true
            else return helper(nums, 0, visited, 0, targetSum, remainSubset-1);  // go into next subset
        }

        // find elements add up to targetSum
        for (int i = startIndex; i < nums.length; i++) {  // startIndex avoid duplicate, O(2^n)
            if (visited[i]) continue;
            visited[i] = true;
            if (helper(nums, i+1, visited, currSum+nums[i], targetSum, remainSubset)) return true;
            visited[i] = false;
        }

        return false;
    }
}
```


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