698. Partition to K Equal Sum Subsets

Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into knon-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.

https://www.youtube.com/watch?v=8XEcEYsG6Ck

cur_num is used to count how many numbers in the current subset

Update: This question has been changed after the contest. It added the special restriction 0 < nums[i] < 10000. My solution here is without that consideration.

Assume sum is the sum of nums[] . The dfs process is to find a subset of nums[] which sum equals to sum/k. We use an array visited[] to record which element in nums[] is used. Each time when we get a cur_sum = sum/k, we will start from position 0in nums[] to look up the elements that are not used yet and find another cur_sum = sum/k.

An corner case is when sum = 0, my method is to use cur_num to record the number of elements in the current subset. Only ifcur_sum = sum/k && cur_num >0, we can start another look up process.

Some test cases may need to be added in:
nums = {-1,1,0,0}, k = 4
nums = {-1,1}, k = 1
nums = {-1,1}, k = 2
nums = {-1,1,0}, k = 2
...

Java version:

    public boolean canPartitionKSubsets(int[] nums, int k) {
        int sum = 0;
        for(int num:nums)sum += num;
        if(k <= 0 || sum%k != 0)return false;
        int[] visited = new int[nums.length];
        return canPartition(nums, visited, 0, k, 0, 0, sum/k);
    }
    
    public boolean canPartition(int[] nums, int[] visited, int start_index, int k, int cur_sum, int cur_num, int target){
        if(k==1)return true;
        if(cur_sum == target && cur_num>0)return canPartition(nums, visited, 0, k-1, 0, 0, target);
        for(int i = start_index; i<nums.length; i++){
            if(visited[i] == 0){
                visited[i] = 1;
                if(canPartition(nums, visited, i+1, k, cur_sum + nums[i], cur_num++, target))return true;
                visited[i] = 0;
            }
        }
        return false;
    }

 带返回值的dfs recursion