LeetCode 698. Partition to K Equal Sum Subsets

原题链接在这里：https://leetcode.com/problems/partition-to-k-equal-sum-subsets/description/

题目：

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.

题解：

首先计算sum, 看sum能否被k整除. 若不能, 铁定不能分成k组. return false.

若能的话，每组的target sum就该是sum/k. 一组一组的减掉. 直到 k = 1. 剩下最后一组, 最后一组的sum肯定是sum/k.

因为这里的已经验证过sum是k的倍数, 而前面已经有k-1组 sum/k找到了. 所以可以直接return true.

This is bottom-up recursion. Set parameters for state first.

It needs count to count number in subarray. Since there may be negative number in nums. If target is 0, there could be [-1, 1] or empty subarray. 

The reason state has both visited and cur starting index is because of trimming dfs tree.

When summing up to target, if index i can't be used, when trying j > i, the next level of DFS, there is no need to try i again. Because if i works, it would be added into res before.

The only case i could be used is to sum up next target.

Note: the question is asking for non-empty, we need to add a count of each sub set. And make sure it is > 0 before accumlating to result.

Time Complexity: exponential.

Space: O(n). stack space.

AC Java:

class Solution {
    public boolean canPartitionKSubsets(int[] nums, int k) {
        if(nums == null || nums.length == 0){
            return false;
        }
        
        int sum = 0;
        for(int num : nums){
            sum += num;
        }
        
        if(sum % k != 0){
            return false;
        }
        
        boolean [] visited = new boolean[nums.length];
        return dfs(nums, visited, 0, 0, sum/k, 0, k);
    }
    
    private boolean dfs(int [] nums, boolean [] visited, int cur, int sum, int target, int count, int k){
        if(sum > target){
            return false;
        }
        
        if(k == 1){
            return true;
        }
        
        if(sum == target && count > 0){
            return dfs(nums, visited, 0, 0, target, 0, k-1);
        }
        
        for(int i = cur; i<nums.length; i++){
            if(!visited[i]){
                visited[i] = true;
                if(dfs(nums, visited, i+1, sum+nums[i], target, count++, k)){
                    return true;
                }
                
                visited[i] = false;
            }
        }
        
        return false;
    }
}

类似Partition Equal Subset Sum, Matchsticks to Square.