LeetCode 473. Matchsticks to Square

原题链接在这里：https://leetcode.com/problems/matchsticks-to-square/

题目：

Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.

Example 1:

Input: [1,1,2,2,2]
Output: true

Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.

Example 2:

Input: [3,3,3,3,4]
Output: false

Explanation: You cannot find a way to form a square with all the matchsticks.

Note:

1. The length sum of the given matchsticks is in the range of 0 to 10^9.
2. The length of the given matchstick array will not exceed 15.

题解：

It is to check if matches could be deivied into 4 equal groups.

Calculate the target sum/4.

When checking if there is combination that hit target, DFS needs states nums, boolean [] visited, current starting index, current sum, target and how many groups already found.

If sum > target, return false.

If sum == target, then check next group, having starting index set back to 0.

The reason state needs both visited and current starting index is to save time. If nums[i] can't be used to sum to current target, then for j>i, there is no need to check nums[i] again. 

Because if nums[i] could be used, it would already be marked as true before.

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

Time Complexity: exponential.

Space: O(nums.length). stack space.

AC Java:

class Solution {
    public boolean makesquare(int[] nums) {
        if(nums == null || nums.length < 4){
            return false;
        }
        
        int sum = 0;
        for(int num : nums){
            sum += num;
        }
        
        if(sum%4 != 0){
            return false;
        }
        
        return dfs(nums, new boolean[nums.length], 0, 0, sum/4, 4);
    }
    
    private boolean dfs(int [] nums, boolean [] visited, int start, int sum, int target, int edges){
        if(edges == 1){
            return true;
        }
        
        if(sum == target){
            return dfs(nums, visited, 0, 0, target, edges-1);
        }
        
        if(sum > target){
            return false;
        }
        
        for(int i = start; i<nums.length; i++){
            if(!visited[i]){
                visited[i] = true;
                if(dfs(nums, visited, i+1, sum+nums[i], target, edges)){
                    return true;
                }
                
                visited[i] = false;
            }
        }
        
        return false;
    }
}

类似Partition to K Equal Sum Subsets.