473. Matchsticks to Square

package LeetCode_473

/**
 * 473. Matchsticks to Square
 *https://leetcode.com/problems/matchsticks-to-square/
 * You are given an integer array matchsticks where matchsticks[i] is the length of the ith matchstick.
 * You want to use all the matchsticks to make one square.
 * You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.
Return true if you can make this square and false otherwise.

Example 1:
Input: matchsticks = [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: matchsticks = [3,3,3,3,4]
Output: false
Explanation: You cannot find a way to form a square with all the matchsticks.

Constraints:
1. 1 <= matchsticks.length <= 15
2. 1 <= matchsticks[i] <= 10^8
 * */
class Solution {
    /*
    * solution: DFS, check each side's sum if equals to target, that's mean we find out the subsequence which the sum if can divided by 4
    * Time complexity: O(2^n), Space: O(2^n)
    * */
    fun makesquare(matchsticks: IntArray): Boolean {
        val sum = matchsticks.sum()
        //base checking
        if (sum % 4 != 0 || matchsticks.size < 4) {
            return false
        }
        val target = sum / 4
        //array to save the sum of every side of square
        val sideSums = IntArray(4)
        return dfs(0, target, sideSums, matchsticks)
    }

    private fun dfs(index: Int, target: Int, sideSums: IntArray, matchsticks: IntArray): Boolean {
        if (index >= matchsticks.size) {
            //check each side's sum if equals to target
            return sideSums[0] == target && sideSums[1] == target && sideSums[2] == target
        }
        //calculate the sum of 4 sides
        for (i in 0 until 4) {
            //do some pruning
            if (sideSums[i] + matchsticks[index] > target) {
                continue
            }
            sideSums[i] += matchsticks[index]
            //and check for next level
            if (dfs(index + 1, target, sideSums, matchsticks)) {
                return true
            }
            //reduce for backtracking
            sideSums[i] -= matchsticks[index]
        }
        return false
    }
}