Leetcode: Matchsticks to Square

Matchsticks to Square

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• Split Array With Same Average
• CheatSheet: Leetcode For Code Interview
• CheatSheet: Common Code Problems & Follow-ups
• Tag: #inspiring, #knapsack, #backtracking

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:

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

// https://code.dennyzhang.com/matchsticks-to-square
// Basic Ideas: backtracking
//
// Complexity: Time O(4^n) Space O(1)
import "sort"
func dfs(pos int, sums []int, nums []int, target int) bool {
    if pos == len(nums) {
        return sums[0] == target && sums[1] == target &&
            sums[2] == target && sums[3] == target
    }

    // backtracking
    for i:=0; i<4; i++ {
        if sums[i]+nums[pos]>target {
            continue
        }
        sums[i]+=nums[pos]
        if dfs(pos+1, sums, nums, target) {
            return true
        }
        sums[i]-=nums[pos]
    }
    return false
}

func makesquare(nums []int) bool {
    if len(nums) < 4 {
        return false
    }
    sum := 0
    for _, num := range nums {
        sum += num
    }
    if sum%4 != 0 {
        return false
    }
    target := sum/4
    sort.Ints(nums)

    sums := []int{0, 0, 0, 0}
    return dfs(0, sums, nums, target)
}