Skip to content

473. Matchsticks to Square

Description

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 <= matchsticks.length <= 15
  • 1 <= matchsticks[i] <= 108

Solutions

Solution 1

Thinking

A square needs a total divisible by \(4\) and no stick longer than a side. Assigning \(n\le 15\) sticks to four sides is \(4^n\) without pruning.

Place large sticks first: try each side that would not overflow, and skip a side equal to the previous one. Fail immediately on a bad total or a too-long stick.

Large sticks overflow sooner; equal-side skipping removes symmetric states.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
class Solution:
    def makesquare(self, matchsticks: List[int]) -> bool:
        def dfs(u):
            if u == len(matchsticks):
                return True
            for i in range(4):
                if i > 0 and edges[i - 1] == edges[i]:
                    continue
                edges[i] += matchsticks[u]
                if edges[i] <= x and dfs(u + 1):
                    return True
                edges[i] -= matchsticks[u]
            return False

        x, mod = divmod(sum(matchsticks), 4)
        if mod or x < max(matchsticks):
            return False
        edges = [0] * 4
        matchsticks.sort(reverse=True)
        return dfs(0)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
class Solution {
    public boolean makesquare(int[] matchsticks) {
        int s = 0, mx = 0;
        for (int v : matchsticks) {
            s += v;
            mx = Math.max(mx, v);
        }
        int x = s / 4, mod = s % 4;
        if (mod != 0 || x < mx) {
            return false;
        }
        Arrays.sort(matchsticks);
        int[] edges = new int[4];
        return dfs(matchsticks.length - 1, x, matchsticks, edges);
    }

    private boolean dfs(int u, int x, int[] matchsticks, int[] edges) {
        if (u < 0) {
            return true;
        }
        for (int i = 0; i < 4; ++i) {
            if (i > 0 && edges[i - 1] == edges[i]) {
                continue;
            }
            edges[i] += matchsticks[u];
            if (edges[i] <= x && dfs(u - 1, x, matchsticks, edges)) {
                return true;
            }
            edges[i] -= matchsticks[u];
        }
        return false;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution {
public:
    bool makesquare(vector<int>& matchsticks) {
        int s = 0, mx = 0;
        for (int& v : matchsticks) {
            s += v;
            mx = max(mx, v);
        }
        int x = s / 4, mod = s % 4;
        if (mod != 0 || x < mx) return false;
        sort(matchsticks.begin(), matchsticks.end(), greater<int>());
        vector<int> edges(4);
        return dfs(0, x, matchsticks, edges);
    }

    bool dfs(int u, int x, vector<int>& matchsticks, vector<int>& edges) {
        if (u == matchsticks.size()) return true;
        for (int i = 0; i < 4; ++i) {
            if (i > 0 && edges[i - 1] == edges[i]) continue;
            edges[i] += matchsticks[u];
            if (edges[i] <= x && dfs(u + 1, x, matchsticks, edges)) return true;
            edges[i] -= matchsticks[u];
        }
        return false;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
func makesquare(matchsticks []int) bool {
    s := 0
    for _, v := range matchsticks {
        s += v
    }
    if s%4 != 0 {
        return false
    }
    sort.Sort(sort.Reverse(sort.IntSlice(matchsticks)))
    edges := make([]int, 4)
    var dfs func(u, x int) bool
    dfs = func(u, x int) bool {
        if u == len(matchsticks) {
            return true
        }
        for i := 0; i < 4; i++ {
            if i > 0 && edges[i-1] == edges[i] {
                continue
            }
            edges[i] += matchsticks[u]
            if edges[i] <= x && dfs(u+1, x) {
                return true
            }
            edges[i] -= matchsticks[u]
        }
        return false
    }
    return dfs(0, s/4)
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
impl Solution {
    pub fn makesquare(matchsticks: Vec<i32>) -> bool {
        let mut matchsticks = matchsticks;

        fn dfs(matchsticks: &Vec<i32>, edges: &mut [i32; 4], u: usize, x: i32) -> bool {
            if u == matchsticks.len() {
                return true;
            }
            for i in 0..4 {
                if i > 0 && edges[i - 1] == edges[i] {
                    continue;
                }
                edges[i] += matchsticks[u];
                if edges[i] <= x && dfs(matchsticks, edges, u + 1, x) {
                    return true;
                }
                edges[i] -= matchsticks[u];
            }
            false
        }

        let sum: i32 = matchsticks.iter().sum();
        if sum % 4 != 0 {
            return false;
        }
        matchsticks.sort_by(|x, y| y.cmp(x));
        let mut edges = [0; 4];

        dfs(&matchsticks, &mut edges, 0, sum / 4)
    }
}

Solution 2

Thinking

Side-by-side backtracking still repeats partitions. A bit mask of used sticks plus the current side sum \(t\) is enough: a legal stick updates \(t\gets (t+v)\bmod s\). After sorting, \(t+v>s\) cuts the rest of the loop.

Memoizing \((\textit{mask},t)\) folds duplicate partitions into \(O(n\,2^n)\).

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
class Solution:
    def makesquare(self, matchsticks: List[int]) -> bool:
        @cache
        def dfs(state, t):
            if state == (1 << len(matchsticks)) - 1:
                return True
            for i, v in enumerate(matchsticks):
                if state & (1 << i):
                    continue
                if t + v > s:
                    break
                if dfs(state | (1 << i), (t + v) % s):
                    return True
            return False

        s, mod = divmod(sum(matchsticks), 4)
        matchsticks.sort()
        if mod:
            return False
        return dfs(0, 0)

Comments