473

473. Matchsticks to Square (Medium)

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

Related Topics:
Array, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask

Solution 1. Bitmask DP on Subsets

// OJ: https://leetcode.com/problems/matchsticks-to-square/
// Author: github.com/lzl124631x
// Time: O(N * 2^N)
// Space: O(2^N)
class Solution {
public:
    bool makesquare(vector<int>& A) {
        int sum = accumulate(begin(A), end(A), 0), N = A.size();
        if (sum % 4 || *max_element(begin(A), end(A)) > sum / 4) return false;
        sum /= 4;
        sort(begin(A), end(A), greater<>()); // Try rocks before sands
        vector<int> dp(1 << N, -1); // -1 unvisited, 0 invalid, 1 valid
        dp[(1 << N) - 1] = 1;
        function<bool(int, int)> dfs = [&](int mask, int target) {
            if (dp[mask] != -1) return dp[mask];
            dp[mask] = 0;
            if (target == 0) target = sum;
            for (int i = 0; i < N && !dp[mask]; ++i) {
                if ((mask >> i & 1) || A[i] > target) continue;
                dp[mask] = dfs(mask | (1 << i), target - A[i]);
            }
            return dp[mask];
        };
        return dfs(0, sum);
    }
};

Solution 2. Backtrack to Fill Buckets

Let target = sum(A) / 4, which is the target length of each edge.

We use DFS to try to fill each A[i] into different edges.

Two optimizations here:

1. The unordered_set<int> seen is used to prevent handling the same edge value again. For example, assume edge = [1, 1, 1, 1], and now we are trying to add a stick of length 2 to it. Adding 2 to either 1 will yield the same result. So we just need to add to a edge with length 1 once.
2. Sorting the sticks in descending order will make it converge faster because it's easy to fill in sands but hard to fill in peddles; filling peddles first will fail faster.

// OJ: https://leetcode.com/problems/matchsticks-to-square/
// Author: github.com/lzl124631x
// Time: O(4^N)
// Space: O(N * SUM(A))
class Solution {
public:
    bool makesquare(vector<int>& A) {
        long v[4] = {}, sum = accumulate(begin(A), end(A), 0L);
        if (sum % 4 || *max_element(begin(A), end(A)) > sum / 4) return false;
        sum /= 4;
        sort(begin(A), end(A), greater<>()); // Try rocks before sands
        function<bool(int)> dfs = [&](int i) {
            if (i == A.size()) return true;
            unordered_set<long> seen;
            for (int j = 0; j < 4; ++j) {
                if (A[i] + v[j] > sum || seen.count(v[j])) continue;
                seen.insert(v[j]);
                v[j] += A[i];
                if (dfs(i + 1)) return true;
                v[j] -= A[i];
            }
            return false;
        };
        return dfs(0);
    }
};

Or

// OJ: https://leetcode.com/problems/matchsticks-to-square/
// Author: github.com/lzl124631x
// Time: O(4^N)
// Space: O(N * SUM(A))
class Solution {
public:
    bool makesquare(vector<int>& A) {
        long v[4] = {}, sum = accumulate(begin(A), end(A), 0L), N = A.size();
        if (sum % 4) return false;
        sum /= 4;
        sort(begin(A), end(A), greater<>()); // Try rocks before sands
        function<bool(int)> dfs = [&](int i) {
            if (i == N) return true;
            for (int j = 0; j < 4; ++j) {
                if (A[i] + v[j] > sum) continue;
                v[j] += A[i];
                if (dfs(i + 1)) return true;
                v[j] -= A[i];
                if (v[j] == 0) break; // Simply don't visit empty bucket again. This takes less space but longer time.
            }
            return false;
        };
        return dfs(0);
    }
};