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Formatted question description: https://leetcode.ca/all/473.html
473. Matchsticks to Square (Medium)
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
0to10^9. - The length of the given matchstick array will not exceed
15.
Related Topics:
Depth-first Search
Solution 1. DP on Subsets
// OJ: https://leetcode.com/problems/matchsticks-to-square/
// Time: O(N * 2^N)
// Space: O(2^N)
class Solution {
vector<int> dp;
bool dfs(vector<int> &A, int used, int todo, int target) {
if (dp[used] != -1) return dp[used];
dp[used] = 0;
int mx = todo % target;
if (mx == 0) mx = target;
for (int i = 0; i < A.size() && !dp[used]; ++i) {
dp[used] = ((used >> i) & 1) == 0 && A[i] <= mx && dfs(A, used | (1 << i), todo - A[i], target);
}
return dp[used];
}
public:
bool makesquare(vector<int>& A) {
if (A.empty()) return false;
int len = accumulate(begin(A), end(A), 0);
if (len % 4 || *max_element(begin(A), end(A)) > len / 4) return false;
dp.assign(1 << A.size(), -1);
dp[(1 << A.size()) - 1] = 1;
return dfs(A, 0, len, len / 4);
}
};
Solution 2. DFS 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:
- The
unordered_set<int> seenis used to prevent handling the same edge value again. For example, assumeedge = [1, 1, 1, 1], and now we are trying to add a stick of length2to it. Adding2to either1will yield the same result. So we just need to add to a edge with length1once. - 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/
// Time: O(4^N)
// Space: O(N * SUM(A))
class Solution {
int edge[4] = {}, target;
bool dfs(vector<int> &A, int i) {
if (i == A.size()) return true;
unordered_set<int> seen;
for (int j = 0; j < 4; ++j) {
if (seen.count(edge[j]) || edge[j] + A[i] > target) continue;
seen.insert(edge[j]);
edge[j] += A[i];
if (dfs(A, i + 1)) return true;
edge[j] -= A[i];
}
return false;
}
public:
bool makesquare(vector<int>& A) {
if (A.size() < 4) return false;
int sum = accumulate(begin(A), end(A), 0);
if (sum % 4 || *max_element(begin(A), end(A)) > sum / 4) return false;
target = sum / 4;
sort(begin(A), end(A), greater<>());
return dfs(A, 0);
}
};
-
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; int sideLength = sum / 4; boolean[] used = new boolean[nums.length]; return backtrack(nums, used, 0, 0, sideLength, 0); } public boolean backtrack(int[] nums, boolean[] used, int index, int sides, int sideLength, int curSideLength) { if (sides == 4) return true; int length = nums.length; boolean flag = false; for (int i = index; i < length; i++) { if (!used[i]) { int num = nums[i]; int curSum = curSideLength + num; if (curSum <= sideLength) { used[i] = true; if (curSum == sideLength) flag = flag || backtrack(nums, used, 0, sides + 1, sideLength, 0); else flag = flag || backtrack(nums, used, i + 1, sides, sideLength, curSum); if (flag) break; used[i] = false; } } } return flag; } } ############ 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; } } -
// OJ: https://leetcode.com/problems/matchsticks-to-square/ // 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); } }; -
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) ############ import collections class Solution(object): def makesquare(self, nums): """ :type nums: List[int] :rtype: bool """ if not nums: return False sumLen = sum(nums) if sumLen % 4 != 0: return False self.sideLen = sideLen = sumLen / 4 for side in nums: if side > sideLen: return False halfLen = 2 * sideLen sticksIdx = set([i for i in range(0, len(nums))]) nums.sort() def backpack(nums, subset): cands = [nums[k] for k in subset] dp = [[False] * (self.sideLen + 1) for _ in range(len(cands))] for i in range(0, len(cands)): dp[i][0] = True for i in range(0, len(cands)): for j in range(1, self.sideLen + 1): dp[i][j] |= dp[i - 1][j] if j - cands[i] >= 0: dp[i][j] |= dp[i - 1][j - cands[i]] return dp[-1][-1] def dfs(nums, start, sticksIdx, halfLen, subSum, subsetIdx): if subSum >= halfLen: if subSum == halfLen and backpack(nums, subsetIdx) and backpack(nums, sticksIdx): return True return False for i in range(start, len(nums)): if i > start and nums[i] == nums[i - 1]: continue if i in sticksIdx: sticksIdx -= {i} subsetIdx |= {i} if dfs(nums, i + 1, sticksIdx, halfLen, subSum + nums[i], subsetIdx): return True subsetIdx -= {i} sticksIdx |= {i} return False return dfs(nums, 0, sticksIdx, halfLen, 0, set()) -
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) } -
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) } }