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 <= matchsticks.length <= 151 <= matchsticks[i] <= 108
Solution
We are looking to build four sides of the same length, and we can easily calculate the target side length by sum(matchsticks)//4.
Then we can solve this problem by trying to put each matchstick on the four different sides, and see if we can reach a solution.
We can apply the backtracking2 template for aggregation, as we are only asked to find whether a solution exists.
Let's fill in the logic.
is_leaf:start_index == len(matchsticks), when we have used all of the matchsticks.get_edges: we can put a match stick on four different sides.is_valid:sides[i] + matchsticks[start_index] <= side_length, the choice is valid only when the newside[i]length will not exceedside_length.- additinal states:
sidesthat stores the current length of the four sides.
Implementation
1def makesquare(self, matchsticks: List[int]) -> bool:
2 if sum(matchsticks) % 4 != 0: return False
3 side_length = sum(matchsticks) // 4
4 matchsticks.sort(reverse=True)
5 sides = [0, 0, 0, 0]
6 def dfs(start_index):
7 if start_index == len(matchsticks):
8 return side_length == sides[0] == sides[1] == sides[2] == sides[3]
9
10 for i in range(4):
11 if sides[i] + matchsticks[start_index] <= side_length:
12 sides[i] += matchsticks[start_index]
13 if dfs(start_index + 1): return True
14 sides[i] -= matchsticks[start_index]
15 return False
16 return dfs(0)
What is the space complexity of the following code?
1int sum(int n) {
2 if (n <= 0) {
3 return 0;
4 }
5 return n + sum(n - 1);
6}
Depth first search is equivalent to which of the tree traversal order?
Solution Implementation
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