Table of Contents
Matchsticks to Square – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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
C++ Matchsticks to Square LeetCode Solution
class Solution {
bool dfs(vector<int> &sidesLength,const vector<int> &matches, int index) {
if (index == matches.size())
return sidesLength[0] == sidesLength[1] && sidesLength[1] == sidesLength[2] && sidesLength[2] == sidesLength[3];
for (int i = 0; i < 4; ++i) {
sidesLength[i] += matches[index];
if (dfs(sidesLength, matches, index + 1))
return true;
sidesLength[i] -= matches[index];
}
return false;
}
public:
bool makesquare(vector<int>& nums) {
if (nums.empty()) return false;
vector<int> sidesLength(4, 0);
return dfs(sidesLength, nums, 0);
}
};
Java Matchsticks to Square LeetCode Solution
public 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;
return dfs(nums, new int[4], 0, sum / 4);
}
private boolean dfs(int[] nums, int[] sums, int index, int target) {
if (index == nums.length) {
if (sums[0] == target && sums[1] == target && sums[2] == target) {
return true;
}
return false;
}
for (int i = 0; i < 4; i++) {
if (sums[i] + nums[index] > target) continue;
sums[i] += nums[index];
if (dfs(nums, sums, index + 1, target)) return true;
sums[i] -= nums[index];
}
return false;
}
}
Python 3 Matchsticks to Square LeetCode Solution
class Solution:
def makesquare(self, nums):
N = len(nums)
basket, rem = divmod(sum(nums), 4)
if rem or max(nums) > basket: return False
@lru_cache(None)
def dfs(mask):
if mask == 0: return 0
for j in range(N):
if mask & 1<<j:
neib = dfs(mask ^ 1<<j)
if neib >= 0 and neib + nums[j] <= basket:
return (neib + nums[j]) % basket
return -1
return dfs((1<<N) - 1) == 0
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Topological Sort-31
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Game Theory-24
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Interactive-18
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Number Theory-16
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Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
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Suffix Array-6
Minimum Spanning Tree-5
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Shell-4
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Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1
