LeetCode Solutions
class Solution {
public:
int numSquarefulPerms(vector<int>& A) {
int ans = 0;
sort(begin(A), end(A));
dfs(A, vector<boool>(A.size()), {}, ans);
return ans;
}
private:
void dfs(vector<int>& A, vector<bool>&& used, vector<int>&& path, int& ans) {
if (path.size() > 1 && !isSquare(path.back() + path[path.size() - 2]))
return;
if (path.size() == A.size()) {
++ans;
return;
}
for (int i = 0; i < A.size(); ++i) {
if (used[i])
continue;
if (i > 0 && A[i] == A[i - 1] && !used[i - 1])
continue;
used[i] = true;
path.push_back(A[i]);
dfs(A, move(used), move(path), ans);
path.pop_back();
used[i] = false;
}
}
bool isSquare(int num) {
const int root = sqrt(num);
return root * root == num;
}
};
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class Solution {
public int numSquarefulPerms(int[] A) {
boolean[] used = new boolean[A.length];
Arrays.sort(A);
dfs(A, used, new ArrayList<>());
return ans;
}
private int ans = 0;
private void dfs(int[] A, boolean[] used, List<Integer> path) {
if (path.size() > 1 && !isSquare(path.get(path.size() - 1) + path.get(path.size() - 2)))
return;
if (path.size() == A.length) {
++ans;
return;
}
for (int i = 0; i < A.length; ++i) {
if (used[i])
continue;
if (i > 0 && A[i] == A[i - 1] && !used[i - 1])
continue;
used[i] = true;
path.add(A[i]);
dfs(A, used, path);
path.remove(path.size() - 1);
used[i] = false;
}
}
private boolean isSquare(int num) {
int root = (int) Math.sqrt(num);
return root * root == num;
}
}
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class Solution:
def numSquarefulPerms(self, A: List[int]) -> int:
ans = 0
used = [False] * len(A)
def isSquare(num: int) -> bool:
root = int(sqrt(num))
return root * root == num
def dfs(path: List[int]) -> None:
nonlocal ans
if len(path) > 1 and not isSquare(path[-1] + path[-2]):
return
if len(path) == len(A):
ans += 1
return
for i, a in enumerate(A):
if used[i]:
continue
if i > 0 and A[i] == A[i - 1] and not used[i - 1]:
continue
used[i] = True
dfs(path + [a])
used[i] = False
A.sort()
dfs([])
return ans
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