# Question

Formatted question description: https://leetcode.ca/all/52.html

The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.

Given an integer n, return the number of distinct solutions to the n-queens puzzle.

Example 1: Input: n = 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown.


Example 2:

Input: n = 1
Output: 1


Constraints:

• 1 <= n <= 9

# Algorithm

In fact, I think the order of the two Queens-question should be reversed. The previous question is a little more complicated than this one. There is no difference in essence between the two. Both have to be solved by backtracking. If you understand the previous question The idea of the question, this question only needs to make a small change, no need to ask for the specific queen’s pendulum, just need to increase the counter by one every time a solution is generated.

# Code

• 
public class N_Queens_II {

// just simply return list.size(), seems costly if only count is needed

public class Solution {

List<List<String>> list = new ArrayList<List<String>>();

public int totalNQueens(int n) {

if (n <= 0)
return 0;

// index is row-number, place[index] is position of queen on that
// row
int[] place = new int[n];

dfs(n, 0, place);

return list.size();
}

public void dfs(int n, int currentRow, int[] place) {
if (currentRow == n) {
// save to list
saveToList(place);
return;
}

for (int i = 0; i < n; i++) {

if (isValid(i, currentRow, place)) {
place[currentRow] = i; // 1 meaning place queen
dfs(n, currentRow + 1, place);
place[currentRow] = i; // undo place queen
}
}
}

public boolean isValid(int currentRowPosition, int currentRow,
int[] place) {
// check all previous rows
for (int i = 0; i < currentRow; i++) {
if (place[i] == currentRowPosition
|| currentRow - i == (int) Math.abs(place[i] - currentRowPosition))
return false;
}

return true;
}

public void saveToList(int[] place) {

int len = place.length;
List<String> one = new ArrayList<String>(len);

// construct string of n dots...
String row = "";
for (int i = 0; i < len; i++) {
row += ".";
}

for (int i = 0; i < len; i++) {
int rowPosition = place[i]; // at position place[i] on row i
+ row.substring(rowPosition + 1));
}

}
}

}


• // OJ: https://leetcode.com/problems/n-queens-ii/
// Time: O(N!)
// Space: O(N)
class Solution {
public:
int totalNQueens(int n) {
vector<bool> col(n), hill(2 * n - 1), dale(2 * n - 1);
function<int(int)> dfs = [&](int i) { // for each i-th row
if (i == n) return 1;
int ans = 0;
for (int j = 0; j < n; ++j) { // try putting on the j-th column
int h = i + j, d = i + n - 1 - j;
if (col[j] || hill[h] || dale[d]) continue;
col[j] = hill[h] = dale[d] = true;
ans += dfs(i + 1);
col[j] = hill[h] = dale[d] = false;
}
return ans;
};
return dfs(0);
}
};

• class Solution:
def totalNQueens(self, n: int) -> int:
def dfs(i):
if i == n:
nonlocal ans
ans += 1
return
for j in range(n):
a, b = i + j, i - j + n
if cols[j] or dg[a] or udg[b]:
continue
cols[j] = dg[a] = udg[b] = True
dfs(i + 1)
cols[j] = dg[a] = udg[b] = False

cols = [False] * 10
dg = [False] * 20
udg = [False] * 20
ans = 0
dfs(0)
return ans

############

class Solution(object):
def totalNQueens(self, n):
"""
:type n: int
:rtype: int
"""

def dfs(path, n):
if len(path) == n:
return 1
res = 0
for i in range(n):
if i not in path and isValidQueen(path, i):
path.append(i)
res += dfs(path, n)
path.pop()
return res

def isValidQueen(path, k):
for i in range(len(path)):
if abs(k - path[i]) == abs(len(path) - i):
return False
return True

return dfs([], n)


• func totalNQueens(n int) (ans int) {
cols := bool{}
dg := bool{}
udg := bool{}
var dfs func(int)
dfs = func(i int) {
if i == n {
ans++
return
}
for j := 0; j < n; j++ {
a, b := i+j, i-j+n
if cols[j] || dg[a] || udg[b] {
continue
}
cols[j], dg[a], udg[b] = true, true, true
dfs(i + 1)
cols[j], dg[a], udg[b] = false, false, false
}
}
dfs(0)
return
}