526 Beautiful Arrangement
1. Question
Suppose you have N integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these N numbers successfully if one of the following is true for the ith position (1 <= i <= N) in this array:
The number at the ith position is divisible by i.
i is divisible by the number at the ith position.
Now given N, how many beautiful arrangements can you construct?
Example 1:
Input: 2
Output: 2
Explanation:
The first beautiful arrangement is [1, 2]:
Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).
Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).
The second beautiful arrangement is [2, 1]:
Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).
Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.
Note:
N is a positive integer and will not exceed 15.
2. Implementation
(1) Backtracking
class Solution {
public int countArrangement(int N) {
boolean[] used = new boolean[N + 1];
int[] count = new int[1];
getArrangement(1, N, used, count);
return count[0];
}
public void getArrangement(int curNum, int N, boolean[] used, int[] count) {
if (curNum == N + 1) {
++count[0];
return;
}
for (int i = 1; i <= N; i++) {
if (!used[i] && (curNum % i == 0 || i % curNum == 0)) {
used[i] = true;
getArrangement(curNum + 1, N, used, count);
used[i] = false;
}
}
}
}
3. Time & Space Complexity
Backtracking: 时间复杂度O(N!), 空间复杂度O(N)
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