982 Triples with Bitwise AND Equal To Zero 按位與為零的三元組
Description:
Given an integer array nums, return the number of AND triples.
An AND triple is a triple of indices (i, j, k) such that:
0 <= i < nums.length
0 <= j < nums.length
0 <= k < nums.length
nums[i] & nums[j] & nums[k] == 0, where & represents the bitwise-AND operator.
Example:
Example 1:
Input: nums = [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Example 2:
Input: nums = [0,0,0]
Output: 27
Constraints:
1 <= nums.length <= 1000
0 <= nums[i] < 2^16
題目描述:
給定一個整數數組 A，找出索引為 (i, j, k) 的三元組，使得：
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0，其中 & 表示按位與（AND）操作符。
示例 :
輸入：[2,1,3]
輸出：12
解釋：我們可以選出如下 i, j, k 三元組：
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
提示:
1 <= A.length <= 1000
0 <= A[i] < 2^16
思路:
狀態壓縮
用一個 count 數組記錄兩個數與的結果
然后查詢能夠和當前數字與的結果為 0 的 count 中的對應值
每次消除最后一位 x = (x - 1) & x
注意需要將本身是 0 的也加上
時間復雜度為 O(n ^ 2), 空間復雜度為 O(m), m 為 max(nums)
代碼:
C++:
class Solution
{
public:
int countTriplets(vector<int>& nums)
{
int max_value = 1 << 16, n = nums.size(), result = 0;
vector<int> count(max_value);
for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) ++count[nums[i] & nums[j]];
for (int i = 0; i < n; i++) for (int cur = max_value - 1 - nums[i], j = cur; j > 0; j = (j - 1) & cur) result += count[j];
for (int i = 0; i < n; i++) result += count.front();
return result;
}
};
Java:
class Solution {
public int countTriplets(int[] nums) {
int maxValue = 1 << 16, n = nums.length, result = 0, count[] = new int[maxValue];
for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) ++count[nums[i] & nums[j]];
for (int i = 0; i < n; i++) for (int cur = maxValue - 1 - nums[i], j = cur; j > 0; j = (j - 1) & cur) result += count[j];
for (int i = 0; i < n; i++) result += count[0];
return result;
}
}
Python:
class Solution:
def countTriplets(self, nums: List[int]) -> int:
result, n, max_value = 0, len(nums), 1 << 16
count = Counter([nums[i] & nums[j] for i in range(n) for j in range(n)])
for i in range(n):
cur = j = max_value - 1 - nums[i]
result += count[0]
while j > 0:
result += count[j]
j = (j - 1) & cur
return result
