LeetCode-in-Java
982. Triples with Bitwise AND Equal To Zero
Hard
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.length0 <= j < nums.length0 <= k < nums.lengthnums[i] & nums[j] & nums[k] == 0, where&represents the bitwise-AND operator.
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 <= 10000 <= nums[i] < 216
Solution
public class Solution {
public int countTriplets(int[] nums) {
int[] arr = new int[1 << 17];
for (int num : nums) {
int mask = 0;
for (int i = 0; i < 16; i++) {
if ((num & (1 << i)) == 0) {
mask |= (1 << i);
}
}
int s = mask;
while (s > 0) {
arr[s]++;
s = (s - 1) & mask;
}
}
int count = 0;
for (int j : nums) {
for (int num : nums) {
int val = j & num;
if (val == 0) {
count += nums.length;
} else {
int mask = 0;
for (int k = 0; k < 16; k++) {
if ((val & (1 << k)) > 0) {
mask |= (1 << k);
}
}
count += arr[mask];
}
}
}
return count;
}
}