Description
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length0 <= j < A.length0 <= k < A.lengthA[i] & A[j] & A[k] == 0, where&represents the bitwise-AND operator.
Example 1:
Input: [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
Note:
1 <= A.length <= 10000 <= A[i] < 2^16
Solutions
DP
class Solution:
def countTriplets(self, A: List[int]) -> int:
n = 1 << 16
m = 3
dp = [[0 for _ in range(n)] for _ in range(m + 1)]
for a in A:
dp[1][a] += 1
for i in range(1, m):
for j in range(n):
if dp[i][j]:
for a in A:
dp[i + 1][j & a] += dp[i][j]
return dp[m][0]
# 25/25 cases passed (11348 ms)
# Your runtime beats 7.14 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (14.9 MB)