62 lines
1.6 KiB
C++
62 lines
1.6 KiB
C++
// Source : https://leetcode.com/problems/triples-with-bitwise-and-equal-to-zero/
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// Author : Hao Chen
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// Date : 2020-07-26
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/*****************************************************************************************************
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*
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* Given an array of integers A, find the number of triples of indices (i, j, k) such that:
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*
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* 0 <= i < A.length
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* 0 <= j < A.length
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* 0 <= k < A.length
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* A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
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*
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* Example 1:
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*
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* Input: [2,1,3]
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* Output: 12
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* Explanation: We could choose the following i, j, k triples:
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* (i=0, j=0, k=1) : 2 & 2 & 1
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* (i=0, j=1, k=0) : 2 & 1 & 2
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* (i=0, j=1, k=1) : 2 & 1 & 1
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* (i=0, j=1, k=2) : 2 & 1 & 3
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* (i=0, j=2, k=1) : 2 & 3 & 1
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* (i=1, j=0, k=0) : 1 & 2 & 2
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* (i=1, j=0, k=1) : 1 & 2 & 1
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* (i=1, j=0, k=2) : 1 & 2 & 3
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* (i=1, j=1, k=0) : 1 & 1 & 2
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* (i=1, j=2, k=0) : 1 & 3 & 2
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* (i=2, j=0, k=1) : 3 & 2 & 1
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* (i=2, j=1, k=0) : 3 & 1 & 2
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*
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* Note:
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*
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* 1 <= A.length <= 1000
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* 0 <= A[i] < 2^16
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*
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******************************************************************************************************/
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class Solution {
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public:
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int countTriplets(vector<int>& A) {
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int n = A.size();
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//using a map to aggregate the duplication
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unordered_map<int, int> rec;
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for (int i=0; i<n; i++) {
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for (int j=0; j<n; j++) {
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rec[A[i] & A[j]]++;
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}
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}
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int result = 0;
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for (auto &r : rec ) {
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for (int k=0; k<n; k++) {
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if ((r.first & A[k]) == 0) result+=r.second;
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}
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}
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return result;
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}
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};
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