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You are given a 0-indexed positive integer array nums
and a positive integer k
.
A pair of numbers (num1, num2)
is called excellent if the following conditions are satisfied:
num1
and num2
exist in the array nums
.num1 OR num2
and num1 AND num2
is greater than or equal to k
, where OR
is the bitwise OR operation and AND
is the bitwise AND operation.Return the number of distinct excellent pairs.
Two pairs (a, b)
and (c, d)
are considered distinct if either a != c
or b != d
. For example, (1, 2)
and (2, 1)
are distinct.
Note that a pair (num1, num2)
such that num1 == num2
can also be excellent if you have at least one occurrence of num1
in the array.
Example 1:
Input: nums = [1,2,3,1], k = 3 Output: 5 Explanation: The excellent pairs are the following: - (3, 3). (3 AND 3) and (3 OR 3) are both equal to (11) in binary. The total number of set bits is 2 + 2 = 4, which is greater than or equal to k = 3. - (2, 3) and (3, 2). (2 AND 3) is equal to (10) in binary, and (2 OR 3) is equal to (11) in binary. The total number of set bits is 1 + 2 = 3. - (1, 3) and (3, 1). (1 AND 3) is equal to (01) in binary, and (1 OR 3) is equal to (11) in binary. The total number of set bits is 1 + 2 = 3. So the number of excellent pairs is 5.
Example 2:
Input: nums = [5,1,1], k = 10 Output: 0 Explanation: There are no excellent pairs for this array.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 109
1 <= k <= 60
class Solution:
def countExcellentPairs(self, nums: List[int], k: int) -> int:
s = set(nums)
ans = 0
cnt = Counter()
for v in s:
cnt[v.bit_count()] += 1
for v in s:
t = v.bit_count()
for i, x in cnt.items():
if t + i >= k:
ans += x
return ans
class Solution {
public long countExcellentPairs(int[] nums, int k) {
Set<Integer> s = new HashSet<>();
for (int v : nums) {
s.add(v);
}
long ans = 0;
int[] cnt = new int[32];
for (int v : s) {
int t = Integer.bitCount(v);
++cnt[t];
}
for (int v : s) {
int t = Integer.bitCount(v);
for (int i = 0; i < 32; ++i) {
if (t + i >= k) {
ans += cnt[i];
}
}
}
return ans;
}
}
class Solution {
public:
long long countExcellentPairs(vector<int>& nums, int k) {
unordered_set<int> s(nums.begin(), nums.end());
vector<int> cnt(32);
for (int v : s) ++cnt[__builtin_popcount(v)];
long long ans = 0;
for (int v : s) {
int t = __builtin_popcount(v);
for (int i = 0; i < 32; ++i) {
if (t + i >= k) {
ans += cnt[i];
}
}
}
return ans;
}
};
func countExcellentPairs(nums []int, k int) int64 {
s := map[int]bool{}
for _, v := range nums {
s[v] = true
}
cnt := make([]int, 32)
for v := range s {
t := bits.OnesCount(uint(v))
cnt[t]++
}
ans := 0
for v := range s {
t := bits.OnesCount(uint(v))
for i, x := range cnt {
if t+i >= k {
ans += x
}
}
}
return int64(ans)
}
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