Medium
Count Number of Nice Subarrays — C++
Full explanation · Time O(n) · Space O(k)
// Time: O(n)
// Space: O(k)
class Solution {
public:
int numberOfSubarrays(vector<int>& nums, int k) {
return atMostK(nums, k) - atMostK(nums, k - 1);
}
private:
int atMostK(const vector<int>& nums, int k) {
int result = 0, left = 0, count = 0;
for (int right = 0; right < nums.size(); ++right) {
count += nums[right] % 2;
while (count > k) {
count -= nums[left] % 2;
++left;
}
result += right - left + 1;
}
return result;
}
};
// Time: O(n)
// Space: O(k)
class Solution2 {
public:
int numberOfSubarrays(vector<int>& nums, int k) {
int result = 0;
deque<int> dq = {-1};
for (int i = 0; i < nums.size(); ++i) {
if (nums[i] % 2) {
dq.emplace_back(i);
}
if (dq.size() > k + 1) {
dq.pop_front();
}
if (dq.size() == k + 1) {
result += dq[1] - dq[0];
}
}
return result;
}
};