Medium
Count Subarrays Where Max Element Appears at Least K Times — C++
Full explanation · Time O(n) · Space O(1)
// Time: O(n)
// Space: O(1)
// two pointers, sliding window
class Solution {
public:
long long countSubarrays(vector<int>& nums, int k) {
const int mx = *max_element(cbegin(nums), cend(nums));
int64_t result = 0;
for (int right = 0, left = 0, cnt = 0; right < size(nums); ++right) {
cnt += static_cast<int>(nums[right] == mx);
while (cnt == k) {
cnt -= static_cast<int>(nums[left++] == mx);
}
result += left;
}
return result;
}
};
// Time: O(n)
// Space: O(1)
// two pointers, sliding window
class Solution2 {
public:
long long countSubarrays(vector<int>& nums, int k) {
const int mx = *max_element(cbegin(nums), cend(nums));
int64_t result = (size(nums) + 1) * size(nums) / 2;
for (int right = 0, left = 0, cnt = 0; right < size(nums); ++right) {
cnt += static_cast<int>(nums[right] == mx);
while (cnt == k) {
cnt -= static_cast<int>(nums[left++] == mx);
}
result -= right - left + 1;
}
return result;
}
};