Medium
Count Alternating Subarrays — C++
Full explanation · Time O(n) · Space O(1)
// Time: O(n)
// Space: O(1)
// dp
class Solution {
public:
long long countAlternatingSubarrays(vector<int>& nums) {
int64_t result = 0;
for (int i = 0, curr = 0; i < size(nums); ++i) {
if (i - 1 >= 0 && nums[i - 1] == nums[i]) {
curr = 0;
}
result += ++curr;
}
return result;
}
};