Medium

Count Alternating SubarraysC++

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;
    }
};