class Solution {
 public:
  long long countSubarrays(vector<int>& nums, int minK, int maxK) {
    long long ans = 0;
    int j = -1;
    int prevMinKIndex = -1;
    int prevMaxKIndex = -1;

    for (int i = 0; i < nums.size(); ++i) {
      if (nums[i] < minK || nums[i] > maxK)
        j = i;
      if (nums[i] == minK)
        prevMinKIndex = i;
      if (nums[i] == maxK)
        prevMaxKIndex = i;
      // any index k in [j + 1, min(prevMinKIndex, prevMaxKIndex)] can be the
      // start of the subarray s.t. nums[k..i] satisfies the conditions
      ans += max(0, min(prevMinKIndex, prevMaxKIndex) - j);
    }

    return ans;
  }
};