Count Subarrays With Fixed Bounds | Leetcode 2444

class Solution {
    #define ll long long
public:
    long long countSubarrays(vector<int>& nums, int minK, int maxK) {
        ll valid_subarrays = 0;
        int invalid_idx = -1;
        int minK_idx = -1;
        int maxK_idx = -1;

        for(int i=0;i<nums.size();++i){
            if(nums[i]<minK or nums[i]>maxK)    invalid_idx = i;
            
            if(nums[i]==minK)   minK_idx = i;
            if(nums[i]==maxK)   maxK_idx = i;

            int count = max((min(minK_idx,maxK_idx)-invalid_idx),0);
            valid_subarrays += count;
        }
        return valid_subarrays;
    }
};
/*
//JAVA
class Solution {
    public long countSubarrays(int[] nums, int minK, int maxK) {
        long validSubarrays = 0;
        int invalidIdx = -1;
        int minKIdx = -1;
        int maxKIdx = -1;

        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] < minK || nums[i] > maxK) {
                invalidIdx = i;
            }
            
            if (nums[i] == minK) {
                minKIdx = i;
            }
            if (nums[i] == maxK) {
                maxKIdx = i;
            }

            int count = Math.max(Math.min(minKIdx, maxKIdx) - invalidIdx, 0);
            validSubarrays += count;
        }
        return validSubarrays;
    }
}

#Python
class Solution:
    def countSubarrays(self, nums: List[int], minK: int, maxK: int) -> int:
        valid_subarrays = 0
        invalid_idx = -1
        minK_idx = -1
        maxK_idx = -1

        for i in range(len(nums)):
            if nums[i] < minK or nums[i] > maxK:
                invalid_idx = i
            
            if nums[i] == minK:
                minK_idx = i
            if nums[i] == maxK:
                maxK_idx = i

            count = max(min(minK_idx, maxK_idx) - invalid_idx, 0)
            valid_subarrays += count

        return valid_subarrays
*/