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+# Leetcode Daily Challenge Solutions
+
+## Today's 31-03-24 [Problem Link](https://leetcode.com/problems/count-subarrays-with-fixed-bounds/description/?envType=daily-question&envId=2024-03-31)
+## 2444. Count Subarrays With Fixed Bounds
+
+# Intuition
+<!-- Describe your first thoughts on how to solve this problem. -->
+The problem requires counting the number of subarrays within a given range [minK, maxK] where the subarray's elements are also within this range. To efficiently count these subarrays, I can iterate through the array once and keep track of the last invalid index, the last occurrence of minK, and the last occurrence of maxK. By doing so, I can determine the valid subarrays that end at each index.
+
+# Approach
+<!-- Describe your approach to solving the problem. -->
+
+- Initialized variables :
+- `count`: to store the count of valid subarrays.
+- `lastInvalidIndex`: to keep track of the index of the last element that violates the range condition.
+- `lastMinIndex`: to store the index of the last occurrence of minK.
+- `lastMaxIndex`: to store the index of the last occurrence of maxK.
+
+- Iterated through the array `nums` :
+- Updated `lastInvalidIndex` if the current element is out of the range [minK, maxK].
+- Updated `lastMinIndex` if the current element is equal to minK.
+- Updated `lastMaxIndex` if the current element is equal to maxK.
+- Calculated the count of valid subarrays up to the current index by taking the maximum of 0 and the difference between the minimum of `lastMinIndex` and `lastMaxIndex` and `lastInvalidIndex`.
+- Added this count to the `count`.
+
+- Returned the total count of valid subarrays.
+
+By following this approach, I can efficiently count the number of valid subarrays within the given range.
+
+---
+Have a look at the code , still have any confusion then please let me know in the comments
+Keep Solving.:)
+# Complexity
+- Time complexity : $O(n)$
+<!-- Add your time complexity here, e.g. $$O(n)$$ -->
+$n$ : length of the input array `nums`
+- Space complexity : $O(1)$
+<!-- Add your space complexity here, e.g. $$O(n)$$ -->
+
+# Code
+```
+class Solution {
+
+// Function to count subarrays within the given range [minK, maxK]
+public long countSubarrays(int[] nums, int minK, int maxK) {
+
+// Variable to store the count of valid subarrays
+long count = 0;
+// Index of the last element that violates the range condition
+int lastInvalidIndex = -1;
+// Index of the last occurrence of minK
+int lastMinIndex = -1;
+// Index of the last occurrence of maxK
+int lastMaxIndex = -1;
+
+// Iterating through the array
+for (int i = 0; i < nums.length; ++i) {
+// Updating lastInvalidIndex if current element is out of range
+if (nums[i] < minK || nums[i] > maxK) {
+lastInvalidIndex = i;
+}
+// Updating lastMinIndex if current element is equal to minK
+if (nums[i] == minK) {
+lastMinIndex = i;
+}
+// Updating lastMaxIndex if current element is equal to maxK
+if (nums[i] == maxK) {
+lastMaxIndex = i;
+}
+// Calculating the count of valid subarrays up to the current index
+count += Math.max(0, Math.min(lastMinIndex, lastMaxIndex) - lastInvalidIndex);
+}
+// Returning the total count of valid subarrays
+return count;
+}
+}
+```