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Java-based LeetCode algorithm problem solutions, regularly updated
package g2401_2500.s2444_count_subarrays_with_fixed_bounds;
// #Hard #Array #Sliding_Window #Queue #Monotonic_Queue
// #2022_12_14_Time_9_ms_(83.94%)_Space_56.5_MB_(97.01%)
/**
* 2444 - Count Subarrays With Fixed Bounds\.
*
* Hard
*
* You are given an integer array `nums` and two integers `minK` and `maxK`.
*
* A **fixed-bound subarray** of `nums` is a subarray that satisfies the following conditions:
*
* * The **minimum** value in the subarray is equal to `minK`.
* * The **maximum** value in the subarray is equal to `maxK`.
*
* Return _the **number** of fixed-bound subarrays_.
*
* A **subarray** is a **contiguous** part of an array.
*
* **Example 1:**
*
* **Input:** nums = [1,3,5,2,7,5], minK = 1, maxK = 5
*
* **Output:** 2
*
* **Explanation:** The fixed-bound subarrays are [1,3,5] and [1,3,5,2].
*
* **Example 2:**
*
* **Input:** nums = [1,1,1,1], minK = 1, maxK = 1
*
* **Output:** 10
*
* **Explanation:** Every subarray of nums is a fixed-bound subarray. There are 10 possible subarrays.
*
* **Constraints:**
*
* * 2 <= nums.length <= 105
* * 1 <= nums[i], minK, maxK <= 106
**/
public class Solution {
public long countSubarrays(int[] nums, int minK, int maxK) {
long ans = 0;
int i = 0;
while (i < nums.length) {
if (nums[i] >= minK && nums[i] <= maxK) {
int a = i;
int b = i;
int mini = 0;
int maxi = 0;
while (i != nums.length && (nums[i] >= minK && nums[i] <= maxK)) {
i++;
}
while (true) {
for (; b != i && (mini == 0 || maxi == 0); b++) {
if (nums[b] == minK) {
mini++;
}
if (nums[b] == maxK) {
maxi++;
}
}
if (mini == 0 || maxi == 0) {
break;
}
for (; mini != 0 && maxi != 0; ans += 1 + (i - b), a++) {
if (nums[a] == minK) {
mini--;
}
if (nums[a] == maxK) {
maxi--;
}
}
}
}
i++;
}
return ans;
}
}
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