2444.Count Subarrays With Fixed Bounds
===

###### tags: `Hard`,`Array`,`Queue`,`Sliding Window`

[2444. Count Subarrays With Fixed Bounds](https://leetcode.com/problems/count-subarrays-with-fixed-bounds/)

### 題目描述

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.

### 範例

**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` <= 10^5^
* 1 <= `nums[i]`, `minK`, `maxK` <= 10^6^

### 解答

#### C++
##### 思路:
1. 找subarray符合最大值等於maxK, 最小值等於minK
2. 看constraint基本上為要求O(n)解法
3. 思考從0遍歷至index i時, 若為合法subarray, 則其前面必有maxK及minK出現, 而此時考慮可以新增的subarray個數為maxK跟minK取最遠的index, 跟不符合條件的index(boundary)做相減算合法的個數, 即為此次可新增的個數, 遍歷累加即可
4. 遍歷時紀錄maxK/minK/boundary index即可

class Solution {
public:
    long long countSubarrays(vector<int>& nums, int minK, int maxK) {
        long long res = 0;
        int minKIndex = -1, maxKIndex = -1, boundary = -1;
        for(int i = 0; i < nums.size(); i++)
        {
            if(nums[i] < minK || nums[i] > maxK)
            {
                boundary = i;
                continue;
            }
            if(nums[i] == minK)
                minKIndex = i;
            if(nums[i] == maxK)
                maxKIndex = i;
            res += max(0, min(minKIndex, maxKIndex) - boundary);
        }
        return res;
    }
};

Time: $O(n)$
Extra Space: $O(1)$

> [name=XD] [time= March 4, 2023]

### Reference

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