# 2444. Count Subarrays With Fixed Bounds

## Description

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

## Solutions

Solution 1: Enumeration of Right Endpoint

From the problem description, we know that all elements of the bounded subarray are in the interval [minK, maxK], and the minimum value must be minK, and the maximum value must be maxK.

We traverse the array $nums$, count the number of bounded subarrays with nums[i] as the right endpoint, and then add all the counts.

The specific implementation logic is as follows:

1. Maintain the index $k$ of the most recent element not in the interval [minK, maxK], initially set to $-1$. Therefore, the left endpoint of the current element nums[i] must be greater than $k$.
2. Maintain the index $j_1$ of the most recent element with a value of minK, and the index $j_2$ of the most recent element with a value of maxK, both initially set to $-1$. Therefore, the left endpoint of the current element nums[i] must be less than or equal to $\min(j_1, j_2)$.
3. In summary, the number of bounded subarrays with the current element as the right endpoint is $\max(0, \min(j_1, j_2) - k)$. Add up all the counts to get the result.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public long countSubarrays(int[] nums, int minK, int maxK) {
long ans = 0;
int j1 = -1, j2 = -1, k = -1;
for (int i = 0; i < nums.length; ++i) {
if (nums[i] < minK || nums[i] > maxK) {
k = i;
}
if (nums[i] == minK) {
j1 = i;
}
if (nums[i] == maxK) {
j2 = i;
}
ans += Math.max(0, Math.min(j1, j2) - k);
}
return ans;
}
}

• class Solution {
public:
long long countSubarrays(vector<int>& nums, int minK, int maxK) {
long long ans = 0;
int j1 = -1, j2 = -1, k = -1;
for (int i = 0; i < nums.size(); ++i) {
if (nums[i] < minK || nums[i] > maxK) k = i;
if (nums[i] == minK) j1 = i;
if (nums[i] == maxK) j2 = i;
ans += max(0, min(j1, j2) - k);
}
return ans;
}
};

• class Solution:
def countSubarrays(self, nums: List[int], minK: int, maxK: int) -> int:
j1 = j2 = k = -1
ans = 0
for i, v in enumerate(nums):
if v < minK or v > maxK:
k = i
if v == minK:
j1 = i
if v == maxK:
j2 = i
ans += max(0, min(j1, j2) - k)
return ans


• func countSubarrays(nums []int, minK int, maxK int) int64 {
ans := 0
j1, j2, k := -1, -1, -1
for i, v := range nums {
if v < minK || v > maxK {
k = i
}
if v == minK {
j1 = i
}
if v == maxK {
j2 = i
}
ans += max(0, min(j1, j2)-k)
}
return int64(ans)
}

• function countSubarrays(nums: number[], minK: number, maxK: number): number {
let res = 0;
let minIndex = -1;
let maxIndex = -1;
let k = -1;
nums.forEach((num, i) => {
if (num === minK) {
minIndex = i;
}
if (num === maxK) {
maxIndex = i;
}
if (num < minK || num > maxK) {
k = i;
}
res += Math.max(Math.min(minIndex, maxIndex) - k, 0);
});
return res;
}


• impl Solution {
pub fn count_subarrays(nums: Vec<i32>, min_k: i32, max_k: i32) -> i64 {
let mut res = 0;
let mut min_index = -1;
let mut max_index = -1;
let mut k = -1;
for i in 0..nums.len() {
let num = nums[i];
let i = i as i64;
if num == min_k {
min_index = i;
}
if num == max_k {
max_index = i;
}
if num < min_k || num > max_k {
k = i;
}
res += (0).max(min_index.min(max_index) - k);
}
res
}
}