Question

2444. Count Subarrays With Fixed Bounds

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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:
  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

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;
  }
};

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
  }
}

#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))

long long countSubarrays(int* nums, int numsSize, int minK, int maxK) {
  long long res = 0;
  int minIndex = -1;
  int maxIndex = -1;
  int k = -1;
  for (int i = 0; i < numsSize; i++) {
    int num = nums[i];
    if (num == minK) {
      minIndex = i;
    }
    if (num == maxK) {
      maxIndex = i;
    }
    if (num < minK || num > maxK) {
      k = i;
    }
    res += max(min(minIndex, maxIndex) - k, 0);
  }
  return res;
}