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Formatted question description: https://leetcode.ca/all/2444.html

2444. Count Subarrays With Fixed Bounds

  • Difficulty: Hard.
  • Related Topics: .
  • Similar Questions: .

Problem

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 <= 10^5

  • 1 <= nums[i], minK, maxK <= 10^6

Solution (Java, C++, Python)

  • 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)
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

  • 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
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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