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795. Number of Subarrays with Bounded Maximum

Description

Given an integer array nums and two integers left and right, return the number of contiguous non-empty subarrays such that the value of the maximum array element in that subarray is in the range [left, right].

The test cases are generated so that the answer will fit in a 32-bit integer.

 

Example 1:

Input: nums = [2,1,4,3], left = 2, right = 3
Output: 3
Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].

Example 2:

Input: nums = [2,9,2,5,6], left = 2, right = 8
Output: 7

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109
  • 0 <= left <= right <= 109

Solutions

Solution 1

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class Solution:
    def numSubarrayBoundedMax(self, nums: List[int], left: int, right: int) -> int:
        def f(x):
            cnt = t = 0
            for v in nums:
                t = 0 if v > x else t + 1
                cnt += t
            return cnt

        return f(right) - f(left - 1)

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class Solution {
    public int numSubarrayBoundedMax(int[] nums, int left, int right) {
        return f(nums, right) - f(nums, left - 1);
    }

    private int f(int[] nums, int x) {
        int cnt = 0, t = 0;
        for (int v : nums) {
            t = v > x ? 0 : t + 1;
            cnt += t;
        }
        return cnt;
    }
}

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class Solution {
public:
    int numSubarrayBoundedMax(vector<int>& nums, int left, int right) {
        auto f = [&](int x) {
            int cnt = 0, t = 0;
            for (int& v : nums) {
                t = v > x ? 0 : t + 1;
                cnt += t;
            }
            return cnt;
        };
        return f(right) - f(left - 1);
    }
};

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func numSubarrayBoundedMax(nums []int, left int, right int) int {
    f := func(x int) (cnt int) {
        t := 0
        for _, v := range nums {
            t++
            if v > x {
                t = 0
            }
            cnt += t
        }
        return
    }
    return f(right) - f(left-1)
}

Solution 2

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class Solution:
    def numSubarrayBoundedMax(self, nums: List[int], left: int, right: int) -> int:
        n = len(nums)
        l, r = [-1] * n, [n] * n
        stk = []
        for i, v in enumerate(nums):
            while stk and nums[stk[-1]] <= v:
                stk.pop()
            if stk:
                l[i] = stk[-1]
            stk.append(i)
        stk = []
        for i in range(n - 1, -1, -1):
            while stk and nums[stk[-1]] < nums[i]:
                stk.pop()
            if stk:
                r[i] = stk[-1]
            stk.append(i)
        return sum(
            (i - l[i]) * (r[i] - i) for i, v in enumerate(nums) if left <= v <= right
        )

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class Solution {
    public int numSubarrayBoundedMax(int[] nums, int left, int right) {
        int n = nums.length;
        int[] l = new int[n];
        int[] r = new int[n];
        Arrays.fill(l<