README_EN.md

795. Number of Subarrays with Bounded Maximum

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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

Python3

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)

Java

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, -1);
        Arrays.fill(r, n);
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            while (!stk.isEmpty() && nums[stk.peek()] <= v) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                l[i] = stk.peek();
            }
            stk.push(i);
        }
        stk.clear();
        for (int i = n - 1; i >= 0; --i) {
            int v = nums[i];
            while (!stk.isEmpty() && nums[stk.peek()] < v) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                r[i] = stk.peek();
            }
            stk.push(i);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (left <= nums[i] && nums[i] <= right) {
                ans += (i - l[i]) * (r[i] - i);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int numSubarrayBoundedMax(vector<int>& nums, int left, int right) {
        int n = nums.size();
        vector<int> l(n, -1);
        vector<int> r(n, n);
        stack<int> stk;
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            while (!stk.empty() && nums[stk.top()] <= v) stk.pop();
            if (!stk.empty()) l[i] = stk.top();
            stk.push(i);
        }
        stk = stack<int>();
        for (int i = n - 1; ~i; --i) {
            int v = nums[i];
            while (!stk.empty() && nums[stk.top()] < v) stk.pop();
            if (!stk.empty()) r[i] = stk.top();
            stk.push(i);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (left <= nums[i] && nums[i] <= right) {
                ans += (i - l[i]) * (r[i] - i);
            }
        }
        return ans;
    }
};

Go

func numSubarrayBoundedMax(nums []int, left int, right int) (ans int) {
	n := len(nums)
	l := make([]int, n)
	r := make([]int, n)
	for i := range l {
		l[i], r[i] = -1, n
	}
	stk := []int{}
	for i, v := range nums {
		for len(stk) > 0 && nums[stk[len(stk)-1]] <= v {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			l[i] = stk[len(stk)-1]
		}
		stk = append(stk, i)
	}
	stk = []int{}
	for i := n - 1; i >= 0; i-- {
		v := nums[i]
		for len(stk) > 0 && nums[stk[len(stk)-1]] < v {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			r[i] = stk[len(stk)-1]
		}
		stk = append(stk, i)
	}
	for i, v := range nums {
		if left <= v && v <= right {
			ans += (i - l[i]) * (r[i] - i)
		}
	}
	return
}

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