README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2900-2999/2962.Count%20Subarrays%20Where%20Max%20Element%20Appears%20at%20Least%20K%20Times/README_EN.md	1700	Weekly Contest 375 Q3	Array Sliding Window

2962. Count Subarrays Where Max Element Appears at Least K Times

中文文档

Description

You are given an integer array nums and a positive integer k.

Return the number of subarrays where the maximum element of nums appears at least k times in that subarray.

A subarray is a contiguous sequence of elements within an array.

 

Example 1:

Input: nums = [1,3,2,3,3], k = 2
Output: 6
Explanation: The subarrays that contain the element 3 at least 2 times are: [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].

Example 2:

Input: nums = [1,4,2,1], k = 3
Output: 0
Explanation: No subarray contains the element 4 at least 3 times.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 106
• 1 <= k <= 105

Solutions

Solution 1: Two Pointers

Let's denote the maximum value in the array as $mx$.

We use two pointers $i$ and $j$ to maintain a sliding window, such that in the subarray between $[i, j)$, there are $k$ elements equal to $mx$. If we fix the left endpoint $i$, then all right endpoints greater than or equal to $j-1$ meet the condition, totaling $n - (j - 1)$.

Therefore, we enumerate the left endpoint $i$, use the pointer $j$ to maintain the right endpoint, use a variable $cnt$ to record the number of elements equal to $mx$ in the current window. When $cnt$ is greater than or equal to $k$, we have found a subarray that meets the condition, and we increase the answer by $n - (j - 1)$. Then we update $cnt$ and continue to enumerate the next left endpoint.

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

Python3

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        mx = max(nums)
        n = len(nums)
        ans = cnt = j = 0
        for x in nums:
            while j < n and cnt < k:
                cnt += nums[j] == mx
                j += 1
            if cnt < k:
                break
            ans += n - j + 1
            cnt -= x == mx
        return ans

Java

class Solution {
    public long countSubarrays(int[] nums, int k) {
        int mx = Arrays.stream(nums).max().getAsInt();
        int n = nums.length;
        long ans = 0;
        int cnt = 0, j = 0;
        for (int x : nums) {
            while (j < n && cnt < k) {
                cnt += nums[j++] == mx ? 1 : 0;
            }
            if (cnt < k) {
                break;
            }
            ans += n - j + 1;
            cnt -= x == mx ? 1 : 0;
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long countSubarrays(vector<int>& nums, int k) {
        int mx = *max_element(nums.begin(), nums.end());
        int n = nums.size();
        long long ans = 0;
        int cnt = 0, j = 0;
        for (int x : nums) {
            while (j < n && cnt < k) {
                cnt += nums[j++] == mx;
            }
            if (cnt < k) {
                break;
            }
            ans += n - j + 1;
            cnt -= x == mx;
        }
        return ans;
    }
};

Go

func countSubarrays(nums []int, k int) (ans int64) {
	mx := slices.Max(nums)
	n := len(nums)
	cnt, j := 0, 0
	for _, x := range nums {
		for ; j < n && cnt < k; j++ {
			if nums[j] == mx {
				cnt++
			}
		}
		if cnt < k {
			break
		}
		ans += int64(n - j + 1)
		if x == mx {
			cnt--
		}
	}
	return
}

TypeScript

function countSubarrays(nums: number[], k: number): number {
    const mx = Math.max(...nums);
    const n = nums.length;
    let [cnt, j] = [0, 0];
    let ans = 0;
    for (const x of nums) {
        for (; j < n && cnt < k; ++j) {
            cnt += nums[j] === mx ? 1 : 0;
        }
        if (cnt < k) {
            break;
        }
        ans += n - j + 1;
        cnt -= x === mx ? 1 : 0;
    }
    return ans;
}