Skip to content

2302. Count Subarrays With Score Less Than K

Description

The score of an array is defined as the product of its sum and its length.

  • For example, the score of [1, 2, 3, 4, 5] is (1 + 2 + 3 + 4 + 5) * 5 = 75.

Given a positive integer array nums and an integer k, return the number of non-empty subarrays of nums whose score is strictly less than k.

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

 

Example 1:

Input: nums = [2,1,4,3,5], k = 10
Output: 6
Explanation:
The 6 subarrays having scores less than 10 are:
- [2] with score 2 * 1 = 2.
- [1] with score 1 * 1 = 1.
- [4] with score 4 * 1 = 4.
- [3] with score 3 * 1 = 3. 
- [5] with score 5 * 1 = 5.
- [2,1] with score (2 + 1) * 2 = 6.
Note that subarrays such as [1,4] and [4,3,5] are not considered because their scores are 10 and 36 respectively, while we need scores strictly less than 10.

Example 2:

Input: nums = [1,1,1], k = 5
Output: 5
Explanation:
Every subarray except [1,1,1] has a score less than 5.
[1,1,1] has a score (1 + 1 + 1) * 3 = 9, which is greater than 5.
Thus, there are 5 subarrays having scores less than 5.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 105
  • 1 <= k <= 1015

Solutions

First, we calculate the prefix sum array $s$ of the array $nums$, where $s[i]$ represents the sum of the first $i$ elements of the array $nums$.

Next, we enumerate each element of the array $nums$ as the last element of the subarray. For each element, we can find the maximum length $l$ such that $s[i] - s[i - l] \times l < k$ by binary search. The number of subarrays with this element as the last element is $l$, and we add all $l$ to get the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        s = list(accumulate(nums, initial=0))
        ans = 0
        for i in range(1, len(s)):
            left, right = 0, i
            while left < right:
                mid = (left + right + 1) >> 1
                if (s[i] - s[i - mid]) * mid < k:
                    left = mid
                else:
                    right = mid - 1
            ans += left
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution {
    public long countSubarrays(int[] nums, long k) {
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        long ans = 0;
        for (int i = 1; i <= n; ++i) {
            int left = 0, right = i;
            while (left < right) {
                int mid = (left + right + 1) >> 1;
                if ((s[i] - s[i - mid]) * mid < k) {
                    left = mid;
                } else {
                    right = mid - 1;
                }
            }
            ans += left;
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
public:
    long long countSubarrays(vector<int>& nums, long long k) {
        int n = nums.size();
        long long s[n + 1];
        s[0] = 0;
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        long long ans = 0;
        for (int i = 1; i <= n; ++i) {
            int left = 0, right = i;
            while (left < right) {
                int mid = (left + right + 1) >> 1;
                if ((s[i] - s[i - mid]) * mid < k) {
                    left = mid;
                } else {
                    right = mid - 1;
                }
            }
            ans += left;
        }
        return ans;
    }
};