README_EN.md

2302. Count Subarrays With Score Less Than K

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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

Python3

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(nums) + 1):
            if nums[i - 1] >= k:
                continue
            left, right = 1, 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

Java

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) {
            if (nums[i - 1] >= k) {
                continue;
            }
            int left = 1, 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;
    }
}

C++

using ll = long long;

class Solution {
public:
    long long countSubarrays(vector<int>& nums, long long k) {
        int n = nums.size();
        vector<ll> s(n + 1);
        for (int i = 0; i < n; ++i) s[i + 1] = s[i] + nums[i];
        ll ans = 0;
        for (int i = 1; i <= n; ++i) {
            if (nums[i - 1] >= k) continue;
            int left = 1, 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;
    }
};

Go

func countSubarrays(nums []int, k int64) int64 {
	n := len(nums)
	s := make([]int64, n+1)
	for i, v := range nums {
		s[i+1] = s[i] + int64(v)
	}
	ans := 0
	for i := 1; i <= n; i++ {
		if s[i]-s[i-1] >= k {
			continue
		}
		left, right := 1, i
		for left < right {
			mid := (left + right + 1) >> 1
			if (s[i]-s[i-mid])*int64(mid) < k {
				left = mid
			} else {
				right = mid - 1
			}
		}
		ans += left
	}
	return int64(ans)
}

TypeScript

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