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Formatted question description: https://leetcode.ca/all/2302.html

# 2302. Count Subarrays With Score Less Than K

• Difficulty: Hard.
• Related Topics: Array, Binary Search, Sliding Window, Prefix Sum.
• Similar Questions: Maximum Subarray, Subarray Product Less Than K, Binary Subarrays With Sum.

## Problem

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

## Solution

• class Solution {
public long countSubarrays(int[] nums, long k) {
long sum = 0;
long count = 0;
int i = 0;
int j = 0;
while (i < nums.length) {
sum += nums[i];
while (sum * (i - j + 1) >= k) {
sum -= nums[j++];
}
count += i++ - j + 1;
}
return count;
}
}

############

class Solution {
public long countSubarrays(int[] nums, long k) {
long ans = 0, s = 0;
for (int i = 0, j = 0; i < nums.length; ++i) {
s += nums[i];
while (s * (i - j + 1) >= k) {
s -= nums[j++];
}
ans += i - j + 1;
}
return ans;
}
}

• class Solution:
def countSubarrays(self, nums: List[int], k: int) -> int:
ans = s = j = 0
for i, v in enumerate(nums):
s += v
while s * (i - j + 1) >= k:
s -= nums[j]
j += 1
ans += i - j + 1
return ans

############

# 2302. Count Subarrays With Score Less Than K
# https://leetcode.com/problems/count-subarrays-with-score-less-than-k

class Solution:
def countSubarrays(self, nums: List[int], k: int) -> int:
res = 0
i = 0
s = 0

for j, x in enumerate(nums):
s += x

while s * (j - i + 1) >= k:
s -= nums[i]
i += 1

res += (j - i + 1)

return res


• class Solution {
public:
long long countSubarrays(vector<int>& nums, long long k) {
long long ans = 0, s = 0;
for (int i = 0, j = 0; i < nums.size(); ++i) {
s += nums[i];
while (s * (i - j + 1) >= k) {
s -= nums[j++];
}
ans += i - j + 1;
}
return ans;
}
};

• func countSubarrays(nums []int, k int64) (ans int64) {
s, j := 0, 0
for i, v := range nums {
s += v
for int64(s*(i-j+1)) >= k {
s -= nums[j]
j++
}
ans += int64(i - j + 1)
}
return
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).