Welcome to Subscribe On Youtube

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;
    }
}

  • Todo

  • 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

Explain:

nope.

Complexity:

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

All Problems

All Solutions