Program to find number of K-Length sublists whose average is greater or same as target in python

Suppose we have a list nums, and two additional values k and target, we have to find the number of sublists whose size is k and its average value ≥ target.

So, if the input is like nums = [1, 10, 5, 6, 7] k = 3 target = 6, then the output will be 2, as the sublist [1, 10, 7] has average value of 6 and [10, 5, 6] has average of 7.

To solve this, we will follow these steps:

• target := target * k
• sum := 0, ans := 0
• for each index i and number n in nums, do
  • if i >= k, then
    • sum := sum - nums[i - k]
  • sum := sum + n
  • if i >=(k - 1) , then
    • if sum >= target, then
      • ans := ans + 1
• return ans

Let us see the following implementation to get better understanding:

Example Code

Live Demo

class Solution:    def solve(self, nums, k, target):       target *= k       sum = 0       ans = 0       for i, n in enumerate(nums):          if i >= k:             sum -= nums[i - k]          sum += n          if i >= (k - 1):             if sum >= target:                ans += 1          return ans ob = Solution() nums = [1, 10, 5, 6, 7] k = 3 target = 6 print(ob.solve(nums, k, target))

Input

[1, 10, 5, 6, 7], 3, 6

Output

2