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Program to find largest average of sublist whose size at least k in Python
Suppose we have a list of numbers called nums and another value k, we have to find the largest average value of any sublist of the list whose length is at least k.
So, if the input is like nums = [2, 10, -50, 4, 6, 6] k = 3, then the output will be 5.33333333, as sublist [4, 6, 6] has the largest average value
To solve this, we will follow these steps −
left := minimum of nums, right := maximum of nums
s := sum of all numbers in nums from index 0 to k − 1
largest_avg := s / k
while left <= right, do
mid :=integer of (left + right) / 2
sum1 := s, avg := s / k, sum2 := 0, cnt := 0
for i in range k to size of nums, do
sum1 := sum1 + nums[i]
sum2 := sum2 + nums[i − k]
cnt := cnt + 1
avg := maximum of avg and (sum1 /(cnt + k))
if sum2 / cnt <= mid, then
sum1 := sum1 − sum2
cnt := 0, sum2 := 0
avg := maximum of avg and (sum1 /(cnt + k))
largest_avg := maximum of largest_avg and avg
if avg > mid, then
left := mid + 1
otherwise,
right := mid − 1
return largest_avg
Let us see the following implementation to get better understanding −
Example
class Solution: def solve(self, nums, k): left, right = min(nums), max(nums) s = sum(nums[:k]) largest_avg = s / k while left <= right: mid = (left + right) // 2 sum1 = s avg = s / k sum2 = 0 cnt = 0 for i in range(k, len(nums)): sum1 += nums[i] sum2 += nums[i − k] cnt += 1 avg = max(avg, sum1 / (cnt + k)) if sum2 / cnt <= mid: sum1 −= sum2 cnt = 0 sum2 = 0 avg = max(avg, sum1 / (cnt + k)) largest_avg = max(largest_avg, avg) if avg > mid: left = mid + 1 else: right = mid − 1 return largest_avg ob = Solution() nums = [2, 10, −50, 4, 6, 6] k = 3 print(ob.solve(nums, k))
Input
[2, 10, −50, 4, 6, 6], k = 3
Output
5.333333333333333
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