Program to find number of sublists containing maximum and minimum after deleting only one element in Python
Suppose we have a list of numbers called nums and we can delete at most one element in the list. We have to find the maximum number of sublists that contain both the maximum and minimum values of the resulting list.
So, if the input is like nums = [3, 2, 6, 2, 4, 10], then the output will be 8, as if we remove 10 we will get [3, 2, 6, 2, 4] and there are eight sublists where it contains both the max and the min −
[2, 6]
[6, 2]
[2, 6, 2]
[3, 2, 6]
[6, 2, 4]
[2, 6, 2, 4]
[3, 2, 6, 2]
[3, 2, 6, 2, 4].
To solve this, we will follow these steps −
Define a function check() . This will take lst
mn := minimum of lst, mx := maximum of lst
min_pos := null, max_pos := null
ret := 0
for each index i and value num in lst, do
if num is same as mn, then
if num is same as mx, then
if min_pos is null or max_pos is null, then
ret := ret + minimum of min_pos and (max_pos + 1)
return ret
From the main method do the following −
if size of nums <= 1, then
ret := check(nums)
for each rem_cand in [minimum of nums , maximum of nums ], do
return ret
Example
Let us see the following implementation to get a better understanding −
Live Demo
class Solution:
def solve(self, nums):
if len(nums) <= 1:
return len(nums)
def check(lst):
mn, mx = min(lst), max(lst)
min_pos, max_pos = None, None
ret = 0
for i, num in enumerate(lst):
if num == mn:
min_pos = i
if num == mx:
max_pos = i
if min_pos is None or max_pos is None:
continue
ret += min(min_pos, max_pos) + 1
return ret
ret = check(nums)
for rem_cand in [min(nums), max(nums)]:
if nums.count(rem_cand) == 1:
idx = nums.index(rem_cand)
ret = max(ret, check(nums[:idx] + nums[idx + 1 :]))
return ret
ob = Solution()
nums = [3, 2, 6, 2, 4, 10]
print(ob.solve(nums))
Input
[3, 2, 6, 2, 4, 10]
Output
8
Published on 23-Dec-2020 10:57:14
