Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
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
min_pos := i
if num is same as mx, then
max_pos := i
if min_pos is null or max_pos is null, then
go for next iteration
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
return size of nums
ret := check(nums)
for each rem_cand in [minimum of nums , maximum of nums ], do
if occurrence of rem_cand is 1, then
idx := index of rem_cand in nums
ret := maximum of ret and check(nums[from index 0 to idx - 1] concatenate nums[from index idx + 1 to end]
return ret
Example
Let us see the following implementation to get a better understanding −
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
133 Views
