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 longest equivalent sublist after K increments in Python
Suppose we have a list of numbers called nums and k. Now, consider an operation where we can increment any one element once. If we can perform operations at most k times, we have to find the longest sublist containing equal elements.
So, if the input is like nums = [3, 5, 9, 6, 10, 7] k = 6, then the output will be 3, as we can increment 9 once and 6 four times to get the sublist [10, 10, 10].
To solve this, we will follow these steps −
if nums is empty, then
return 0
wMax := a double ended queue of size same as nums. and insert a pair (nums[0], 0)
i := 0, inc := 0
for j in range 1 to size of nums, do
while wMax is not empty and wMax[0, 1] < i, do
delete left element of wMax
pMax := wMax[0, 0]
while wMax is not empty and first element of last item of wMax <= nums[j], do
delete right element from wMax
insert (nums[j], j) at the end of wMax
if pMax < wMax[0, 0], then
inc = inc + (j - i) * (wMax[0, 0] - pMax)
otherwise,
inc := inc + pMax - nums[j]
if inc > k, then
inc := inc - wMax[0, 0] - nums[i]
while wMax is not empty and wMax[0, 1] <= i, do
delete left element of wMax
if wMax[0, 0] < nums[i], then
inc = inc - (nums[i] - wMax[0, 0]) * (j - i)
i := i + 1
return size of nums - i
Let us see the following implementation to get better understanding −
Example
from collections import deque class Solution: def solve(self, nums, k): if not nums: return 0 wMax = deque([(nums[0], 0)], maxlen=len(nums)) i = 0 inc = 0 for j in range(1, len(nums)): while wMax and wMax[0][1] < i: wMax.popleft() pMax = wMax[0][0] while wMax and wMax[-1][0] <= nums[j]: wMax.pop() wMax.append((nums[j], j)) if pMax < wMax[0][0]: inc += (j - i) * (wMax[0][0] - pMax) else: inc += pMax - nums[j] if inc > k: inc -= wMax[0][0] - nums[i] while wMax and wMax[0][1] <= i: wMax.popleft() if wMax[0][0] < nums[i]: inc -= (nums[i] - wMax[0][0]) * (j - i) i += 1 return len(nums) - i ob = Solution() nums = [3, 5, 9, 6, 10, 7] k = 6 print(ob.solve(nums, k))
Input
[3, 5, 9, 6, 10, 7], 6
Output
3
91 Views
