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 the longest subsequence where the absolute difference between every adjacent element is at most k in Python.
Suppose we are given a list of numbers and another value k. This time our task is to find the length of the longest subsequence where the absolute difference between every adjacent element is at most k.
So, if the input is like nums = [5, 6, 2, 1, −6, 0, −1, k = 4, then the output will be 6.
To solve this, we will follow these steps −
Define a function update() . This will take i, x
i := i + n
while i is non−zero, do
segtree[i] := maximum of segtree[i], x
i := i / 2
Define a function query() . This will take i, j
ans := −infinity
i := i + n
j := j + n + 1
while i < j is non−zero, do
if i mod 2 is same as 1, then
ans := maximum of ans, segtree[i]
i := i + 1
if j mod 2 is same as 1, then
j := j − 1
ans := maximum of ans, segtree[j]
i := i / 2
j := j / 2
return ans
Now, in the main function, do the following −
nums = [5, 6, 2, 1, −6, 0, −1]
k = 4
n = 2 to the power (logarithm base 2 of (length of (nums) + 1) + 1)
segtree := [0] * 100000
snums := sort the list nums
index := make a collection where x: i for each index i and element x in (snums)
ans := 0
for each x in nums, do
lo := return the leftmost position from snums, where (x − k) can be inserted while maintaining the sorted order
hi := (the leftmost position from snums, where (x + k) can be inserted while maintaining the sorted order) − 1
count := query(lo, hi)
update(index[x], count + 1)
ans := maximum of ans, count + 1
return ans
Let us see the following implementation to get better understanding −
Example
import math, bisect
class Solution:
def solve(self, nums, k):
n = 2 ** int(math.log2(len(nums) + 1) + 1)
segtree = [0] * 100000
def update(i, x):
i += n
while i:
segtree[i] = max(segtree[i], x)
i //= 2
def query(i, j):
ans = −float("inf")
i += n
j += n + 1
while i < j:
if i % 2 == 1:
ans = max(ans, segtree[i])
i += 1
if j % 2 == 1:
j −= 1
ans = max(ans, segtree[j])
i //= 2
j //= 2
return ans
snums = sorted(nums)
index = {x: i for i, x in enumerate(snums)}
ans = 0
for x in nums:
lo = bisect.bisect_left(snums, x − k)
hi = bisect.bisect_right(snums, x + k) − 1
count = query(lo, hi)
update(index[x], count + 1)
ans = max(ans, count + 1)
return ans
ob = Solution()
print(ob.solve([5, 6, 2, 1, −6, 0, −1], 4))Input
[5, 6, 2, 1, −6, 0, −1], 4
Output
6
332 Views
