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 check subarrays can be rearranged from arithmetic sequence or not in Python
Suppose we have sequence of numbers nums, and another two arrays l and r of size m, these l and r are representing range queries like [l[i], r[i]]. We have to find a Boolean sequence ans, where ans[i] is true when the subarray nums[l[i]], nums[l[i] + 1], ... nums[r[i] - 1], nums[r[i]] can be arranged to generate an arithmetic sequence, otherwise false.
A sequence is said to be arithmetic, if it consists of at least two elements, and the difference between every two consecutive elements is the same. For example, some arithmetic sequences are: [2, 4, 6, 8, 10], [5, 5, 5, 5], [4, -2, -8, -14], but not [2, 2, 3, 6, 9].
So, if the input is like nums = [6,8,7,11,5,9], l = [0,0,2], r = [2,3,5], then the output will be [True, False, True] because −
for query [0, 2], the sequence is [6,8,7], can be rearranged as [6,7,8], this is valid
for query [0, 3], the sequence is [6,8,7,11], cannot be rearranged in arithmetic sequence
for query [2, 5], the sequence is [7,11,5,9], can be rearranged as [5,7,9,11], this is valid
To solve this, we will follow these steps −
new := a list of size same as l and fill with all true values
for i in range 0 to size of l - 1, do
data := sublist of nums from index l[i] to r[i]
sort the list data
d := a new list
for j in range 0 to size of data - 2, do
insert data[j+1] - data[j] at the end of d
d := a new list from a new set from d
if size of d is not same as 1, then
new[i] := False
return new
Example
Let us see the following implementation to get better understanding −
def solve(nums, l, r): new = [True]*len(l) for i in range(len(l)): data = nums[l[i]:r[i]+1] data.sort() d = [] for j in range(len(data) - 1): d.append(data[j+1] - data[j]) d = list(set(d)) if len(d) != 1: new[i] = False return new nums = [6,8,7,11,5,9] l = [0,0,2] r = [2,3,5] print(solve(nums, l, r))
Input
[6,8,7,11,5,9], [0,0,2], [2,3,5]
Output
[True,False,True]
88 Views
