Easy Array Counting II#12269

Easy Array Counting II (Editoryal)

Problem Author:	Hasan Saleh (MathModel)
Prepared by:	Hasan Saleh (MathModel)
Edited by:	Hasan Saleh (MathModel)

Solution

Claim. $n$ doesn't matter.

Observation : If any $a_{i} \neq = a_{i + 1}$ a_i \neq a_{i+1} then there exists at least one subarray such that it's not a palindrome.

Thus from that we've a sufficient condition that $a_{1} = a_{2} = a_{3} = ... = a_{n}$ a_1=a_2=a_3=...=a_n to hold , thus we've $r - l + 1$ r-l+1 choices which are $[l, l, .., l]$ [l,l,..,l] ,... , $[l + 1, l + 1, l + 1, ..., l + 1]$ [l+1,l+1,l+1,...,l+1] , and so on upto .. $[r, r, ..., r]$ [r,r,...,r].

Solution is $r - l + 1$ r-l+1 , Complexity $O (1)$ .

Complexity $O (1)$ .

for _ in range(int(input())):
   n,l,r=map(int,input().split())
   print(r-l+1)