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Chef Likes Good Sequences Codechef October Lunchtime Problem solution

 Chef Likes Good Sequences Codechef October Lunchtime Problem solution -

Lets read the problem statement carefully. 
Problem statement- 

Chef calls a sequence good if it does not contain any two adjacent identical elements.

Initially, Chef has a sequence a1,a2,,aN. He likes to change the sequence every now and then. You should process Q queries. In each query, Chef chooses an index x and changes the x-th element of the sequence to y. After each query, can you find the length of the longest good subsequence of the current sequence?

Note that the queries are not independent.

Input

  • The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
  • The first line of each test case contains two space-separated integers N and Q.
  • The second line contains N space-separated integers a1,a2,,aN.
  • Q lines follow. Each of these lines contains two space-separated integers x and y describing a query.

Output

After each query, print a single line containing one integer ― the length of the longest good subsequence.

Constraints

  • 1T5
  • 1N,Q105
  • 1ai109 for each valid i
  • 1xN
  • 1y109

Subtasks

Subtask #1 (35 points): Q=1

Subtask #2 (65 points): original constraints

Example Input

1
5 2
1 1 2 5 2
1 3
4 2

Example Output

5
3

Chef Likes Good Sequences Codechef October Lunchtime Problem solution


Solution-

First of all do not confuse with subarray and subsequence. 
subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. A subarray will always be contiguous but a subsequence need not be contiguous.

It's easy problem follow some simply step-
step 1- Calculate the longest subsequence in the initial array without changes.

Now think that if you made changes at xth position of array it will affect initial length if y!= a[x]
if y!=a[x] there occur some cases  for which initial length may affect. You have to observe these cases and handle initial length accordingly. 

  • Case 1 : (a_x == a_{x-1}) and (a_x == a_{x+1}), then answer will increase by 2 as it will make 1 segment into 3 segments.
  • Case 2 : (a_x \neq a_{x-1}) and (a_x == a_{x+1}), then the answer will increase by 1 as earlier a_x == a_{x+1} and y will introduce an extra segment. If y = a_{x-1} then we will reduce the answer by 1 as earlier we added 1 as changing to y increased the segment count but now it will be merged with previous segment. Example : let a_x = a_{x+1} = 1 and a_{x-1} = 2, so If y = 3 answer will increase by 1, and if y = 2 then there will no change to the answer.
  • Case 3 : (a_x == a_{x-1}) and (a_x \neq a_{x+1}), similar to above case, we will increase the answer by 1. if y=a_{x+1} then we will reduce the answer by 1.
  • Case 4 : (a_x \neq a_{x-1}) and (a_x \neq a_{x+1}), then if y=a_{x-1} we will reduce answer by 1. if y = a_{x+1} we will reduce answer by 1.(Note that you might reduce the answer by 2 when y=a_{x-1}=a_{x+1}).

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