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  1. program mountain;
  2. Uses Math;
  3. const
  4.     MAXN = 100005;
  5.  
  6. var
  7.     ANS, N, i, j, maxMountainLength : LongInt;
  8.     P, leftLIS, rightLIS, rimosso  : Array[0..MAXN-1] of LongInt;
  9.  
  10. begin
  11.  
  12.     (*assign(input,  'input.txt');  reset(input);
  13.     assign(output, 'output.txt'); rewrite(output);*)
  14.  
  15.     ReadLn(N);
  16.  
  17.     for i:=0 to N-1 do
  18.         Read(P[i]);
  19.     ReadLn();
  20.  
  21.     ANS := 0;
  22.  
  23. 	(*leftLIS[i] stores the length of longest increasing subsequence ending at index i*)
  24. 	(*rightLIS[i] stores the length of longest decreasing subsequence starting at index i*)
  25.  
  26.    for i:=0 to  N-1 do begin leftLIS[i]:=1; rightLIS[i]:=1; rimosso[i]:=0; end;
  27.  
  28. 	(*Calculate LIS from left to right for each position*)
  29.    for i:=1 to N-2 do if  (P[i]<P[i-1]) and (P[i]<P[i+1]) then rimosso[i]:=1; 
  30.    for i := 1 to N-1 do
  31.          if (rimosso[i]=0) then 
  32.              for j:= 0 to i-1 do
  33.                 begin
  34.                   if (rimosso[j]=0) and (P[i] > P[j]) then leftLIS[i] := max(leftLIS[i], leftLIS[j] + 1);                          
  35.                 end;  
  36.  
  37. 	(* Calculate LIS from right to left (decreasing subsequence) for each position*)
  38.  
  39.    for i := N - 2 downto 0 do
  40.         if (rimosso[i]=0) then 
  41.             for  j := i + 1 to N-1 do
  42.                begin
  43.                 if (rimosso[j]=0) and (P[i] > P[j]) then rightLIS[i] := max(rightLIS[i], rightLIS[j] + 1);                
  44. 	       end;
  45.  
  46.         (* Find the maximum length of mountain subsequence*)
  47.    maxMountainLength := 0;
  48.    for i := 0 to N-1 do
  49. 	(*A valid mountain peak must have at least one element on both sides*)
  50. 	(*leftLIS[i] > 1 ensures there's at least one element before peak*)
  51. 	(*rightLIS[i] > 1 ensures there's at least one element after peak*)
  52.    if (leftLIS[i] >= 1) and (rightLIS[i] >= 1)  then
  53.   	(*Total mountain length with peak at i Subtract 1 because peak is counted in both leftLIS and rightLIS*)
  54.          maxMountainLength := max(maxMountainLength, leftLIS[i] + rightLIS[i] - 1);
  55.  	(* Minimum removals = total elements - maximum mountain length*)
  56.    ANS:= N - maxMountainLength; 
  57.    WriteLn(ans);
  58. end.
Success #stdin #stdout 0s 5320KB
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https://ideone.com/BEC10B
language:
Pascal (fpc 3.0.4)
created:
23 hours ago
visibility:
public

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