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  1. program mountain;
  2. Uses Math;
  3. const
  4.     MAXN = 100005;
  5.  
  6. var
  7.     ANS, N, i, j, maxMountainLength, count : LongInt;
  8.     P, leftLIS, rightLIS, valli, scartato  : 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.      valli[0]:=0; valli[N-1]:=0;
  22.      for i:=1 to N-1 do 
  23.                       if (P[i]<P[i-1]) and (P[i]<P[i+1]) then valli[i]:=1;
  24.  
  25.     ANS := 0;
  26.  
  27. 	(*leftLIS[i] stores the length of longest increasing subsequence ending at index i*)
  28. 	(*rightLIS[i] stores the length of longest decreasing subsequence starting at index i*)
  29.  
  30.    for i:=0 to  N-1 do begin leftLIS[i]:=1; rightLIS[i]:=1; scartato[i]:=0; end;
  31.  
  32. 	(*Calculate LIS from left to right for each position*)
  33.  
  34.    for i := 1 to N-1 do
  35.             for j:= 0 to i-1 do
  36.                 begin
  37.                   if (P[i] > P[j]) then leftLIS[i] := max(leftLIS[i], leftLIS[j] + 1);
  38.                   if (i>0) and (i<N-1) and (P[i]<P[i-1]) then scartato[i]:=1;          
  39.                 end;  
  40.  
  41. 	(* Calculate LIS from right to left (decreasing subsequence) for each position*)
  42.  
  43.    for i := N - 2 downto 0 do
  44.             for  j := i + 1 to N-1 do
  45.                 if (P[i] > P[j]) then rightLIS[i] := max(rightLIS[i], rightLIS[j] + 1);
  46.                 if (i<N-1) and (i>0) and (P[i]<P[i+1]) then scartato[i]:=1; 
  47.  
  48. 	(* Find the maximum length of mountain subsequence*)
  49. 	count:=0;
  50.     for i:=0 to N-1 do  if ((scartato[i]=1) and (valli[i]=1)) then count:=count+1;
  51.    maxMountainLength := 0;
  52.    for i := 0 to N-1 do
  53. 	(*A valid mountain peak must have at least one element on both sides*)
  54. 	(*leftLIS[i] > 1 ensures there's at least one element before peak*)
  55. 	(*rightLIS[i] > 1 ensures there's at least one element after peak*)
  56.    if (leftLIS[i] >= 1) and (rightLIS[i] >= 1)  then
  57.   	(*Total mountain length with peak at i Subtract 1 because peak is counted in both leftLIS and rightLIS*)
  58.          maxMountainLength := max(maxMountainLength, leftLIS[i] + rightLIS[i] - 1);
  59.  	(* Minimum removals = total elements - maximum mountain length*)
  60.    ANS:=max (N - maxMountainLength,N - maxMountainLength - count) ; 
  61.    WriteLn(ANS);
  62. end.
Success #stdin #stdout 0s 5324KB
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https://ideone.com/2EivOF
language:
Pascal (fpc 3.0.4)
created:
21 hours ago
visibility:
public

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