program mountain;
Uses Math;
const
    MAXN = 100005;

var
    ANS, N, i, j, maxMountainLength : LongInt;
    P, leftLIS, rightLIS, scartato  : Array[0..MAXN-1] of LongInt;
    
begin

    (*assign(input,  'input.txt');  reset(input);
    assign(output, 'output.txt'); rewrite(output);*)

    ReadLn(N);

    for i:=0 to N-1 do
        Read(P[i]);
    ReadLn();
                    
    ANS := 0;

	(*leftLIS[i] stores the length of longest increasing subsequence ending at index i*)
	(*rightLIS[i] stores the length of longest decreasing subsequence starting at index i*)

   for i:=0 to  N-1 do begin leftLIS[i]:=1; rightLIS[i]:=1; scartato[i]:=0; end;

	(*Calculate LIS from left to right for each position*)

   for i := 1 to N-1 do
            for j:= 0 to i-1 do
                begin
                  if (i<N-1) and (P[i]<P[i-1]) and (P[i]<P[i+1]) then scartato[i]:=1; 
                  if (j>0) and (P[j]<P[j-1]) and (P[j]<P[j+1]) then scartato[j]:=1; 
                  if (scartato[i]=0) and (scartato[j]=0) and (P[i] > P[j]) then leftLIS[i] := max(leftLIS[i], leftLIS[j] + 1);                          
                end;  
  
	(* Calculate LIS from right to left (decreasing subsequence) for each position*)

   for i := N - 2 downto 0 do
            for  j := i + 1 to N-1 do
               begin
                if (i>0) and (P[i]<P[i-1]) and (P[i]<P[i+1]) then scartato[i]:=1; 
                if (j<N-1) and (P[j]<P[j-1]) and (P[j]<P[j+1]) then scartato[j]:=1; 
                if (scartato[i]=0) and (scartato[j]=0) and (P[i] > P[j]) then rightLIS[i] := max(rightLIS[i], rightLIS[j] + 1);                
	       end;
	
       (* Find the maximum length of mountain subsequence*)
   maxMountainLength := 0;
   for i := 0 to N-1 do
	(*A valid mountain peak must have at least one element on both sides*)
	(*leftLIS[i] > 1 ensures there's at least one element before peak*)
	(*rightLIS[i] > 1 ensures there's at least one element after peak*)
   if (leftLIS[i] >= 1) and (rightLIS[i] >= 1)  then
  	(*Total mountain length with peak at i Subtract 1 because peak is counted in both leftLIS and rightLIS*)
         maxMountainLength := max(maxMountainLength, leftLIS[i] + rightLIS[i] - 1);
 	(* Minimum removals = total elements - maximum mountain length*)
   ANS:= N - maxMountainLength; 
   WriteLn(ANS);
end.