Maximum lenght bitonic subarray

#1

I am just able to pass 1 testcase . I cant figure out , where i went wrong

#2

@1706157
Your algorithm is of N*N complexity, Try to do it in O(N).
here is the algorithm to do so.
0. Create inc[n] and dec[n]

  1. Initialize inc[0] to 1 and dec[n-1] to 1
    a. Till end of the array ie, i=1 to n, if arr[i] > arr[i-1] then inc[i] = inc[i-1] + 1. else, inc[i] = 1
    b. From the end of the array ie, i = n-2 till i =0, if arr[i] > arr[i+1] then dec[i] = dec[i +1] +1 else, dec[i] = 1
    then final ans is
for (int i = 1; i < n; i++) 
            if (inc[i] + dec[i] - 1 > max) 
                max = inc[i] + dec[i] - 1;
#3

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