Not able to solve

#1

my code is only working for testcase 1

#2

hello @sindhu_21
a) ur array is small , check maximum value of n , and declare ur array with size greater than that.
b) ur solution time complexity is O(tnn) which will not work for this problem.

another approach
Let us consider the array {12, 4, 78, 90, 45, 23} to understand the solution.

  1. Construct an auxiliary array inc[] from left to right such that inc[i] contains length of the nondecreaing subarray ending at arr[i].
    For A[] = {12, 4, 78, 90, 45, 23}, inc[] is {1, 1, 2, 3, 1, 1}

  2. Construct another array dec[] from right to left such that dec[i] contains length of nonincreasing subarray starting at arr[i].
    For A[] = {12, 4, 78, 90, 45, 23}, dec[] is {2, 1, 1, 3, 2, 1}.

  3. Once we have the inc[] and dec[] arrays, all we need to do is find the maximum value of (inc[i] + dec[i] ā€“ 1).
    For {12, 4, 78, 90, 45, 23}, the max value of (inc[i] + dec[i] ā€“ 1) is 5 for i = 3.

    int bitonic(int arr[], int n)
    {
    int inc[n]; // Length of increasing subarray ending at all indexes
    int dec[n]; // Length of decreasing subarray starting at all indexes
    int i, max;

    // length of increasing sequence ending at first index is 1
    inc[0] = 1;

    // length of increasing sequence starting at first index is 1
    dec[n-1] = 1;

    // Step 1) Construct increasing sequence array
    for (i = 1; i < n; i++)
    inc[i] = (arr[i] >= arr[i-1])? inc[i-1] + 1: 1;

    // Step 2) Construct decreasing sequence array
    for (i = n-2; i >= 0; iā€“)
    dec[i] = (arr[i] >= arr[i+1])? dec[i+1] + 1: 1;

    // Step 3) Find the length of maximum length bitonic sequence
    max = inc[0] + dec[0] - 1;
    for (i = 1; i < n; i++)
    if (inc[i] + dec[i] - 1 > max)
    max = inc[i] + dec[i] - 1;

    return max;
    }