SDE - 1

One approach to solve the above question would be to traverse the array and check for the strictly increasing sequence and then check for strictly decreasing subsequence or vice-versa. 
Steps : 
1. If arr[1] > arr[0], then check for strictly increasing then strictly decreasing as: 
• Check for every consecutive pair until at any index i arr[i + 1] is less than arr[i].
• Now from index i + 1 check for every consecutive pair check if arr[i + 1] is less than arr[i] till the end of the array or not. If at any index i, arr[i] < arr[i + 1] then break the loop.
• If the end is reached, the array forms an increasing-decreasing sequence. 
2. If arr[1] < arr[0], then check for strictly decreasing then strictly increasing as: 
• Check for every consecutive pair until at any index i arr[i + 1] is greater than arr[i].
• Now from index i + 1 check for every consecutive pair check if arr[i + 1] is greater than arr[i] till the end of the array or not. If at any index i, arr[i] > arr[i + 1] then break the loop.
• If the end is reached, the array forms an decreasing-increasing sequence.

For any number ‘num’ all its divisors are always less than and equal to ‘num/2’ and all prime factors are always less than and equal to sqrt(num). 
So, iterate through ‘i’ till i<=sqrt(num) and for any ‘i’ if it divides ‘num’ , then we get two divisors ‘i’ and ‘num/i’ , continuously add these divisors but for some numbers divisors ‘i’ and ‘num/i’ will be the same, in this case just add only one divisor. Finally, add one as one is divisor of all natural numbers.