Code

def CheckIncrease(lst):
    '''
    This function returns the number of times an integer is followed immediately
    by a bigger integer in lst.
    For example, if lst = [1, 2, 1, 3, 4] then this function returns 3 since
    1 followed by 2 is the first time an integer is immediately followed by a bigger integer
    1 followed by 3 is the second time this hapens
    and 3 followed by 4 is the last and final time an integer is followed by a bigger integer
    Hence this function returns 3.

    More test cases listed below in main function along with
    their expected output (inside comments)
    '''
    count = 0
    for i in range(0,len(lst)-1):
        if(lst[i+1] > lst[i]):
            count +=1
    return count


def main():
    print(CheckIncrease([1, 2, 3, 4, 5])) ## 4
    print(CheckIncrease([1, 2, 2, 2, 2])) ## 1
    print(CheckIncrease([2, 2, 2, 2, 2, 2])) ## 0
    print(CheckIncrease([1, 2, 2, 2, 2, 3, 3, 3, 3])) ## 2
    print(CheckIncrease([1, 2, 2, 2, 1, 3, 3, 3, 1])) ## 2
    print(CheckIncrease([100, 50, 25, 12, 6, 3, 1])) ## 0
    print(CheckIncrease([1, 2, 1, 3, 4])) ## 3
    print(CheckIncrease([])) ## 0

main()