Question

How do I approach this? I've tried different ways and this is what I've so far . ex: [1, 2, 5, 7] returns True and [1, 1, 6, 9] returns False. The second example doesn't work. It returns true even though first two elements are equal. What am I doing wrong?

def increasing(L):
    n = len(L)
    i = 0
    while i <= n-1:
        if L[i] > L[i+1]:
            return False
        else:
            i+=1
    return True

Answer 1

Problem lies here

while i <= n:
    if L[i] > L[i+1]:

In your example, n=4. Since array index starts at 0, list index would be 0 1 2 3. You are checking till n=4 which is incorrect.

Then, you are doing [i+1] which checks for 5th element.

Replace with these lines

while i < n-1:
    if L[i] > L[i+1]:

This will take care of your index out of range error

Do I HAVE TO also use a for loop here?

No. Check Python - How to check list monotonicity

Answer 2

Replace

while i <= n:

to

while i <= n-1:

Modified code :

def increasing(L):
    n = len(L)
    i = 0
    while i < n-1:
        if L[i] >= L[i+1]: #Equal to is also included as we want increasing sequence
            return False
        else:
            i+=1
    return True

Answer 3

This fixes your code. Your bound was actually 2 too high. I also removed some redundancy.

def increasing(L):
    i = 0
    while i < len(L) - 1:
        if L[i] > L[i+1]:
            return False
        i+=1
    return True

Boolean function to check if a list of ints is in increasing order- Must use a while loop

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