# Python: Check if element is less than the next elements

I want to check each element in the list, if it is smaller than the next after it and print all the numbers that met the condition.

Example:

``````A = [3,1,2,4]

A < A

A < A

A < A

A < A
``````

And I don’t want to check the last element.

What is the best way to do that?

I tried to do that with for loop:

``````A = [3,1,2,4]
for i in range(0,len(A)-1):
for j in range(i,len(A)-1):
if A[j] < A[j + 1]:
print(A[j],A[j + 1])
A[j] = A[j + 1]
``````

And the result is:

``````1 2
2 4
1 4
``````

While the desired results is:

``````3 4
1 2
1 4
2 4
``````

### >Solution :

You should fix some indices in comparison as well as the ranges. The first counter `i` should iterate from `0` to `len(A)-1` and the second counter should iterate from `i+1` to `len(A)`.

The comparison should happen between `A[i]`, `A[j]` instead of `A[j]`,`A[j+1]`. No need for the last line `A[j] = A[j + 1]` because the counters get incremented automatically.

``````A = [3, 1, 2, 4]
for i in range(0,len(A)-1):
for j in range(i+1,len(A)):
if A[i] < A[j]:
print(A[i],A[j])
3 4
1 2
1 4
2 4
``````