Python: Check if element is less than the next elements

Question

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[0] < A[3]

A[1] < A[2] 

A[1] < A[3]

A[2] < A[3]

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

Answer 1

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