While loop doesn't work as I designed

Question

0.00/5 (No votes)

See more:

when the program receive input for "cubes", I want to first check if the maximium value is on one of the end of the sequence, if yes, delete the max value, if not exit and print no, as long as 'd' is not empty, keep on looping, till 'd' is empty then print yes, or exit the program in one of the loop.

I can input 'T' number of test sequences.
suppose I give a sequense [4, 3 ,1, 3 ,4]

I use debugger and found out , while loop delete '4' in the first iteration, and directly goes to outside "for loop". what is the reason? thank you very much in advance!

from collections import deque
T=int(input())#number of test cases
for i in range(T):
    n=int(input())
    cubes=map(int, input().split())
    d=deque(cubes)
    
    while len(d)!=0:
        a=max(d)
        if d.index(a)==0 or d.index(a)==-1:
            d.remove(a)
        else:
            break
            print(no)
    print('yes')

What I have tried:

tried different input, the result is the same

Posted 2-Sep-21 0:59am

Zhengxi Jiang

Updated 2-Sep-21 2:13am

v3

Add a Solution

Comments

Maciej Los 2-Sep-21 7:33am

Why to loop over entire "cube"?

inp = [4, 3 ,1, 3 ,4]
if inp[0] == inp[len(inp)-1]:
    print('yes')
else:
    print('no')

Zhengxi Jiang 2-Sep-21 7:39am

it is a game about put cubes together. there are rules: 1. you can only take the leftmost or the rightmost of the cubes. these integers reprenst the sidelength of the cubes. the cubes i on top of cube j has to smaller than the cubes j. if according to the rules i can build it, answer is yes, otherwise it is no, means not possible to build. but here the problem is the code, why the while loop not keep on delete 4, after that 3 and 1, it goes to the higher 'for loop'. in fact it works as if is no while loop ,

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Richard MacCutchan · Accepted Answer · 2021-09-02T02:13:00

Solution 1

Python

if d.index(a)==0 or d.index(a)==-1: d.remove(a) else: break # this terminates the while loop print(no) # this statement will never be executed

Move the print statement before the break.

[edit]
Change the while loop to the following:

Python

while len(d)!=0:
    a=max(d)
    print(F'max: {a}, index: {d.index(a)}')
    # check if the max value is at the beginning, or end of the list
    if d.index(a)==0 or d.index(a)==len(d) - 1:
        d.remove(a)
        print(F'{a} removed')
    else:
        print('no')
        break
print('yes: ', d)

Note: the added print statements show exactly what happens.

[/edit]

Posted 2-Sep-21 2:13am

Richard MacCutchan

Updated 2-Sep-21 2:25am

v3

Comments

jsc42 2-Sep-21 12:58pm

I'm not a Python expert but ...
Doesn't d.remove(a) remove the item in position 'a', not the item with value 'a'?
So, if you have [4, 3, 1, 3, 4]
First Iteration: a = 4 (max n); d.index(a) == 0 and d.index(a) also == len(d) - 1; so you do d.remove(a) meaning remove item 4 which (using zero based subscripts) just coincidentally is one of the items that has a value of 4.
Second Iteration: a = 4 (because there were 2 '4's in the original list); d.index(a) == 0 but not == len(d) - 1; d.remove(a) once again tried to remove item 4, but your list now only has indicies 0 to 3 so the remove will fail.

If I understand the task, then I think the simplest way to implement want you need is to split the tests for d.index(a) into two 'if' branches and when one is true, remove it using the found index (which will be 0 or len(d) - 1)

Richard MacCutchan 3-Sep-21 3:29am

No, it removes the item with the value a: collections — Container datatypes — Python 3.9.7 documentation[^].

And on the second iteration, the expression len(d) - 1, still returns the correct length as d is now one item shorter, so it removes the 4 at the end. And I did run a few tests on the code before I posted it here.

Richard MacCutchan 3-Sep-21 4:00am

Here are the results for the list [4, 3, 1, 3, 4]

        python cp.py
max: 4, index: 0
4 removed
max: 4, index: 3
4 removed
max: 3, index: 0
3 removed
max: 3, index: 1
3 removed
max: 1, index: 0
1 removed
yes:  deque([])

jsc42 3-Sep-21 6:13am

Thanks!