Identifying blocks (of variable length) of numbers that must occur in a consecutive sequence

Question

Problem Statement: An array is accepted if and only if it has the following structure:

First a1 elements equal 1.
Next a2 elements equal 2.
Next a3 elements equal 3.
Next a4 elements equal 4.
Next a5 elements equal 5.
Next a6 elements equal 6.
Next a7 elements equal 7.

Where:

a(i) can be any non-zero positive integer.
There are no other elements in array.

Even though the algorithm for this problem seems pretty easy to implement, I am having some difficulty with the code. Here is what I have written.

print("Enter list: ")
arr = [int(x) for x in input().split()]
print(arr)
i = 0

if arr[0] == 1:  # to check whether input starts with a 1. 
    if arr[i] == 1:  #If next number is also 1, then
        while arr[i] == 1:  #while numbers are equal to 1, increment i
            i = i + 1
            if arr[i] == 2: #If next number is 2, then
                while arr[i] == 2: #same procedure as above ... and so on
                    i = i + 1
                    if arr[i] == 3:
                        while arr[i] == 3:
                            i = i + 1
                            if arr[i] == 4:
                                while arr[i] == 4:
                                    i = i + 1
                                    if arr[i] == 5:
                                        while arr[i] == 5:
                                            i = i + 1
                                            if arr[i] == 6:
                                                while arr[i] == 6:
                                                    i = i + 1
                                                    if arr[i] == 7:
                                                        while arr[i] == 7:
                                                            i = i + 1
                                                        if arr[-1] == 7: #to check if last number is a 7
                                                            print("Accepted")
                                                        else:
                                                            print("not")
                                                    else:
                                                        print("not")
                                            else:
                                                print("not")
                                    else:
                                        print("not")
                            else:
                                print("not")
                    else:
                        print("not")
            else:
                print("not")
    else:
        print("not")
else:
    print("not")

I seem to be getting some kind of indexing error where it says:

   while arr[i] == 7:
IndexError: list index out of range

I don't understand why I am encountering this error. As far as I can tell, I am not exceeding the list index.

Welcome to Stack Overflow - nice to have you. Please read [How do I ask a good question?](https://stackoverflow.com/help/how-to-ask) and [How to create a Minimal, Complete, and Verifiable example](https://stackoverflow.com/help/mcve) to help keeping Stack Overflows content on the highest possible level and increase your chances getting an appropriate answer. — Axel, Oct 11 '17 at 00:51
I'd have to say that this looks like highly redundant code. I am pretty sure you can simplify all of this in just a few lines. To give you a hint, the number you are expecting to find correlates very close to the position in the array you expect it to be. — , Oct 11 '17 at 10:02
Possible duplicate of [How to prevent ArrayIndexOutOfBoundsException in Java?](https://stackoverflow.com/questions/5554734/how-to-prevent-arrayindexoutofboundsexception-in-java) — EJoshuaS - Stand with Ukraine, Jan 03 '19 at 04:31
Dupe target is for Java, but the solution isn't all that different in Python. — EJoshuaS - Stand with Ukraine, Jan 03 '19 at 04:32

Ajax1234 · Answer 1 · 2017-10-10T20:03:44.053

1

You can try this:

def is_accepted(arr):
    return arr[0] == 1 and all(arr[i] -1 == arr[i-1] or arr[i] == arr[i-1] for i in range(1, len(arr)))

print(is_accepted([1, 2, 3, 4, 5, 6, 7]))

Output:

True

Second List:

s = [1, 2, 3, 4, 5, 5, 4, 3, 5, 6, 7]
print(is_accepted(s))

Output:

False

Last input:

s = [1, 1, 2, 2, 3, 3, 3, 4, 5, 6, 6, 7]
print(is_accepted(s))

Output:

True

edited Oct 10 '17 at 20:03

answered Oct 10 '17 at 19:05

Ajax1234

69,937
8
61
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`Enter list: 1 2 3 4 5 5 4 3 5 6 7 [1, 2, 3, 4, 5, 5, 4, 3, 5, 6, 7] True` It shows True for the wrong pattern too. – Sakshi Jain Oct 10 '17 at 19:37
@SakshiJain When I run that list you just posted, the output is False. – Ajax1234 Oct 10 '17 at 19:38
It doesn't work on redundant numbers. Say if I input a string `s = [1, 1, 2, 2, 3, 3, 3, 4, 5, 6, 6, 7]` according to the problem statement, it is a valid input. But the program says otherwise. – Sakshi Jain Oct 10 '17 at 19:52
@SakshiJain So you are saying that redundant numbers are acceptable in this problem? – Ajax1234 Oct 10 '17 at 19:53
I think the solution to that would be adding an if statement like `if arr[i] == arr[i-1] or arr[i] == arr[i-1]+1` instead of your original condition. – Sakshi Jain Oct 10 '17 at 19:53
Yes. It says in the problem statement that a(i) elements equal to number. Where a(i) could be any non-zero number. – Sakshi Jain Oct 10 '17 at 19:55

Oluwafemi Sule · Answer 2 · 2017-10-11T18:44:21.177

0

A fun exercise in recursion.

Write cases the following conditions:

list is empty
first item of list less than last_checked as a short-circuit
match recursively.

def match(lst, last_checked=1):
  if lst == []: return True # base case

  if lst[0] < last_checked:
    return False

  return True and match(lst[1:], lst[0])

edited Oct 11 '17 at 18:44

answered Oct 11 '17 at 16:38

Oluwafemi Sule

36,144
1
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Identifying blocks (of variable length) of numbers that must occur in a consecutive sequence

2 Answers2