Tjyylli

Contest math problem about crossing out numbers in the table

Why don't American passenger airlines operate dedicated cargo flights any more?

How to prevent cleaner from hanging my lock screen in Ubuntu 16.04

How to convert a ListContourPlot into primitive usable with Graphics3D?

Disable the ">" operator in Rstudio linux terminal

Is there some relative to Dutch word "kijken" in German?

Why would the Pakistan airspace closure cancel flights not headed to Pakistan itself?

Can I write a book of my D&D game?

Does Windows 10's telemetry include sending *.doc files if Word crashed?

Why did other German political parties disband so fast when Hitler was appointed chancellor?

Checking for the existence of multiple directories

What is this metal M-shaped device for?

Jumping Numbers

Broken patches on a road

How to deal with an incendiary email that was recalled

What flying insects could re-enter the Earth's atmosphere from space without burning up?

Would these multi-classing house rules cause unintended problems?

Groups acting on trees

A universal method for left-hand alignment of a sequence of equalities

Chess tournament winning streaks

How to avoid being sexist when trying to employ someone to function in a very sexist environment?

What makes the Forgotten Realms "forgotten"?

How do you funnel food off a cutting board?

If I delete my router's history can my ISP still provide it to my parents?

Check if the digits in the number are in increasing sequence in python

How do I check if a list is empty?How do I check whether a file exists without exceptions?Calling an external command in PythonWhat are metaclasses in Python?How do I check if an array includes an object in JavaScript?Does Python have a ternary conditional operator?Check if a given key already exists in a dictionaryHow to get the number of elements in a list in Python?Does Python have a string 'contains' substring method?Easy interview question got harder: given numbers 1..100, find the missing number(s)

6

I was working on a problem that determines whether the digits in the numbers are in the increasing sequence. Now, the approach I took to solve the problem was, For instance, consider the number 5678.

To check whether 5678 is an increasing sequence, I took the first digit and the next digit and the last digit which is 5,6,8 and substitute in range function range(first,last,(diff of first digit and the next to first digit)) i.e range(5,8+1,abs(5-6)).The result is the list of digits in the ascending order

To this problem, there is a constraint saying

For incrementing sequences, 0 should come after 9, and not before 1, as in 7890. Now my program breaks at the input 7890. I don't know how to encode this logic. Can someone help me, please?.

The code for increasing sequence was

  len(set(['5','6','7','8']) - set(map(str,range(5,8+1,abs(5-6))))) == 0 

python algorithm

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Does each digit have to be exactly one bigger than the last?
  
  – John Gordon
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes @JohnGordon
  
  – s326280
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The accepted answer currently seems to fail for 78901.
  
  – גלעד ברקן
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry, i didn't notice that !!.
  
  – s326280
  1 hour ago
  

  
  
  
  

add a comment | 

6

I was working on a problem that determines whether the digits in the numbers are in the increasing sequence. Now, the approach I took to solve the problem was, For instance, consider the number 5678.

To check whether 5678 is an increasing sequence, I took the first digit and the next digit and the last digit which is 5,6,8 and substitute in range function range(first,last,(diff of first digit and the next to first digit)) i.e range(5,8+1,abs(5-6)).The result is the list of digits in the ascending order

To this problem, there is a constraint saying

For incrementing sequences, 0 should come after 9, and not before 1, as in 7890. Now my program breaks at the input 7890. I don't know how to encode this logic. Can someone help me, please?.

The code for increasing sequence was

  len(set(['5','6','7','8']) - set(map(str,range(5,8+1,abs(5-6))))) == 0 

python algorithm

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Does each digit have to be exactly one bigger than the last?
  
  – John Gordon
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes @JohnGordon
  
  – s326280
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The accepted answer currently seems to fail for 78901.
  
  – גלעד ברקן
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry, i didn't notice that !!.
  
  – s326280
  1 hour ago
  

  
  
  
  

add a comment | 

I was working on a problem that determines whether the digits in the numbers are in the increasing sequence. Now, the approach I took to solve the problem was, For instance, consider the number 5678.

To check whether 5678 is an increasing sequence, I took the first digit and the next digit and the last digit which is 5,6,8 and substitute in range function range(first,last,(diff of first digit and the next to first digit)) i.e range(5,8+1,abs(5-6)).The result is the list of digits in the ascending order

To this problem, there is a constraint saying

For incrementing sequences, 0 should come after 9, and not before 1, as in 7890. Now my program breaks at the input 7890. I don't know how to encode this logic. Can someone help me, please?.

The code for increasing sequence was

  len(set(['5','6','7','8']) - set(map(str,range(5,8+1,abs(5-6))))) == 0 

python algorithm

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

  len(set(['5','6','7','8']) - set(map(str,range(5,8+1,abs(5-6))))) == 0 

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

share|improve this question

edited 3 hours ago

s326280

asked 3 hours ago

s326280s326280

776

asked 3 hours ago

s326280s326280

776

asked 3 hours ago

s326280s326280

776

5 Answers
5

active

oldest

votes

3

you could zip the string representation of the number with a shifted self and iterate on consecutive digits together. Use all to check that numbers follow, using a modulo 10 to handle the 0 case.

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  you can avoid the double str'ing and slicing by using zip(*[iter(str(num))] * 2) instead but I imagine that's got way more overhead for such use in this case anyway... just throwing it out there...
  
  – Jon Clements♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This would incorrectly return True for 78901, as the OP says "0 should come after 9, and not before 1".
  
  – blhsing
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JonClements nice but in that case I would create a string beforehand.
  
  – Jean-François Fabre
  1 hour ago
  

  
  
  
  

add a comment | 

4

You can simply check if the number, when converted to a string, is a substring of '1234567890':

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

add a comment | 

2

I would create a cycling generator and slice that:

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

This is faster than solutions that check differences between individual digits. Of course, you can sacrifice the length to make it faster:

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

add a comment | 

0

Since you already have the zip version, here is an alternative solution:

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  well, it doesn't work. and it doesn't handle the case 7890 either
  
  – Jean-François Fabre
  3 hours ago
  

  
  
  
  

add a comment | 

0

Here's my take that just looks at the digits and exits as soon as there is a discrepancy:

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54954713%2fcheck-if-the-digits-in-the-number-are-in-increasing-sequence-in-python%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

you could zip the string representation of the number with a shifted self and iterate on consecutive digits together. Use all to check that numbers follow, using a modulo 10 to handle the 0 case.

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  you can avoid the double str'ing and slicing by using zip(*[iter(str(num))] * 2) instead but I imagine that's got way more overhead for such use in this case anyway... just throwing it out there...
  
  – Jon Clements♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This would incorrectly return True for 78901, as the OP says "0 should come after 9, and not before 1".
  
  – blhsing
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JonClements nice but in that case I would create a string beforehand.
  
  – Jean-François Fabre
  1 hour ago
  

  
  
  
  

add a comment | 

3

you could zip the string representation of the number with a shifted self and iterate on consecutive digits together. Use all to check that numbers follow, using a modulo 10 to handle the 0 case.

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  you can avoid the double str'ing and slicing by using zip(*[iter(str(num))] * 2) instead but I imagine that's got way more overhead for such use in this case anyway... just throwing it out there...
  
  – Jon Clements♦
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  This would incorrectly return True for 78901, as the OP says "0 should come after 9, and not before 1".
  
  – blhsing
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JonClements nice but in that case I would create a string beforehand.
  
  – Jean-François Fabre
  1 hour ago
  

  
  
  
  

add a comment | 

you could zip the string representation of the number with a shifted self and iterate on consecutive digits together. Use all to check that numbers follow, using a modulo 10 to handle the 0 case.

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

num = 7890



result = all((int(y)-int(x))%10 == 1 for x,y in zip(str(num),str(num)[1:]))

share|improve this answer

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

answered 3 hours ago

Jean-François FabreJean-François Fabre

105k955112

4

You can simply check if the number, when converted to a string, is a substring of '1234567890':

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

add a comment | 

4

You can simply check if the number, when converted to a string, is a substring of '1234567890':

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

add a comment | 

You can simply check if the number, when converted to a string, is a substring of '1234567890':

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

str(num) in '1234567890'

share|improve this answer

answered 2 hours ago

blhsingblhsing

36.2k41639

answered 2 hours ago

blhsingblhsing

36.2k41639

answered 2 hours ago

blhsingblhsing

36.2k41639

2

I would create a cycling generator and slice that:

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

This is faster than solutions that check differences between individual digits. Of course, you can sacrifice the length to make it faster:

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

add a comment | 

2

I would create a cycling generator and slice that:

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

This is faster than solutions that check differences between individual digits. Of course, you can sacrifice the length to make it faster:

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

add a comment | 

I would create a cycling generator and slice that:

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

This is faster than solutions that check differences between individual digits. Of course, you can sacrifice the length to make it faster:

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

from itertools import cycle, islice



num = 5678901234



num = tuple(str(num))

print(num == tuple(islice(cycle(map(str, range(10))), int(num[0]), int(num[0]) + len(num))))

def digits(num):

    while num:

        yield num % 10

        num //= 10



def check(num):

    num = list(digits(num))

    num.reverse()

    for i, j in zip(islice(cycle(range(10)), num[0], num[0] + len(num)), num):

        if i != j:

          return False

    return True

share|improve this answer

edited 3 hours ago

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

answered 3 hours ago

Tomothy32Tomothy32

7,2721627

0

Since you already have the zip version, here is an alternative solution:

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  well, it doesn't work. and it doesn't handle the case 7890 either
  
  – Jean-François Fabre
  3 hours ago
  

  
  
  
  

add a comment | 

0

Since you already have the zip version, here is an alternative solution:

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  well, it doesn't work. and it doesn't handle the case 7890 either
  
  – Jean-François Fabre
  3 hours ago
  

  
  
  
  

add a comment | 

Since you already have the zip version, here is an alternative solution:

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

import sys





order = dict(enumerate(range(10)))

order[0] = 10



def increasing(n):

    n = abs(n)

    o = order[n % 10] + 1

    while n:

        r = n % 10

        n = n / 10

        if o - order[r] != 1:

            return False

        o = order[r]

    return True





for n in sys.argv[1:]:

    print n, increasing(int(n))

share|improve this answer

edited 3 hours ago

answered 3 hours ago

khachikkhachik

21.1k54381

answered 3 hours ago

khachikkhachik

21.1k54381

answered 3 hours ago

khachikkhachik

21.1k54381

0

Here's my take that just looks at the digits and exits as soon as there is a discrepancy:

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

add a comment | 

0

Here's my take that just looks at the digits and exits as soon as there is a discrepancy:

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

add a comment | 

Here's my take that just looks at the digits and exits as soon as there is a discrepancy:

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

def f(n):

  while (n):

    last = n % 10

    n = n / 10

    if n == 0:

      return True

    prev = n % 10

    print last, prev

    if prev == 0 or prev != (10 + last - 1) % 10:

      return False



print f(1234)

print f(7890)

print f(78901)

print f(1345)

share|improve this answer

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542

answered 1 hour ago

גלעד ברקןגלעד ברקן

13k21542