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

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

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7

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 4 hours ago

s326280

asked 4 hours ago

s326280s326280

826

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes @JohnGordon
  
  – s326280
  4 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 | 

7

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 4 hours ago

s326280

asked 4 hours ago

s326280s326280

826

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes @JohnGordon
  
  – s326280
  4 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 4 hours ago

s326280

asked 4 hours ago

s326280s326280

826

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

share|improve this question

edited 4 hours ago

s326280

asked 4 hours ago

s326280s326280

826

share|improve this question

edited 4 hours ago

s326280

asked 4 hours ago

s326280s326280

826

asked 4 hours ago

s326280s326280

826

asked 4 hours ago

s326280s326280

826

5 Answers
5

active

oldest

votes

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 | 

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 | 

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

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 | 

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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

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

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

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

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