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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 









share|improve this question




















  • 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
















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 









share|improve this question




















  • 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














7












7








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 









share|improve this question
















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















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago







s326280

















asked 4 hours ago









s326280s326280

826




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














  • 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








1




1





Does each digit have to be exactly one bigger than the last?

– John Gordon
4 hours ago







Does each digit have to be exactly one bigger than the last?

– John Gordon
4 hours ago















yes @JohnGordon

– s326280
4 hours ago





yes @JohnGordon

– s326280
4 hours ago




1




1





The accepted answer currently seems to fail for 78901.

– גלעד ברקן
1 hour ago







The accepted answer currently seems to fail for 78901.

– גלעד ברקן
1 hour ago















Sorry, i didn't notice that !!.

– s326280
1 hour ago





Sorry, i didn't notice that !!.

– s326280
1 hour ago












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































    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



















    • 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



















    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

































      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

































        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























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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          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




























            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


























              4












              4








              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













              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












              share|improve this answer



              share|improve this answer










              answered 2 hours ago









              blhsingblhsing

              36.2k41639




              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



















                  • 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
















                  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



















                  • 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














                  3












                  3








                  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













                  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












                  share|improve this answer



                  share|improve this answer










                  answered 3 hours ago









                  Jean-François FabreJean-François Fabre

                  105k955112




                  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














                  • 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








                  1




                  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







                  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




                  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







                  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





                  @JonClements nice but in that case I would create a string beforehand.

                  – Jean-François Fabre
                  1 hour ago











                  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






























                    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




























                      2












                      2








                      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















                      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














                      share|improve this answer



                      share|improve this answer








                      edited 3 hours ago

























                      answered 3 hours ago









                      Tomothy32Tomothy32

                      7,2721627




                      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






























                            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




























                              0












                              0








                              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















                              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














                              share|improve this answer



                              share|improve this answer








                              edited 3 hours ago

























                              answered 3 hours ago









                              khachikkhachik

                              21.1k54381




                              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




























                                    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


























                                      0












                                      0








                                      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













                                      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)






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                                      answered 1 hour ago









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