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


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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 









share|improve this question




















  • 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
















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 









share|improve this question




















  • 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














6












6








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 









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 3 hours ago







s326280

















asked 3 hours ago









s326280s326280

776




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














  • 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








1




1





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

– John Gordon
3 hours ago







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

– John Gordon
3 hours ago















yes @JohnGordon

– s326280
3 hours ago





yes @JohnGordon

– s326280
3 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


















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



















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































    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


























      • well, it doesn't work. and it doesn't handle the case 7890 either

        – Jean-François Fabre
        3 hours ago



















      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









        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













        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























                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


























                        • well, it doesn't work. and it doesn't handle the case 7890 either

                          – Jean-François Fabre
                          3 hours ago
















                        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


























                        • well, it doesn't work. and it doesn't handle the case 7890 either

                          – Jean-François Fabre
                          3 hours ago














                        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













                        • well, it doesn't work. and it doesn't handle the case 7890 either

                          – Jean-François Fabre
                          3 hours ago



















                        • well, it doesn't work. and it doesn't handle the case 7890 either

                          – Jean-François Fabre
                          3 hours ago

















                        well, it doesn't work. and it doesn't handle the case 7890 either

                        – Jean-François Fabre
                        3 hours ago





                        well, it doesn't work. and it doesn't handle the case 7890 either

                        – Jean-François Fabre
                        3 hours ago











                        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)






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 1 hour ago









                            גלעד ברקןגלעד ברקן

                            13k21542




                            13k21542






























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