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  1. Practice
  2. Algorithms
  3. Implementation
  4. Birthday Chocolate
  5. Discussions

Birthday Chocolate

  • rakesht2499 + 4 comments

    The shortest method for the above python code

    return sum(1 for x in range(len(s)) if sum(s[x:x+m]) == d)

    • lifebalance + 0 comments

      Well done!

    • wrutka + 1 comment

      Im not sure of this solution. This will not be working with this input:

      5
1 2 1 3 3
3 2

      The output should be 2 but will be 3 because it will count last digit of the series 3. This solution does not consider what if solution will be shorter than a month digit.

      Sure, this will pass all validators but still... :).

      • daniel_augusto11 + 0 comments

        return sum(1 for x in range(len(s)) if sum(s[x:x+m]) == d and len(s[x:x+m]) == m )

    • DaniRed_ + 1 comment

      That't beutiful thank you for sharing : i did this way but we can make way shorter by yours..

      1. num=0
      2. for i in range(len(s)):
      3. if sum(s[i:i+m])==d:
      4. num+=1
      5. return(num)
      • DaniRed_ + 1 comment

        List comprehension is amazing

        • arshiyasoleimani + 0 comments

          I agree but also his method is n * m time complexity sadly.

    • rusaka + 0 comments

      a minor optimization:

      return sum(1 for x in range(len(s)-m+1) if sum(s[x:x+m]) == d)