[generallyfalse]

I should like to see some of the problems that take you 2-5 hours to solve. Show your solution, and text about the way you tackle the problem.

Perhaps you need a mentor, someone that can give you some tips about your coding abilities and what you should do to improve.

If you need more than 10 minutes to sum the cubes of numbers from 1 to n in python, then I think you have a problem as a programmer.

Can you point to a leetcode problem that takes you 2-5 hours to solve?

Edited: Two weeks ago you submitted a similar post and in the comments you give this example:

>> To give a concrete example of some of the above, just this morning I tried to code the brute force solution following problem: Give two sorted arrays return the median. I spent over an hour trying to implement the brute force solution involving linearly going through each array and couldn’t code it in Python.

As a hobby programmer, my ten minutes approach:

  def median(a,b):
    ...:     ab = len(a)+len(b)
    ...:     pos = ab/2
    ...:     ia = 0
    ...:     ib = 0
    ...:     for ic in range(pos):
    ...:         if a[ia] <= b[ib]:
    ...:             x = a[ia]
    ...:             ia = ia + 1
    ...:         else:
    ...:             x = b[ib]
    ...:             ib = ib + 1
    ...:     if ab % 2 == 1 :
    ...:        return x
    ...:     else:
    ...:        return (x + max(a[ia],b[ib]))/2
    ...:

 But it is not correct

[Tmp1234]

For the median question, I did not allow myself to combine the array. My idea was to increment in each array based on which value would come next in a sequence (as if they were on array), stop when I've incremented n times where n = length of both arrays / 2 (the median position) and then return. Here's the code:

class Solution: def findMedianSortedArrays(self, nums1, nums2): """ :type nums1: List[int] :type nums2: List[int] :rtype: float """

        length = len(nums1) + len(nums2)
        medianIndex = length // 2
        ptr1 = 0
        ptr2 = 0
        curr = 0
        flag = 0
        
        if nums1[ptr1] < nums2[ptr2]:
            val = nums1[ptr1]
        else:
            val = nums2[ptr2]
        
        while curr < medianIndex:
            if nums1[ptr1] == val:
                if nums2[ptr2] < val:  
                    curr += 1
                    ptr1 +=1
                    val = nums1[ptr1]
                    flag = 0
                else:
                    if ptr1+1 < len(nums1) and abs(nums1[ptr1+1]) - abs(val) < abs(nums2[ptr2]) - abs(val):
                        curr+=1
                        ptr1+=1
                        val = nums1[ptr1]
                        flag = 0
                    else:
                        curr+=1
                        val = nums2[ptr2]
                        flag = 1
            else:
                if nums1[ptr1] < val:  
                    curr += 1
                    ptr2 +=1
                    val = nums2[ptr2]
                    flag = 1
                else:
                    if ptr2+1 < len(nums2) and abs(nums2[ptr2+1]) - abs(val) < abs(nums1[ptr1]) - abs(val):
                        curr+=1
                        ptr2+=1
                        val = nums2[ptr2]
                        flag = 1
                    else:
                        curr+=1
                        val = nums1[ptr1]
                        flag = 0
        
        if flag == 1:
            med = nums2[ptr2]
        else:
            med = nums1[ptr2]
            
        if length % 2 == 0:
            if med - nums1[ptr1 + 1] < med - nums2[ptr2 +1]:
                return (med + nums1[ptr1+1])/2
            else:
                return (med + nums2[ptr2+1])/2
        else:
            return med

Here's what I wrote for your sum the cubes. Took me 1 minute. I hope I didn't misunderstand the question.

def sumCubes(n): sumC = 0 for i in range(n+1): sumC += i 3 return sumC

I'll give you 2 more leetcode examples. Neither passed all test cases. The first is 2-3 hours. The latter 3. I didn't post the last submission for the latter one because I believe the one I posted is more telling of my thought process and coding ability imo so that solution is between 1 and 2 hours of work, but I wasn't able to get it in the end.

I honestly don't recall my idea here. I think it was to see if the start and end of the array had the same letter and if it didn't cut off the first and last letter and continue analyzing the array until the center is reached.

https://leetcode.com/problems/longest-palindromic-substring/...

class Solution: def longestPalindrome(self, s): """ :type s: str :rtype: str """ curr = "" big = ""

        for c in s:
            curr += c
            if curr != curr[::-1]:
                if len(curr[:len(curr)-1]) >= len(big):
                    big = curr[:len(curr)-1]
                    print(big)
                while curr != curr[::-1]:
                    curr = curr[1:]
        
        if len(curr) > len(big):
            return curr
        else:
            return big

I thought there were 3 cases in this problem. You hit a character, you hit a ? or you hit . The first one meant an exact match had to be made. The second that there only had to be some character in the string to take a position (the string couldn't end if there was a ?). The last could mean 0 to n matches so there had to be a lot of checks in my mind. I also thought about what happens if there's "?" which I wrote as a case inside the "" if case. When test cases didn't work I would try and code for that test case and it kept going wrong so I eventually gave up and looked at the solution. My main flaw is that it didn't occur to me that backtracking was the way to go or a way to check all the potential combinations. I thought you just had to keep moving forward.

https://leetcode.com/problems/wildcard-matching/description/

class Solution: def isMatch(self, s, p): """ :type s: str :type p: str :rtype: bool """ if len(p) == 0 and len(s) == 0: return True if len(p) == 0: return False

        i = 0
        j = 0
        while i < len(s):
            if j >= len(p) and p[len(p) - 1] != '*':
                return False
            
            if p[j] == '*':
                while j < len(p) and p[j] == '*':
                    j+=1
                if j == len(p):
                    break
                elif p[j] == '?':
                    x = len(p)- j
                    while len(s) - i > x:
                        i+=1
                    while j < len(p) and p[j] == '?':
                        i+=1
                        j+=1
                    if i > len(s) or j >= len(p) and i < len(s):
                        print(i)
                        return False
                    if i == len(s) and j == len(p):
                        break
                else:
                    while i < len(s) and p[j] != s[i]:
                        i+=1
                    

                    if i == len(s):
                        return False
            elif p[j] == '?':
                i+=1
                j+=1
            else:
                if p[j] != s[i]:
                    print(i, s[i])
                    print(j, p[j])
                    return False
                i+=1
                j+=1
        
        if j < len(p):
            while j < len(p) and p[j] =='*':
                j+=1
                
            
            if j == len(p) and p[j-1] == '*':
                return True
            else:
                return False
        return True

[generallyfalse]

This is what I tried, recursion taking care of puting the simplest cases first. I haven't checked the solution. I think this kind of problem is like a state machine. In each call of the recursion the length of the pattern or the lengh of the string should decrease, so finally one get to the easy cases of length in (0,1,2).

This kind of problems are designed to test you can define a grammar and apply recursion. Another problem with the same flavor could be to evaluate string expressions like "3+25" using a grammar. My one-hour solution (not checked). But I think I have done similar problems.

         def pat(s,p):
    ...:     if s == "":
    ...:         if len(p)>1 and p[1] in "?*":
    ...:             return pat(s,p[2:])
    ...:     if p == "": return s == p
    ...:     if len(p) == 1:
    ...:         if p == "*": return True
    ...:         return s == p
    ...:     if len(p) == 2:
    ...:         if p[1] == "?":
    ...:             return s == p[0] or s==""
    ...:         if p[1] == "*":
    ...:             if s[0] != p[0]: return False
    ...:             if s == p[0]: return True
    ...:             return pat(s[1:],p)
    ...:     if p[1] != "?" and p[1] != "*":
    ...:         return s[0] == p[0] and pat(s[1:],p[1:])
    ...:     if p[1] == "?":
    ...:         if (s[0] != p[0]):
    ...:              return pat(s,p[2:]))
    ...:         if (s[0] == p[0]):
    ...:             return pat(s[1:],p[2:]) or pat(s,p[2:])
    ...:     if p[1] =="*":
    ...:         if (s[0] != p[0]): return pat(s,p[2:])
    ...:         return pat(s[1:],p) or pat(s,p[2:])
    ...:
    ...:

[Tmp1234]

Your solution looks better than mine. Is there anything you've gleaned from the leetcode examples I posted?

[generallyfalse]

I see that you prefer not to use recurrency in Python. That decision is correct since python is not optimized for recurrency but it also can make the code more complex. See https://realpython.com/python-thinking-recursively/

On the other hand, using iterators in python is nice. You didn't use iterators to compute the sum of cubes, many problems can be solved in one line using iterators. https://www.datacamp.com/community/tutorials/python-iterator...

Perhaps your code should try to go from the easy cases to the complex cases in pattern matching.

As a hobby programmer I program without stress, if I was to make a living by programming I think I shouldn't enjoy so much.

I think you should read norvig post: http://norvig.com/21-days.html, also to learn to program in python: https://github.com/norvig/pytudes

I believe that in a year or two you could be prepared to judge your progress

[generallyfalse]

In one of your comments you say you have improved a lot in two months or so, so perhaps you need a little more practice and I think you could improve a lot more (less time for solving puzzles and writing better code). Anyway, I have tried to solve some of the puzzles. Perhaps other people in this thread can give more valuable information about your coding style. I wish you the best.

Sum of cubes:

  sum(i**3 for i in range(n+1))

findMedianSortedArrays using Haskell:

  let  m a [] n = a !! n
       m [] a n = a !! n
       m (x:_) (y:_) 0 = if x<=y then x else y
       m (x:xs) (y:ys) n = if x<=y then m xs (y:ys) (n-1) 
                         else m (x:xs) ys (n-1)

  let m1 a b = if odd n1 then m a b n12 
               else (m a b n12 + m a b (n12-1))/2 
               where n1 = length a + length b; n12 = div n1 2

  let findMedianSortedArrays = m1
 I haven't checked the code.