My below solution to the following problem is exceeding the maximum recursion depth. I am looking for any improvements which i can make.


Alice and Bob need to send secret messages to each other and are discussing ways to encode their messages:

Alice: “Let’s just use a very simple code: We’ll assign ‘A’ the code word 1, ‘B’ will be 2, and so on down to ‘Z’ being assigned 26.”

Bob: “That’s a stupid code, Alice. Suppose I send you the word ‘BEAN’ encoded as 25114. You could decode that in many different ways!” Alice: “Sure you could, but what words would you get? Other than ‘BEAN’, you’d get ‘BEAAD’, ‘YAAD’, ‘YAN’, ‘YKD’ and ‘BEKD’. I think you would be able to figure out the correct decoding. And why would you send me the word ‘BEAN’ anyway?” Bob: “OK, maybe that’s a bad example, but I bet you that if you got a string of length 5000 there would be tons of different decodings and with that many you would find at least two different ones that would make sense.” Alice: “How many different decodings?” Bob: “Jillions!”

For some reason, Alice is still unconvinced by Bob’s argument, so she requires a program that will determine how many decodings there can be for a given string using her code.


Input will consist of multiple input sets. Each set will consist of a single line of at most 5000 digits representing a valid encryption (for example, no line will begin with a 0). There will be no spaces between the digits. An input line of ‘0’ will terminate the input and should not be processed.


For each input set, output the number of possible decodings for the input string. All answers will be within the range of a 64 bit signed integer.






The following is my solution -

def helper_dp(input_val, k, memo):

    if k == 0:
        return 1

    s = len(input_val) - k

    if input_val[s] == '0':
        return 0

    if memo[k] != -1:
        return memo[k]

    result = helper_dp(input_val, k - 1, memo)

    if k >= 2 and int(input_val[s:s+2]) <= 26:
        result += helper_dp(input_val, k - 2, memo)

    memo[k] = result
    return memo[k]

def num_ways_dp(input_num, input_len):
    memo = [-1] * (len(input_num) + 1)
    return helper_dp(input_num, input_len, memo)

if __name__ == '__main__':
    number = input()
    while number != '0':
        print(num_ways_dp(number, len(number)))
        number = input()

Problem Link - https://www.spoj.com/problems/ACODE/

  • 3
    \$\begingroup\$ What exactly is the purpose of MAX_LIMIT = 5001? \$\endgroup\$
    – Linny
    Jan 14, 2020 at 11:20
  • 3
    \$\begingroup\$ @pacmaninbw If it can be fixed by raising the limit (and on local machines, that's trivial to do by orders of magnitude), that's about as bad as a Time-limit exceeded. And we haven't outlawed those. \$\endgroup\$
    – Mast
    Jan 14, 2020 at 20:31
  • \$\begingroup\$ @Linny That's a hold over from my previous code. I will remove that line. Thanks for pointing it out \$\endgroup\$ Jan 15, 2020 at 6:07

1 Answer 1


Your code performs one recursion per character and recursion depth is limited. For long inputs, your code will raise RecursionError: maximum recursion depth exceeded. Replacing recursion with an explicit loop solves the issue.

Also, if you start the calculation at k = 1 with memo[-1] and memo[0] properly initialized and then work upwards to k = len(s), the code will become significantly simpler. When calculating memo[k], you only ever access memo[k-2] and memo[k-1], so you can store only those. I renamed them old and new. Here is how it could look like:

def num_ways_dp(s):
    old = new = 1
    for i in range(len(s)-1):
        if int(s[i:i+2]) > 26: old = 0 # no two-digit solutions
        if int(s[i+1]) == '0': new = 0 # no one-digit solutions
        (old, new) = (new, old + new)
    return new

if __name__ == '__main__':
    number = input()
    while number != '0':
        number = input()
  • \$\begingroup\$ Your inner loop could be old, new = new, (old + new) if int(s[i:i+2]) <= 26 else new. The parenthesis on the right are not needed and on the left you can constrain the ternary to the part where it actually matters. \$\endgroup\$
    – Graipher
    Jan 14, 2020 at 15:03
  • \$\begingroup\$ Almost, but you forgot to properly handle 0. In particular, the string "101" has value 1, not 3. \$\endgroup\$
    – David G.
    Jan 14, 2020 at 19:29
  • 1
    \$\begingroup\$ Do you have a good reason for the two not operators? I find the code easier to understand when the first condition is negated and the second condition is simply s[i+1] == '0'. \$\endgroup\$ Jan 15, 2020 at 7:25
  • 1
    \$\begingroup\$ @RainerP. I didn't quite understand the logic for the (old, new) = (new, old + new) line \$\endgroup\$ Jan 15, 2020 at 8:17
  • 1
    \$\begingroup\$ @strawhatsai - The (old, new) = (new, old + new) is equivalent to (memo[k-2], memo[k-1]) = (memo[k-1], memo[k]). I store only the latest two memo-entries so I need to rename them each time I increment the loop counter. \$\endgroup\$
    – Rainer P.
    Jan 15, 2020 at 9:07

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