#dailyCoding

AcodeAlphacode.py

# This is a "dynamic programming" (aka dp) question and the solution is "recursive"
# Using loop to make the code run until provided non-acceptable input
while True:
    n = str(input())
    # check if given no is zero
    if n == "0":
        break
    dp = []
    # Make an array with a length len(n)+1 with each element as zero
    # We will use this array 
    for i in range(len(n) + 1):
        dp.append(0)
    # print(dp)
    # dp = [0, 0, 0, 0,....., (n+1)th 0]
    dp[0] = 1
    dp[1] = 1
    # 1. Loop over the given input 
    # 2. If it is a valid single digit number, Copy the previous element's value to the current element (DP[i] = DP[i-1])
    # 3. If it is a valid two digit number, Add the previous to previous element's value to the current element (DP[i] += DP[i-2])

    # In one line : dp[i] = dp[i-1] {if valid single digit number} + dp[i-2] {if current and previous elements make a valid two digit number}
    for i in range(2,len(n)+1):
        if 0 < int(n[i-2]+n[i-1]) <= 26 and n[i-2] != "0":
            if n[i-1] == "0":
                dp[i] = dp[i-2]
                # print(dp)
                # break
            else:
                dp[i] = dp[i-1] + dp[i-2]
                # print(dp)
                # break
        # print(dp)
        # break
        elif n[i-1] != "0" :
            dp[i] = dp[i-1] 
            # print(dp)
            # break
    # print(dp) 
    #Taking last digit from dp array
    print(dp[-1])