# Answers to the String/List Test on Friday 4/12:

# Problem 1

def MyLower(s):
    answer = ''
    for c in answer:
        if ord('A') <= ord(c) <= ord('Z'):
            answer += chr(ord(c) - ord('A') + ord('a'))
        else:
            answer += c
    return answer

def countString(target,lookfor):
    count = 0
    target_low = MyLower(target)
    lookfor_low = MyLower(lookfor)
    ln_target = len(target)
    ln_lookfor = len(lookfor)
    
    for i in range(len(target)-len(lookfor)+1):
        if target_low[i:i+ln_lookfor] == lookfor_low:
            count += 1
    return count

# Problem 2

def IsInteger(word):
    if len(word) == 0:
        return False
    if word[0] == '-' or word[0] == '+':
        word = word[1:]
    return word.isdigit()

def addIntegers(s):
    words = s.split(';')
    answer = 0
    for word in words:
        if IsInteger(word):
            answer += int(word)
    return answer

# Problem 3

def Eng2Kling(s):
    # create 2 lists, first of English words and second of Klingon words
    Eng = []
    Kling = []
    pairs = KL.split('\n')
    for pair in pairs:
        words = pair.split('=')
        Eng.append(words[0])
        Kling.append(words[1])

    # now translate s
    answer = ''
    PhraseWords = s.split()
    for word in PhraseWords:

        # now, let's find the position of the English word in the English list
        # Way #1:
        pos = Eng.index(word)

        # or Way #2:
        for pos in range(len(Eng)):
            if Eng[pos] == word:
                break
            
        # or Way #3:
        pos = 0
        while Eng[pos] != word:
            pos += 1

        KlingWord = Kling[pos]
        answer += KlingWord + ' '

    return answer[:-1]

# Problem 4
def Interleave(A,B):
    answer = []
    for i in range(len(A)):
        answer.append(A[i])
        if len(B) > i:
            answer.append(B[i])
        else:
            # no more B's, so add all the leftover A's
            answer += A[i+1:]
            return answer
    # add any leftover B's
    answer += B[len(A):]
    return answer

# Problem 5
def countStringRecursive(target, lookfor):
    if len(target) < len(lookfor):
        return 0
    my_answer = 0
    if target[:len(lookfor)] == lookfor:
        my_answer = 1
    return my_answer + countStringRecursive(target[1:],lookfor)