indeed interview2

indeed2.py

def summmaryranges(nums):
    ranges = []
    for num in nums:
        if not ranges or num - ranges[-1][-1] > 1:
            ranges.append([])
        ranges[-1][1:] = [num]
    return ['->'.join(map(str, r)) for r in ranges]

def reverse(l, r, cl):
    while l < r:
        cl[l], cl[r] = cl[r], cl[l]
        l += 1
        r -= 1

def reversestring(s):
    n = len(s)
    s = list(s)
    reverse(0, n - 1, s)
    l = 0
    for r in range(1, n + 1):
        if r == n or s[r] == ' ':
            reverse(l, r - 1, s)
            l = r + 1
    return ''.join(s)

def reverseHTML(h):
    n = len(h)
    h = list(h[::-1])
    l = 0
    while l < n - 1:
        if l == ';':
            while l < n and h[l] == h[l - 1]:
                l += 1
            r = l + 1
            while r < n and h[r] != '&':
                r += 1
            reverse(l, r, h)
            l = r + 1
        l += 1
    return ''.join(h)

def findduplicate(s):
    d = {}
    for word in s.split():
        d[word] = d.get(word, 0) + 1
    return [item[0] for item in filter(lambda x:x[1] > 1, d.items())]

def findfirstdup(s):
    d = {}
    words = s.split()
    min_idx = len(words)
    for idx, word in enumerate(words):
        if word not in d:
            d[word] = idx
        else:
            min_idx = min(min_idx, d[word])
    if min_idx == len(words):
        return None
    return words[min_idx]

def a2num(s):
    '''
    A 1     AA 26 + 1   
    B 2     AB 26 + 2
    Z 26    AZ 26 + 26
    '''
    res = 0
    for idx, c in enumerate(s[::-1]):
        res += (ord(c) - ord('A') + 1) * 26 ** idx
    return res

def num2a(n):
    res = ''
    while n > 0:
        res += chr((n - 1) % 26 + ord('A'))
        n = (n - 1) // 26
    return res[::-1]

def findkbucket(nums, k):
    d = {}
    res = []
    for num in nums:
        d[num] = d.get(num, 0) + 1
    freqdict = [[] for _ in range(len(nums) + 1)]
    for key in d:
        freqdict[d[key]] += [key]
    for item in freqdict[::-1]:
        res += item
    return res[:k]