slide window & +Counter() : leetcode 438. Find All Anagrams in a String

438. Find All Anagrams in a String

Input:
s: “cbaebabacd” p: “abc”

Output:
[0, 6]

Explanation:
The substring with start index = 0 is “cba”, which is an anagram of “abc”.
The substring with start index = 6 is “bac”, which is an anagram of “abc”.

def findAnagrams_leetcode(self, s: str, p: str) -> List[int]:
    if len(s) < len(p):
        return []

    p_chars = Counter(p)
    i = 0
    j = len(p)
    window = Counter(s[i:j])
    results = []
    while j <= len(s):
        if +window == p_chars:
            results.append(i)
        window[s[i]] -= 1
        if j < len(s):
            window[s[j]] += 1
        i += 1
        j += 1
    return results

在上面的算法中，"+windows" 的意思是 複製一個和windows相同的Counter實例，並且過濾掉其中 包含 0和負數的項。

b = {"a": 1, "b": 3, "c": 0, "d": -1}
a = + Counter(b)
print("a:{}".format(a))
# a:Counter({'b': 3, 'a': 1})

“+Counter()” 之所以生效，是因爲Counter 重載了 pos 方法, 實現了一元(unary)操作符 ‘+’ 。

class Counter(Dict[_T, int], Generic[_T]):
...
    def __add__(self, other: Counter[_T]) -> Counter[_T]: ...
...
    def __pos__(self) -> Counter[_T]: ...
    def __neg__(self) -> Counter[_T]: ...
...

python doc about counter

Common patterns for working with Counter objects:

sum(c.values())                 # total of all counts
c.clear()                       # reset all counts
list(c)                         # list unique elements
set(c)                          # convert to a set
dict(c)                         # convert to a regular dictionary
c.items()                       # convert to a list of (elem, cnt) pairs
Counter(dict(list_of_pairs))    # convert from a list of (elem, cnt) pairs
c.most_common()[:-n-1:-1]       # n least common elements
c += Counter()                  # remove zero and negative counts