Leetcode 2268. Minimum Number of Keypresses [Python]

把字母出现的次数统计出来，从多到少排序，出现次数多的字母尽可能排在第一轮就可以按到，也就是press1次，以此类推到press 2次，press 3次。

class Solution:
    def minimumKeypresses(self, s: str) -> int:
        Counter = collections.Counter(s)
        newCounter = []
        for c,num in Counter.items():
            newCounter.append((c,num))
        newCounter.sort(key = lambda x:x[1], reverse = True)
        flag = 0
        ans = 0
        for c,num in newCounter:
            flag += 1
            if flag <=9:
                ans += num * 1
            elif flag >18:
                ans += num *3
            else:
                ans += num *2
        return ans