【算法】力扣第 266场周赛

5918. 统计字符串中的元音子字符串

看数据范围，暴力过

class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        res=0
        n=len(word)
        for a in range(n):
            for b in range(a+5,n+1):
                i=word[a:b]
                if any(x not in 'aeiou' for x in i):continue
                if 'a' in i and 'e' in i and 'i' in i and 'o' in i and 'u' in i:
                    res+=1
        return res

5919. 所有子字符串中的元音

第一次考虑前缀和做，子字符串word[a:b]内的元音字符数为pre[b+1]-pre[a]

class Solution:
    def countVowels(self, word: str) -> int:
        n=len(word)
        res=0
        dic=[0]*n
        for i in range(n):
            if word[i] in ('a','e','i','o','u'):
                dic[i]=1
        
        pre=[0,*accumulate(dic)]

        for a in range(n):
            for b in range(a,n):
                res+=pre[b+1]-pre[a]
        return res
        

果然超时， 找一下规律，可以发现pre[i]的个数为2*i-n

class Solution:
    def countVowels(self, word: str) -> int:
        n=len(word)
        res=0
        dic=[0]*n
        for i in range(n):
            if word[i] in 'aeiou':
                dic[i]=1
        
        pre=[0,*accumulate(dic)]

        for i in range(n+1):
            res+=(2*i-n)*pre[i]
        return res
        

整理下变成python二行

class Solution:
    def countVowels(self, word: str) -> int:
        pre=[0,*accumulate([1 if word[i] in 'aeiou' else 0  for i in range(len(word))])]
        return sum((2*i-len(word))*pre[i] for i in range(len(word)+1))

5921. 最大化一张图中的路径价值

这次的T4比较简单，看下数据范围：

• n == values.length
• 1 <= n <= 1000
• 10 <= timej, maxTime <= 100
• 每个节点 至多有四条 边

直接暴搜+回溯

class Solution:
    def maximalPathQuality(self, values: List[int], edges: List[List[int]], maxTime: int) -> int:
        n=len(values)
        dic={i:list() for i in range(n)}
        for e in edges:
            a,b,c=e
            dic[a].append([b,c])
            dic[b].append([a,c])
        

        def dfs(cur,tc,res,nodes):
            if tc>maxTime:
                return 
            
            if cur==0:
                vc=0
                for i in set(nodes):
                    vc+=values[i]
                res[0]=max(res[0],vc)
                    
            
            for x in dic[cur]:
                nodes.append(x[0])
                dfs(x[0],tc+x[1],res,nodes)
                nodes.pop()
        
        res=[0]
        dfs(0,0,res,[0])
        
        return res[0]

5920. 分配给商店的最多商品的最小值

最小化最大值问题，首先可见是贪心，其次可以注意到单调性，而后二分求解

class Solution:
    def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
        l,r=1,max(quantities)
        while l<=r:
            mid=l+(r-l>>1)
            need=sum(ceil(q/mid) for q in quantities)
            if need>n:
                l=mid+1
            elif need<=n:
                r=mid-1
        return l

875. 爱吃香蕉的珂珂

周赛P3换皮题，用来巩固下二分板子

class Solution:
    def minEatingSpeed(self, piles: List[int], h: int) -> int:
        l,r=1,max(piles)
        while l<=r:
            mid=l+(r-l>>1)
            need=sum(ceil(p/mid) for p in piles)
            if need>h:
                l=mid+1
            elif need<=h:
                r=mid-1
        return l

复盘总结

学习一下六弦爷的最短写法！

P1 统计字符串中的元音子字符串

class Solution:
    def countVowelSubstrings(self, word: str) -> int:
        return sum(set(word[i: j + 1]) == set('aeiou') for i in range(len(word)) for j in range(i + 1, len(word)))

P2 所有子字符串中的元音

class Solution:
    def countVowels(self, word: str) -> int:
        return sum((i + 1) * (len(word) - i) for i, c in enumerate(word) if c in 'aeiou')

P3 分配给商店的最多商品的最小值

class Solution:
    def minimizedMaximum(self, n: int, Q: List[int]) -> int:
        class A:
            def __getitem__(self, x):
                return sum(q // x + (1 if q % x else 0) for q in Q) <= n
        return bisect_left(A(), True, 1, max(Q) + 1)