Leetcode 3714. Longest Balanced Substring II

• Leetcode 3714. Longest Balanced Substring II
  • 1. 解题思路
  • 2. 代码实现

• 题目链接：3714. Longest Balanced Substring II

1. 解题思路

这一题我没有什么很好的思路，最后是看了答案之后给出了一个最暴力的解法，即分别考察一个字符的情况下能够构成的最长的balanced string的长度，只有两个字符的情况下能够构成的最长的balanced string的长度，以及三个字符的情况下能够构成的最长的balanced string的长度。

2. 代码实现

给出python代码实现如下：

class Solution:
    def longestBalanced(self, s: str) -> int:
        n = len(s)
        chars = set(s)

        def longest_balanced_1(s):
            ans = 0
            prev, cnt = "", 0
            for ch in s:
                if ch == prev:
                    cnt += 1
                else:
                    prev, cnt = ch, 1
                ans = max(ans, cnt)
            return ans

        def longest_balanced_2(s, a, b):
            ans = 0
            cache = defaultdict(list)
            diff = 0
            cache[diff].append(0)
            for i, ch in enumerate(s):
                if ch == a:
                    diff += 1
                elif ch == b:
                    diff -= 1
                else:
                    ans = max(ans, max(loc[-1]-loc[0] for loc in cache.values()))
                    cache = defaultdict(list)
                    diff = 0
                    cache[diff].append(i+1)
                    continue
                cache[diff].append(i+1)
            return max(ans, max(loc[-1]-loc[0] for loc in cache.values()))

        def longest_balanced_3(s):
            cache = defaultdict(list)
            diff_ab, diff_ac = 0, 0
            cache[(diff_ab,diff_ac)].append(0)
            for i, ch in enumerate(s):
                if ch == 'a':
                    diff_ab += 1
                    diff_ac += 1
                elif ch == "b":
                    diff_ab -= 1
                else:
                    diff_ac -= 1
                cache[(diff_ab,diff_ac)].append(i+1)
            return max(loc[-1]-loc[0] for loc in cache.values())

        if len(chars) == 1:
            return n
        elif len(chars) == 2:
            a, b = list(chars)
            return max(longest_balanced_2(s, a, b), longest_balanced_1(s))
        else:
            return max(longest_balanced_3(s), longest_balanced_2(s, "a", "b"), longest_balanced_2(s, "a", "c"), longest_balanced_2(s, "b", "c"), longest_balanced_1(s))

提交代码评测得到：耗时1669ms，占用内存48.55MB。