LeetCode 2767. Partition String Into Minimum Beautiful Substring
Given a binary string s, partition the string into one or more substrings such that each substring is beautiful.
A string is beautiful if:
It doesn't contain leading zeros.
It's the binary representation of a number that is a power of
5.
Return the minimum number of substrings in such partition. If it is impossible to partition the string s into beautiful substrings, return -1.
A substring is a contiguous sequence of characters in a string.
Example 1:
Input: s = "1011"
Output: 2
Explanation:
We can paritition the given string into ["101", "1"].
- The string "101" does not contain leading zeros and is the binary representation of integer 51 = 5.
- The string "1" does not contain leading zeros and is the binary representation of integer 50 = 1.
It can be shown that 2 is the minimum number of beautiful substrings that s can be partitioned into.
Example 2:
Input: s = "111"Output: 3
Explanation:
We can paritition the given string into ["1", "1", "1"].
- The string "1" does not contain leading zeros and is the binary representation of integer 50 = 1.
It can be shown that 3 is the minimum number of beautiful substrings that s can be partitioned into.
Example 3:
Input: s = "0"
Output: -1
Explanation: We can not partition the given string into beautiful substrings.
Constraints:
1 <= s.length <= 15s[i]is either'0'or'1'.
Hide Hint 1
To check if number x is a power of 5 or not, we will divide x by 5 while x > 1 and x mod 5 == 0. After iteration if x == 1, then it was a power of 5.
Hide Hint 2
Since the constraint of s.length is small, we can use recursion to find all the partitions.
--------------------感觉是个经典题目；
1：先写一个函数判断是否是5的幂，
2：然后去遍历字符串，先从最长的开始，当前数是5的幂，那么就可以去处理剩下的字符串。
剩下的字符串如果不含前导0，那么就返回1+当前函数处理的字符串的长度，返回即可；
递归无止境，还是多学习；
Runtime: 4 ms, faster than 25.00% of Java online submissions for Partition String Into Minimum Beautiful Substrings.
Memory Usage: 42 MB, less than 25.00% of Java online submissions for Partition String Into Minimum Beautiful Substrings.
