2405. Optimal Partition of String
Medium
Given a string
s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once.Return the minimum number of substrings in such a partition.
Note that each character should belong to exactly one substring in a partition.
Example 1:
Input: s = "abacaba"
Output:
4
Explanation:
Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba").
It can be shown that 4 is the minimum number of substrings needed.
Example 2:
Input: s = "ssssss"
Output:
6
Explanation:
The only valid partition is ("s","s","s","s","s","s").
Constraints:
1 <= s.length <= 105sconsists of only English lowercase letters.
Map 的解法
func partitionString(s string) int {
m := make(map[byte]bool)
ans := make([]string, 0)
pre := 0
for i:=0; i<len(s); i++ {
if m[s[i]] {
ans = append(ans, s[pre: i])
pre = i
m = make(map[byte]bool)
}
m[s[i]] = true
}
return len(ans)+1
}