Class Solution
- java.lang.Object
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- g1801_1900.s1830_minimum_number_of_operations_to_make_string_sorted.Solution
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public class Solution extends Object
1830 - Minimum Number of Operations to Make String Sorted.Hard
You are given a string
s( 0-indexed ). You are asked to perform the following operation onsuntil you get a sorted string:- Find the largest index
isuch that1 <= i < s.lengthands[i] < s[i - 1]. - Find the largest index
jsuch thati <= j < s.lengthands[k] < s[i - 1]for all the possible values ofkin the range[i, j]inclusive. - Swap the two characters at indices
i - 1andj. - Reverse the suffix starting at index
i.
Return the number of operations needed to make the string sorted. Since the answer can be too large, return it modulo
109 + 7.Example 1:
Input: s = “cba”
Output: 5
Explanation: The simulation goes as follows: Operation 1: i=2, j=2. Swap s[1] and s[2] to get s=“cab”, then reverse the suffix starting at 2. Now, s=“cab”.
Operation 2: i=1, j=2. Swap s[0] and s[2] to get s=“bac”, then reverse the suffix starting at 1. Now, s=“bca”.
Operation 3: i=2, j=2. Swap s[1] and s[2] to get s=“bac”, then reverse the suffix starting at 2. Now, s=“bac”.
Operation 4: i=1, j=1. Swap s[0] and s[1] to get s=“abc”, then reverse the suffix starting at 1. Now, s=“acb”.
Operation 5: i=2, j=2. Swap s[1] and s[2] to get s=“abc”, then reverse the suffix starting at 2. Now, s=“abc”.
Example 2:
Input: s = “aabaa”
Output: 2
Explanation: The simulation goes as follows:
Operation 1: i=3, j=4. Swap s[2] and s[4] to get s=“aaaab”, then reverse the substring starting at 3. Now, s=“aaaba”.
Operation 2: i=4, j=4. Swap s[3] and s[4] to get s=“aaaab”, then reverse the substring starting at 4. Now, s=“aaaab”.
Constraints:
1 <= s.length <= 3000sconsists only of lowercase English letters.
- Find the largest index
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Constructor Summary
Constructors Constructor Description Solution()
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Method Detail
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makeStringSorted
public int makeStringSorted(String s)
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