java.lang.Object
g1801_1900.s1830_minimum_number_of_operations_to_make_string_sorted.Solution

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 on s until you get a sorted string:

  1. Find the largest index i such that 1 <= i < s.length and s[i] < s[i - 1].
  2. Find the largest index j such that i <= j < s.length and s[k] < s[i - 1] for all the possible values of k in the range [i, j] inclusive.
  3. Swap the two characters at indices i - 1 and j.
  4. 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 <= 3000
  • s consists only of lowercase English letters.
  • Constructor Details

    • Solution

      public Solution()
  • Method Details

    • makeStringSorted

      public int makeStringSorted(String s)