Minimum Number of Operations to Make String Sorted

Problem

You are given a string s (0-indexed). You are asked to perform the following operation on s until you get a sorted string:

Find the largest index i such that 1 <= i < s.length and s[i] < s[i - 1].
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.
Swap the two characters at indices i - 1 and j.
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 10^9 + 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.

Solution

class Solution {
    public int makeStringSorted(String s) {
        int n = s.length();
        int[] count = new int[26];
        for (int i = 0; i < n; i++) {
            count[s.charAt(i) - 'a']++;
        }

        long[] fact = new long[n + 1];
        fact[0] = 1;
        int mod = 1000000007;
        for (int i = 1; i <= n; i++) {
            fact[i] = (fact[i - 1] * i) % mod;
        }
        int len = n;
        long ans = 0;
        for (int i = 0; i < n; i++) {
            len--;
            int bound = s.charAt(i) - 'a';
            int first = 0;
            long rev = 1;
            for (int k = 0; k < 26; k++) {
                if (k < bound) {
                    first += count[k];
                }
                rev = (rev * fact[count[k]]) % mod;
            }
            ans =
                    ans % mod
                            + (((first * fact[len]) % mod)
                                            * (modPow(rev, (long) mod - 2, mod))
                                            % mod)
                                    % mod;
            ans = ans % mod;
            count[bound]--;
        }
        return (int) ans;
    }

    private long modPow(long x, long n, int m) {
        long result = 1;
        while (n > 0) {
            if ((n & 1) != 0) {
                result = (result * x) % m;
            }
            x = (x * x) % m;
            n = n >> 1;
        }
        return result;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).